Analysis II notes

David Pierce

June 20, 2003

These notes are for Math 272 at METU. I intend to edit them and add to them from time to time; the latest version is in the course directory, <http:

//www.math.metu.edu.tr/~dpierce/courses/272/>. The catalog description

of the course is:

Riemann–Stieltjes Integral. Infinite series and products. Sequences of functions. Inverse Function Theorem. Multiple Integrals.

These topics are covered in [1, chs 7, 8, 9, 12, 13 and 14]. This book will be my main reference, although [2] may also be useful. I shall also cover functions of bounded variation (in [1, ch. 6]), a topic left over from Math 271. My proof of Theorem 4.13 below is based on that of [3, ch. 1, § 1, Proposition 6]; my §§ 5 and 6 are influenced by [4, chs 2 and 3].

I prepare the notes, first of all, for my own use. They are only an outline of what is to be discussed in class. In particular, for the student, reading these notes is probably not an adequate substitute for coming to class. Your own notes, taken properly in class, will be richer and more complete than these notes.

The parts of these notes labelled ‘proof’ are generally only sketches of proofs. I leave it to the reader both:

• to recognize where details are missing, and

• to supply those details.

I might give the details in class, especially if I am asked to. I myself might ask for the details on an exam.

Some proofs are omitted entirely and are thus left to be given in class or to be done as exercises. These proofs too might be asked for on an exam.

I do intend to write my proofs (or proof-sketches) in complete sentences, with the usual sorts of punctuation (commas, semicolons, full stops/periods). Like any other writing (in English and Turkish and many other languages), the proofs are to be read left to right, top to bottom. Students should follow this example in writing their own complete proofs.

Every time a new class of functions (for example) is introduced, one should ask: What are some examples of functions that belong to this class? Which functions do not belong to this class?

Likewise, when a theorem is introduced, one should ask: What sorts of functions does the theorem apply to? What does the theorem not tell us? For example, if the theorem is an implication (an if-then statement), then can we prove the converse, or is there a counter-example?

Examples and counter-examples could be requested on an exam.

In studying the proof of a theorem, one should ask: What previous lemmas and theorems does the proof rely on? Is the proof in the style of earlier proofs; does it use familiar techniques; or does it introduce a new approach? Will a similar proof work to prove something else? Is there an alternative proof?

References

[1] Tom M. Apostol. Mathematical analysis. Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills, Ont., second edition, 1974.

[2] Robert G. Bartle. The elements of real analysis. John Wiley & Sons, New York-London-Sydney, second edition, 1976.

[3] Raghavan Narasimhan. Complex analysis in one variable. Birkhäuser Boston Inc., Boston, MA, 1985.

[4] Michael Spivak. Calculus on manifolds. A modern approach to classical theorems of advanced calculus. W. A. Benjamin, Inc., New York-Amsterdam, 1965.

Contents

1 Variation 1

2 The Riemann–Stieltjes Integral 4

3 Infinite series and products 12

4 Sequences of functions 20

5 Differentiation in several dimensions 29

6 Integration in several dimensions 37

1 Variation

Here are some conventions, definitions and notations to be used throughout this section:

Let I be a compact interval of ℝ; so I is [a, b] for some numbers a and b. Let f be a real-valued function on I. A partition of I is a subset that:

• is finite, and

• contains the endpoints a and b.

Let P be a partition of I. We may also write P as

{x_{0}, x_{1}, …, x_{n}}

for some positive integer n, where

a = x_{0} < x_{1} < ⋯ < x_{n} = b.

If 0 < k ⩽ n, then for x_{k} − x_{k−1} we may write

△x_{k},

and for f(x_{k}) − f(x_{k−1}) we may write

△f_{k}.

For ∑_{k=1}^{n} |△f_{k}| we may write

∑(P).

The set of partitions of I can be denoted

PI.

(This is not the power-set of I!) Suppose there is a number M such that for all P in PI we have ∑(P) ⩽ M. Then f is said to be of bounded variation on I. In this case, the supremum of the set {∑(P) : P ∈ PI} exists; this supremum is called the total variation of f on I, and can be denoted

V_{f}I

or just V_{f}. Note that total variation is always non-negative. We may say that f itself is bounded on I by A if |f(x)| ⩽ A for all x in I.

What sorts of functions are of bounded variation? How bad can a function be while still being of bounded variation?

Theorem 1.1. Suppose f is monotone on I. Then f is of bounded variation on I, and in fact

V_{f}I = |f(b) − f(a)| .

In particular, constant functions x ↦ a and the identity-function x ↦ x are of bounded variation on any compact interval.

Recall that monotonicity does not imply continuity. So, functions of bounded variation need not be continuous.

The converse of Theorem 1.1 fails: consider x ↦ x^{2} : [−1, 1] → ℝ.

Theorem 1.2. If f is of bounded variation on I, then f is bounded on I, and in fact

|f(x)| ⩽ V_{f} + |f(a)|

for all x in I.

Proof. Suppose f is of bounded variation. If x ∈ I, let P_{x} be the partition {a, x, b} of I. Then

|f(x)| − |f(a)| ⩽ |f(x) − f(a)| ⩽ |f(x) − f(a)| + |f(b) − f(x)| = ∑(P_{x}) ⩽ V_{f},

which yields the claim.

The converse fails: consider x ↦ \begin{cases}x sin \frac{1}{x}, & if x ∈ (0, 1]; \\ 0, & if x = 0. \end{cases}

Theorem 1.3. Suppose f and g are of bounded variation on I. Then so are f + g and f − g, and in fact

V_{f±g} ⩽ V_{f} + V_{g}.

Moreover, f · g is also of bounded variation on I, and if A bounds f and B bounds g on I, then

V_{f·g} ⩽ B · V_{f} + A · V_{g}.

Consequently, all polynomial functions are of bounded variation on compact intervals.

Lemma 1.4. If P and Q are partitions of I such that P ⊆ Q, then

∑(P) ⩽ ∑(Q).

Theorem 1.5. Suppose x ∈ (a, b). Then f is of bounded variation on [a, b] if and only if it is of bounded variation on both [a, x] and [x, b]. In either case,

V_{f}[a, b] = V_{f}[a, x] + V_{f}[x, b].

Proof. Say f is of bounded variation on [a, b]. Let Q be a partition of [a, x], and R be a partition of [x, b]. Then Q ∪ R is a partition of [a, b]. Hence

∑(Q) + ∑(R) = ∑(Q ∪ R) ⩽ V_{f}[a, b].

Therefore f is of bounded variation on [a, x] and on [x, b], and

V_{f}[a, x] + V_{f}[x, b] ⩽ V_{f}[a, b].  (i)

Conversely, say f is of bounded variation both on [a, x] and on [x, b]. If P is a partition of [a, b], then so is P ∪ {x}, and the latter is Q ∪ R for some partitions Q and R of [a, x] and [x, b] respectively. Hence (by Lemma 1.4)

∑(P) ⩽ ∑(P ∪ {x}) = ∑(Q) + ∑(R) ⩽ V_{f}[a, x] + V_{f}[x, b].

Therefore f is of bounded variation on [a, b], and

V_{f}[a, b] ⩽ V_{f}[a, x] + V_{f}[x, b].  (ii)

Either of the foregoing hypotheses now yields the other and hence yields both (i) and (ii).

Theorem 1.5 (with Theorem 1.1) gives another proof that polynomial functions are of bounded variation on compact intervals.

Theorem 1.6. Suppose that f is continuous on [a, b] and differentiable on (a, b) and that f′ is bounded on (a, b). Then f is of bounded variation on [a, b].

Proof. For every partition {x_{0}, …, x_{n}} in P[a, b], for every k in {1, …, n}, by the Mean-Value Theorem there is t_{k} in (x_{k−1}, x_{k}) such that

\frac{△f_{k}}{△x_{k}} = f′(t_{k}).

Hence, if A is an upper bound of {f′(x) : x ∈ (a, b)}, then

∑_{k=1}^{n} |△f_{k}| = ∑_{k=1}^{n} |f′(t_{k})| △x_{k} ⩽ ∑_{k=1}^{n} A△x_{k} = A(b − a),

so f is of bounded variation by definition.

Hence for example the function

x ↦ \begin{cases}x sin log x, & if x ∈ (0, 1]; \\ 0, & if x = 0 \end{cases}

is of bounded variation. The converse of Theorem 1.6 fails: consider x ↦ √x : [0, 1] → ℝ.

Theorem 1.7. Suppose f is of bounded variation on I. Then the functions x ↦ V_{f}[a, x] and x ↦ V_{f}[a, x] − f(x) are increasing on I.

Proof. Suppose a ⩽ x ⩽ y ⩽ b. By Theorem 1.5, we have

V_{f}[a, x] ⩽ V_{f}[a, y].

Hence x ↦ V_{f}[a, x] is increasing. We also have

V_{f}[a, x] + f(y) − f(x) ⩽ V_{f}[a, x] + |f(y) − f(x)| ⩽ V_{f}[a, x] + V_{f}[x, y] = V_{f}[a, y],

whence V_{f}[a, x] − f(x) ⩽ V_{f}[a, y] − f(y). Therefore x ↦ V_{f}[a, x] − f(x) is increasing.

In the proof, note that if, for example, x = y, then [x, y] is not in fact an interval, although we can still understand the total variation V_{f}[x, y] to be 0.

Theorem 1.8. A function is of bounded variation if and only if it is the difference of two increasing functions.

Lemma 1.9. Suppose f is of bounded variation on [a, b], and c ∈ [a, b]. Then f is continuous at c if and only if x ↦ V_{f}[a, x] is continuous at c.

Proof (not given in class). Let V be the increasing function x ↦ V_{f}[a, x]. If a ⩽ x ⩽ y ⩽ b, then

0 ⩽ |f(y) − f(x)| ⩽ V_{f}[x, y] = V (y) − V (x).

Hence continuity of V implies that of f.

Suppose conversely that f is continuous at c. Let ε > 0. It is enough to find positive numbers δ_{ℓ} and δ_{r} such that, for all x in [a, b], we have the two implications

c − δ_{ℓ} < x ⩽ c ⟹ V (c) − V (x) < ε,
c ⩽ x < c + δ_{r} ⟹ V (x) − V (c) < ε.

Now, by continuity of f at c, we can pick a positive δ so that, for all x in [a, b],

c − δ < x < c + δ′ ⟹ |f(c) − f(x)| < \frac{ε}{2}.

There is a partition P of [a, c] such that, for all finer partitions {u_{0}, …, u_{m}}, we have

V (c) ⩽ ∑_{k=1}^{m} |△f_{k}| + ε/2.

Let y be the greatest element of P ∖ {c}, and let

δ_{ℓ} = min(c − y, δ).

Suppose c − δ_{ℓ} < x < c. We may assume u_{m−1} = x; hence

V (c) ⩽ ∑_{k=1}^{m−1} |△f_{k}| + ε ⩽ V (x) + ε.

Thus we have the first implication above; the second is obtained similarly.

Theorem 1.10. A continuous function is of bounded variation if and only if it is the difference of two continuous increasing functions.

2 The Riemann–Stieltjes Integral

Let us continue to use the notation of the previous section, letting also g be a real-valued function on I. If g is in fact differentiable, and if one writes g′(x) as d g/ d x, then one might write g′(x) d x as d g, and ∫_{a}^{b} f(x)g′(x) d x as ∫_{a}^{b} f d g. If f is also differentiable, then the rule of integration by parts is

∫_{a}^{b} f d g = f(b)g(b) − f(a)g(a) − ∫_{a}^{b} g d f.  (iii)

We shall show that this equation is meaningful and true even if f and g are not differentiable.

If P, Q ∈ PI, and P ⊆ Q, let us say that Q is finer than P (also, P is coarser than Q).

Theorem 2.1. There is at most one number A such that, for every positive ε, there is a partition P of I such that, for all partitions {u_{0}, …, u_{m}} of I that are finer than P, and for all t_{k} in [u_{k−1}, u_{k}] (where k ∈ {1, 2, …, m}),

|∑_{k=1}^{m} f(t_{k})△g_{k} − A| < ε.

The sum in the theorem is a Riemann–Stieltjes sum of f with respect to g for the partition P. If the number A exists, it is denoted

∫_{a}^{b} f d g.

This is a Riemann–Stieltjes integral, and in particular it is the Riemann integral of f on I with respect to g. The set of all functions f on I whose Riemann integrals with respect to g exist is denoted

R(g).

If g is differentiable, with constant derivative 1, then ∫_{a}^{b} f d g is just ∫_{a}^{b} f, the ordinary Riemann integral.

Theorem 2.2. If f ∈ R(g), then g ∈ R(f), and also equation (iii) holds.

Proof. Let {x_{0}, …, x_{n}} be a partition P as in Theorem 2.1. Choose some t_{k} in [x_{k−1}, x_{k}]. Then

f(b)g(b) − f(a)g(a) − ∑_{k=1}^{n} g(t_{k})△f_{k} =

∑_{k=1}^{n} f(x_{k})g(x_{k}) − ∑_{k=1}^{n} f(x_{k−1})g(x_{k−1}) − ∑_{k=1}^{n} g(t_{k})f(x_{k}) + ∑_{k=1}^{n} g(t_{k})f(x_{k−1}) =

∑_{k=1}^{n} f(x_{k})(g(x_{k}) − g(t_{k})) + ∑_{k=1}^{n} f(x_{k−1})(g(t_{k}) − g(x_{k−1})).

This is a Riemann–Stieltjes sum for f with respect to g corresponding to the partition {x_{0}, t_{1}, x_{1}, …, x_{n−1}, t_{n}, x_{n}}. This partition is finer than P, so the sum is within ε of ∫_{a}^{b} f d g. Therefore the Riemann–Stieltjes sum ∑_{k=1}^{n} g(t_{k})△f_{k} is within ε of f(b)g(b) − f(a)g(a) − ∫_{a}^{b} f d g.

Theorem 2.3. R(g) is a vector-space, that is, it contains the 0-function and is also closed under the taking of linear combinations. In fact, if f_{i} ∈ R(g) and a_{i} ∈ ℝ (where i < n), then

∫_{a}^{b} ∑_{i<n} a_{i}f_{i} d g = ∑_{i<n} a_{i} ∫_{a}^{b} f_{i} d g.

Hence also

⋂_{i<n} R(f_{i}) ⊆ R( ∑_{i<n} a_{i}f_{i}),

and if g ∈ ⋂_{i<n} R(f_{i}), then

∫_{a}^{b} g d ∑_{i<n} a_{i}f_{i} = ∑_{i<n} a_{i} ∫_{a}^{b} g d f_{i}.

If ∫_{a}^{b} f d g exists, then we may define

∫_{b}^{a} f d g = − ∫_{a}^{b} f d g.

Also, ∫_{a}^{a} f d g = 0.

Theorem 2.4. If x ∈ I, and any two of the three integrals

∫_{a}^{x} f d g,  ∫_{x}^{b} f d g,  ∫_{b}^{a} f d g

exist, then they all exist, and their sum is 0.

Theorem 2.5. Let h be an increasing bijection from an interval [c, d] onto [a, b]. If f ∈ R(g), then f ∘ h ∈ R(g ∘ h), and

∫_{c}^{d} f ∘ h d(g ∘ h) = ∫_{h(c)}^{h(d)} f d g.

Proof. Pick P as in Theorem 2.1. This determines a partition h^{−1}(P) of [c, d]. A Riemann–Stieltjes sum of f ∘ h on [c, d] with respect to g ∘ h for a partition finer than h^{−1}(P) can be understood as a Riemann–Stieltjes sum of f on [a, b] with respect to g for a partition finer than P; hence it is within ε of ∫_{a}^{b} f d g.

Theorem 2.6. If f is bounded on I, and f ∈ R(g), and g has a continuous derivative on I, then fg′ is Riemann-integrable on I, and

∫_{a}^{b} fg′ = ∫_{a}^{b} f d g.

Proof. For any partition {x_{0}, …, x_{n}} of I, by the Mean-Value Theorem, there are u_{k} in (x_{k−1}, x_{k}) such that △g_{k} = g′(u_{k})△x_{k}. Hence, for any t_{k} in [x_{k−1}, x_{k}], the Riemann–Stieltjes sum

∑_{k=1}^{n} f(t_{k})△g_{k}

is precisely

∑_{k=1}^{n} f(t_{k})g′(u_{k})△x_{k}.

We have

|∑_{k=1}^{n} f(t_{k})g′(u_{k})△x_{k} − ∑_{k=1}^{n} f(t_{k})g′(t_{k})△x_{k}| = ∑_{k=1}^{n} f(t_{k}) |g′(u_{k}) − g′(t_{k})| △x_{k}.

Since f is bounded on I, and g′—being continuous on a compact interval—is uniformly continuous on I, we conclude that, in the last equation, the right member can be made smaller than ε/2, provided that the partition {x_{0}, …, x_{n}} is sufficiently fine. This partition can also be fine enough that the sum ∑_{k=1}^{n} f(t_{k})△g_{k} is within ε/2 of ∫_{a}^{b} f d g. In this case, we have

|∫_{a}^{b} f d g − ∑_{k=1}^{n} f(t_{k})g′(t_{k})△x_{k}| < ε,

which establishes the claim.

Lemma 2.7. Suppose u ∈ (a, b], and f is continuous from the left at u, that is, for every positive ε, there is a positive δ such that for all t in I,

u − δ < t ⩽ u ⟹ |f(u) − f(t)| < ε.

Let g be defined so that g(x) = \begin{cases}0, & if a ⩽ x < u; \\ 1, & if u ⩽ x ⩽ b. \end{cases} Then f ∈ R(g) and

∫_{a}^{b} f d g = f(u).

Proof. Suppose s ∈ (a, u). For every partition that is finer than {a, s, u, b}, every Riemann–Stieltjes sum for f with respect to g is

f(t)(g(u) − g(s)),

that is, f(t), for some t in [s, u]. With ε and δ as in the statement, if we choose s in (u − δ, u), then |f(u) − f(t)| < ε.

Theorem 2.8. Every finite sum can be expressed as a Riemann–Stieltjes integral. In fact, if f is continuous on [a, b], and a < x_{1} < ⋯ < x_{n} = b, then

∑_{k=1}^{n} f(x_{k}) = ∫_{a}^{b} f d g

for some function g.

Proof. Let g = max{k : x_{k} ⩽ x}.

In the theorem and its proof, if x_{k} = k for each k, then g is the greatest-integer function, x ↦ ⌊x⌋. Hence:

Theorem 2.9 (Euler’s summation-formula). If f is continuously differentiable on [0, n], then

∑_{k=1}^{n} f(k) = ∫_{0}^{n} f + ∫_{0}^{n}(x − ⌊x⌋)f′(x) d x;

∑_{k=0}^{n} f(k) = ∫_{0}^{n} f + ∫_{0}^{n} (x − ⌊x⌋ − \frac{1}{2}) f′(x) d x + \frac{1}{2}(f(n) + f(0)).

Proof. Just calculate:

∑_{k=1}^{n} f(k) = ∫_{0}^{n} f(x) d ⌊x⌋ = ∫_{0}^{n} f − ∫_{0}^{n} f(x) d(x − ⌊x⌋) =

∫_{0}^{n} f − ∫_{0}^{n}(x − ⌊x⌋) d f(x) = ∫_{0}^{n} f + ∫_{0}^{n}(x − ⌊x⌋)f′(x) d x,

by Theorems 2.8, 2.3, 2.2 and 2.6 respectively. Note however that, to have the second equation, we must know that ∫_{a}^{b} f does in fact exist. Since f is continuous, its Riemann-integrability will be a consequence of Theorem 2.18. For the second equation, add f(0) to both sides, and use

∫_{0}^{n} f′ = f(n) − f(0),

which will be Theorem 2.22.

As noted in the last proof, we need to be able to say which functions are Riemann-integrable with respect to a given function. Towards this end, we make some definitions—but we must assume that f is bounded:

m_{k}(f) = inf{f(x) : x ∈ [x_{k−1}, x_{k}]},  L(P, f, g) = ∑_{k=1}^{n} m_{k}(f)△g_{k},

M_{k}(f) = sup{f(x) : x ∈ [x_{k−1}, x_{k}]},  U(P, f, g) = ∑_{k=1}^{n} M_{k}(f)△g_{k}.

Lemma 2.10. Assume g is increasing. For all partitions P and Q of I,

L(P, f, g) ⩽ L(P ∪ Q, f, g) ⩽ U(P ∪ Q, f, g) ⩽ U(Q, f, g).

If g is increasing, then by the lemma we can define

\underline{∫{b \atop a}}f d g = sup\{L(P, f, g) : P ∈ PI\}, \overline{∫{b \atop a}}f d g = inf\{U(P, f, g) : P ∈ PI\}.

Moreover:

Lemma 2.11. Assume g is increasing. Then

L(P, f, g) ⩽ \underline{∫{b \atop a}}f d g ⩽ \overline{∫{b \atop a}}f d g ⩽ U(P, f, g)

for all partitions P of I.

Suppose g is increasing. Let us say that f satisfies Riemann’s condition with respect to g on I if for each positive ε there is a partition P of I such that

U(P, f, g) − L(P, f, g) < ε.

Lemma 2.12. Assume g is increasing. Then the following conditions are equivant:

(0) f satisfies Riemann’s condition with respect to g on I.

(1) \underline{∫{b \atop a}}f d g = \overline{∫{b \atop a}}f d g.

(2) f ∈ R(g).

Proof. If condition (0) holds, then \overline{∫{b \atop a}}f d g − \underline{∫{b \atop a}}f d g < ε for all positive ε, so condition (1) holds.

Suppose ε > 0. By Lemma 2.10, we can let P be a partition of I such that L(P, f, g) is within ε of \underline{∫{b \atop a}}f d g, and U(P, f, g) is within ε of \overline{∫{b \atop a}}f d g. But if P = {x_{0}, …, x_{n}}, for any t_{k} in [x_{k−1}, x_{k}], we have

L(P, f, g) ⩽ ∑_{i=1}^{n} f(t_{k})△g_{k} ⩽ U(P, f, g);

if condition (1) holds, then by Lemma 2.11, the sum ∑_{i=1}^{n} f(t_{k})△g_{k} must be

within ε of the common value of \underline{∫{b \atop a}}f d g and \overline{∫{b \atop a}}f d g. Hence this common value

is ∫_{a}^{b} f d g, and condition (2) holds.

Finally, suppose (2) holds. We can choose P so that

|∑_{i=1}^{n} f(t_{k})△g_{k} − ∫_{a}^{b} f d g| < \frac{ε}{4}

for all t_{k} in [x_{k−1}, x_{k}]. But also, since g is increasing, we can choose the t_{k} so that, in addition,

|∑_{i=1}^{n} f(t_{k})△g_{k} − L(P, f, g)| < \frac{ε}{4}.

Then |L(P, f, g) − ∫_{a}^{b} f d g| < ε/2. Likewise, |U(P, f, g) − ∫_{a}^{b} f d g| < ε/2. So condition (0) holds.

Theorem 2.13. Assume g is increasing. If f_{0}, f_{1} ∈ R(g), and f_{0} ⩽ f_{1} (that is, f_{0}(x) ⩽ f_{1}(x) for all x in I), then

∫_{a}^{b} f_{0} d g ⩽ ∫_{a}^{b} f_{1} d g.

Lemma 2.14. Assume g is increasing. If f ∈ R(g), then |f| ∈ R(g) and

|∫_{a}^{b} f d g| ⩽ ∫_{a}^{b} |f| d g.

Proof. Since | |f(x)| − |f(y)| | ⩽ |f(x) − f(y)|, we have

M_{k}(|f|) − m_{k}(|f|) ⩽ M_{k}(f) − m_{k}(f),

hence U(P, |f| , g) − L(P, |f| , g) ⩽ U(P, f, g) − L(P, f, g). Now use Lemma 2.12.

Lemma 2.15. Assume g is increasing. If f ∈ R(g), then f^{2} ∈ R(g).

Proof. Suppose M bounds |f|. Then

M_{k}(f^{2}) − m_{k}(f^{2}) =

(M_{k}(|f|) + m_{k}(|f|))(M_{k}(|f|) − m_{k}(|f|)) ⩽

2M(M_{k}(|f|) − m_{k}(|f|)).

Now use Lemma 2.12.

If g is merely of bounded variation, then g is the difference g_{0} − g_{1} of increasing functions, by Theorem 1.8, and R(g_{0}) ∩ R(g_{1}) ⊆ R(g), by Theorem 2.3. But what about the reverse inclusion?

Lemma 2.16. Assume g is of bounded variation. Let V be the function x ↦ V_{g}[a, x]. If f is bounded, and f ∈ R(g), then f ∈ R(V).

Proof (not given in class). Since V is increasing, we can use Lemma 2.12. We have

U(P, f, V) − L(P, f, V) = ∑_{k=1}^{n}(M_{k}(f) − m_{k}(f))△V_{k} =

∑_{k=1}^{n}(M_{k}(f) − m_{k}(f))(△V_{k} − |△g_{k}|) + ∑_{k=1}^{n}(M_{k}(f) − m_{k}(f)) |△g_{k}| ,

and we can make each of the last summations less than ε/2. Indeed, if |f| ⩽ M, then

∑_{k=1}^{n}(M_{k}(f) − m_{k}(f))(△V_{k} − |△g_{k}|) ⩽ 2M(V (b) − ∑_{k=1}^{n} |△g_{k}|),

which is less than ε/2 if P is fine enough. We can also choose t_{k} and t′_{k} in [x_{k−1}, x_{k}] so that

M_{k}(f) − m_{k}(f) < |f(t_{k}) − f(t′_{k})| + \frac{ε}{4V (b)}.

We may assume also that f(t_{k}) − f(t′_{k}) has the same sign as △g_{k}. Then

∑_{k=1}^{n}(M_{k}(f) − m_{k}(f)) |△g_{k}| < ∑_{k=1}^{n}(f(t_{k}) − f(t′_{k}))△g_{k} + \frac{ε}{4}.

If P is fine enough, then ∑_{k=1}^{n}(f(t_{k}) − f(t′_{k}))△g_{k} < ε/4.

Theorem 2.17. Assume g is of bounded variation. Then there are increasing functions g_{0} and g_{1} such that g = g_{0} − g_{1} and every bounded function in R(g) is also in R(g) = R(g_{0}) ∩ R(g_{1}).

Theorem 2.18. If g is of bounded variation, then R(g) contains all continuous functions.

Proof. It is enough to prove the theorem in case g is increasing and g(a) < g(b). Suppose f is continuous (on I); then f is uniformly continuous. Let ε > 0. Then there is a positive δ such that for all x and y (in I),

|x − y| < δ ⟹ |f(x) − f(y)| < \frac{ε}{g(b) − b(a)}.

Let P be a partition fine enough that △x_{k} < δ in each case. Since f is continuous, we have

M_{k}(f) − m_{k}(f) = |f(t_{k}) − f(t′_{k})| < \frac{ε}{g(b) − b(a)}

for some t_{k} and t′_{k} in [x_{k−1}, x_{k}], for each k; this means

U(P, f, g) − L(P, f, g) < ∑_{k=1}^{n} \frac{ε}{g(b) − g(a)}△g_{k} = ε.

Thus, Riemann’s condition is met.

Corollary 2.19. All functions of bounded variation are Riemann-integrable.

Lemma 2.20 (Mean-Value Theorem for Integrals). Suppose f is continuous, and g is increasing. Then

∫_{a}^{b} f d g = f(t)(g(b) − g(a)) = f(t) ∫_{a}^{b} d g

for some t in (a, b).

Proof. The second equation is clear from the definition of the Riemann–Stieltjes integral, or from Theorem 2.2. For the first equation, work with the partition {a, b} of I. We have

m_{1}(f)(g(b) − g(a)) ⩽ ∫_{a}^{b} f d g ⩽ M_{1}(f)(g(b) − g(a)),

so (∫_{a}^{b} f d g)/(g(b) − g(a)) is between m_{1}(f) and M_{1}(f). Now use the intermediate-value theorem for continuous functions.

Theorem 2.21 (First Fundamental, of Calculus). If g is increasing, then the function

x ↦ ∫_{a}^{x} f d g

is well-defined on I, and is differentiable, with derivative f(u)g′(u), at every point u of I where f is continuous and g is differentiable.

Proof. If u is such a point, and u + h ∈ I, then

\frac{∫_{a}^{u+h} f d g − ∫_{a}^{u} f d g}{h} = \frac{∫_{u}^{u+h} f d g}{h} = f(t)\frac{g(u + h) − g(u)}{h}

for some t between u and u + h inclusive.

Theorem 2.22 (Second Fundamental, of Calculus). For any function g that is continuous on I and differentiable on (a, b),

∫_{a}^{b} g′ = g′(b) − g′(a),

provided the integral exists.

Proof. For any partition {x_{0}, …, x_{n}} of I, we can pick t_{k} in (x_{k−1}, x_{k}) such that g′(t_{k})△x_{k} = △g′_{k}, whence the Riemann sum ∑_{k=1}^{n} g′(t_{k})△x_{k} is just g′(b) − g′(a).

3 Infinite series and products

An infinite sequence is a function with domain {n ∈ ℤ : k ⩽ n} for some k in ℤ. Usually the domain is ℕ (the set {0, 1, 2, … } of natural numbers) or ℤ_{+} (the set {1, 2, 3, … } of positive integers).

Informally, a sequence (a_{n}) has a real number b as a limit, provided that a_{n} is close to b whenever n is sufficiently large. We can understand ‘sufficiently large’ to mean ‘close to ∞’. Then we can allow the limit b to be ∞ or −∞ as well.

To be more precise: Recall that a neighborhood of a real number x is a set of real numbers that has, as a subset, an open interval that contains x. We can define a neighborhood of ∞ to be a set of real numbers that has, as a subset, an interval of the form (a, ∞). Similarly, neighborhoods of −∞ have subsets (∞, a).

Now we can say that the sequence (a_{n}) of real numbers has the limit b (where b ∈ {−∞}∪ℝ∪{∞}) if, for every neighborhood U of b, there is a neighborhood U′ of ∞ such that, for all n in the domain of the sequence,

n ∈ U′ ⟹ b_{n} ∈ U.

In this case, we write

lim_{n→∞} a_{n} = b.

If b is finite (that is, b ∈ ℝ), then the sequence is said to converge to b. If b is infinite, then the sequence diverges to b. If the sequence has no limit, then the sequence simply diverges.

The extended real number system, ℝ^{∗}, consists of the elements of the set

{−∞} ∪ ℝ ∪ {∞}.

This can be ordered in the obvious way, so that −∞ < x < ∞ for all x in ℝ. Defining open and closed intervals of ℝ^{∗} is somewhat easier than for ℝ. If a, b ∈ ℝ^{∗}, then

(a, b) = {x ∈ ℝ^{∗} : a < x < b}; [a, b] = {x ∈ ℝ^{∗} : a ⩽ x ⩽ b}.

An open interval of ℝ^{∗} is an interval (a, b); a closed, [a, b]. Note that ℝ is an open, but not a closed, interval of ℝ^{∗}.

Let f be the function from [−1, 1] to ℝ^{∗} given by

f(x) = \begin{cases}−∞, & if x = −1; \\ \frac{x}{1−x^{2}} , & if − 1 < x < 1; \\ ∞, & if x = 1. \end{cases}

Then f is a bijective, order-preserving function. This means that both f^{−1} and f send open sets to open sets; that is, both f and f^{−1} are continuous. However, we cannot extend the usual metric on ℝ so as to make ℝ^{∗} into a metric space; ℝ^{∗} is simply a topological space.

If a subset A of ℝ has no upper bound in ℝ, then we may write

sup A = ∞;

if no lower bound, then inf A = −∞. Also, sup ∅ = inf ℝ = −∞, and inf ∅ = sup ℝ = ∞. Thus, every subset of ℝ has a supremum and in infimum in ℝ^{∗}.

Every monotone sequence has a limit in ℝ^{∗}.

A sequence (a_{n}) has the limit superior b in ℝ if for all positive ε:

∃M ∀n (M < n ⟹ a_{n} < b + ε);

and

∀M ∃n (M < n & b − ε < a_{n}).

In this case, we write

lim sup_{n→∞} a_{n} = b.

Theorem 3.1. The following are equivalent:

(1) lim sup_{n→∞} a_{n} = b.

(2) lim_{n→∞} sup{a_{m} : n < m} = b.

(3) (a_{n}) has a subsequence converging to b, but not a subsequence converging to a greater limit.

If {a_{n}} has no upper bound, then

lim sup_{n→∞} a_{n} = ∞;

if there is an upper bound, but no finite limit superior, then

lim sup_{n→∞} a_{n} = −∞.

So every sequence has a limit superior in ℝ^{∗}. The limit inferior is defined similarly; in fact,

lim inf_{n→∞} a_{n} = − lim sup_{n→∞} −a_{n}.

Examples 3.2.

a_{n}  lim sup_{n→∞} a_{n} lim inf_{n→∞} a_{n}
(−1)^{n}  1  −1
\frac{n + 1}{n}(1 + (−1)^{n})  2  0
n(1 + (−1)^{n})  ∞  0
log n  ∞  ∞

For a sequence (a_{p}, a_{p+1}, …), the sequence of its partial sums is

(∑_{i=p}^{n} a_{i} : n = p, p + 1, …) .

Considered with respect to its sequence of partial sums, a sequence is a series; the series converges to a limit b if the sequence of partial sums does; in this case, we write

∑_{i=p}^{∞} a_{i} = b.

A sequence (a_{n}), as a series, can be written

∑ a_{n}.

Theorem 3.3. The set of convergent series is a vector-space: if ∑ a_{n} and ∑ b_{n} converge, and c, d ∈ ℝ, then ∑(ca_{n} + db_{n}) converges, and

∑_{k=p}^{∞}(ca_{n} + db_{n}) = c ∑_{k=p}^{∞} a_{n} + d ∑_{k=p}^{∞} b_{n}.

Lemma 3.4 (Cauchy condition). The series ∑ a_{n} converges if and only if, for every positive ε, there is M such that, for all k in ℕ,

n > M ⟹ | ∑_{m=n}^{n+k} a_{m}| < ε.

Theorem 3.5. If ∑ a_{n} converges, then lim_{n→∞} a_{n} = 0.

Proof. Use the Cauchy condition when k = 1.

It is clear that, if ∑ a_{n} converges, then so does ∑(a_{2n} + a_{2n+1}).

Lemma 3.6. If ∑(a_{2n} + a_{2n+1}) converges, and lim_{n→∞} a_{n} = 0, then ∑ a_{n} converges.

Proof. Let ∑_{k=p}^{∞}(a_{2k} + a_{2k+1}) = b. Let M be such that

n > M ⟹ |∑_{k=p}^{p+n}(a_{2k} + a_{2k+1}) − b| , |a_{2(p+n)+1}| < \frac{ε}{2}.

Then also n > M ⟹ |∑_{k=p}^{2(p+n)} a_{k} − b| , |∑_{k=p}^{2(p+n)+1} a_{k} − b| < ε.

A series ∑ a_{n} is alternating if (−1)^{n}a_{n} ⩾ 0 for each n, or (−1)^{n}a_{n} ⩽ 0 for each n.

Theorem 3.7. An alternating series ∑ a_{n} converges if lim_{n→∞} a_{n} = 0.

Proof. Assume (−1)^{n}a_{n} ⩾ 0 in each case. Then a_{2n} − a_{2n+1} ⩾ 0, and

∑_{k=0}^{n}(a_{2k} − a_{2k+1}) = a_{0} − ∑_{k=0}^{n−1}(a_{2k+1} − a_{2k+2}) − a_{2n+1},

so partial sums of ∑(a_{2n} − a_{2n+1}) are bounded. Therefore the latter series converges. Now use Lemma 3.6.

Example 3.8. ∑(−1)^{n}/n converges.

Theorem 3.9. If |x| < 1, then

∑_{k=0}^{∞} x^{k} = \frac{1}{1 − x}.

Proof. ∑_{k=0}^{n} x^{k} = (1−x^{n+1})/(1−x). If |x| < 1, then the partial sums converge to 1/(1 − x).

Lemma 3.10 (Comparison Test). If 0 ⩽ a_{n} ⩽ cb_{n} when n is large enough, for some non-zero c, and ∑ b_{n} converges, then ∑ a_{n} converges.

Theorem 3.11 (Limit Comparison Test). If lim_{n→∞} a_{n}/b_{n} is finite and non-zero, then

∑ a_{n} converges ⟺ ∑ b_{n} converges.

Theorem 3.12 (Integral Test). If f is monotone on [0, ∞), then

∑ f(n) converges ⟺ lim_{x→∞} ∫_{0}^{x} f is finite.

Proof. Assume f is decreasing. If f(x) < 0 for some x, then each member of the equivalence fails. Suppose f(x) ⩾ 0 for all x. Since f(k + 1) ⩽ f(x) ⩽ f(k) when k ⩽ x ⩽ k + 1, we have

∑_{k=1}^{n} f(k) ⩽ ∫_{0}^{n} f ⩽ ∑_{k=0}^{n−1} f(k).

If ∫_{0}^{∞} f exists, then so does ∑_{k=1}^{∞} a_{k}; if not, not.

Examples 3.13. We have

∫_{1}^{x} \frac{d t}{t^{s}} = \begin{cases}log x, & if s = 1; \\ \frac{1}{s − 1} (1 − \frac{1}{x^{s−1}} ) , & if s ≠ 1. \end{cases}

Hence ∑ 1/n^{s} converges if and only if s > 1.

A series ∑ a_{n} is absolutely convergent if ∑ |a_{n}| converges. Absolute convergence implies convergence, by the Cauchy criterion. Theorem 3.14 (Ratio Test). Let

r = lim inf_{n→∞} |\frac{a_{n+1}}{a_{n}} | and R = lim sup_{n→∞} |\frac{a_{n+1}}{a_{n}} | .

Then • R < 1 ⟹ ∑ a_{n} converges absolutely;

• 1 < r ⟹ ∑ a_{n} diverges. In case r ⩽ 1 ⩽ R, no conclusion is possible. Proof. Suppose R < x < 1. Then, for some M,

n ⩾ M ⟹ |\frac{a_{n+1}}{a_{n}} | < x,

so n ⩾ M ⟹ |a_{n+1}| < |a_{M}| x^{n−M}. By the Comparison Test, ∑ a_{n} converges. Suppose 1 < r. Then, for some M,

n ⩾ M ⟹ |\frac{a_{n+1}}{a_{n}} | > 1,

which means a_{n} cannot converge to 0.

Finally, lim_{n→∞} \frac{1/(n + 1)}{1/n} = 1 = \frac{1/(n + 1)^{2}}{1/n^{2}} .

Theorem 3.15 (Root Test). Let ρ = lim sup_{n→∞} \sqrt[n]{|a_{n}|}. Then • ρ < 1 ⟹ ∑ a_{n} converges absolutely;

• 1 < ρ ⟹ ∑ a_{n} diverges. In case ρ = 1, no conclusion is possible. Proof. If ρ < x < 1, and n is large enough, then

\sqrt[n]{|a_{n}|} < x,

whence |a_{n}| < x^{n}. If 1 < ρ, then a_{n} does not converge to 0. Finally, lim_{n→∞} n^{1/n} = exp lim_{n→∞} log n/n = e^{0} = 1; hence lim_{n→∞}(n^{2})^{1/n} = 1. Theorem 3.16. If ∑ a_{n} is absolutely convergent, and f : ℕ → ℕ is a bijection, then ∑ a_{f(n)} converges, and

∑_{k=0}^{∞} |a_{f(k)}| = ∑_{k=0}^{∞} |a_{k}| .

Proof. ∑_{k=0}^{n} |a_{f(k)}| ⩽ ∑_{k=0}^{∞} |a_{k}|, so ∑ a_{f(n)} converges absolutely.

We have

|∑_{k=0}^{n} a_{f(k)} − ∑_{k=0}^{∞} a_{k}| ⩽ |∑_{k=0}^{n} a_{f(k)} − ∑_{k=0}^{m} a_{k}| + | ∑_{k=m+1}^{∞} a_{k}| .

Let ε > 0. Let m be large enough that |∑_{k=m+1}^{∞} a_{k}| ⩽ ∑_{k=m+1}^{∞} |a_{k}| ⩽ ε/2. Then n can be large enough that

{0, 1, …, m} ⊆ {f(0), f(1), …, f(n)}.

Then |∑_{k=0}^{n} a_{f(k)} − ∑_{k=0}^{m} a_{k}| < ε/2, so |∑_{k=0}^{n} a_{f(k)} − ∑_{k=0}^{∞} a_{k}| < ε.

Theorem 3.17. If ∑ a_{n} is convergent, but not absolutely convergent, and −∞ ⩽ b ⩽ c ⩽ ∞, then there is a bijection f : ℕ → ℕ such that

lim inf_{n→∞} ∑_{k=0}^{n} a_{f(k)} = b and lim sup_{n→∞} ∑_{k=0}^{n} a_{f(k)} = c.

Proof. For any x in ℝ, we have

x = max(0, x) + min(0, x) and |x| = max(0, x) − min(0, x).

Hence (by Theorem 3.3), if ∑ a_{n} and one of ∑ max(0, a_{n}) and ∑ min(0, a_{n}) converge, then ∑ a_{n} converges absolutely.

Suppose ∑ a_{n} converges, but not absolutely, and −∞ < b ⩽ c < ∞. Then both ∑ max(0, a_{n}) and ∑ min(0, a_{n}) diverge, to ∞ and −∞ respectively. So it is possible to find a strictly increasing sequence (g(n) : n ∈ ℕ) and a bijection f : ℕ → ℕ such that:

• g(0) = 0;

• f is increasing on {k ∈ ℕ : (∃n ∈ ℕ) g(2n) ⩽ k < g(2n + 1)} and on its complement;

• a_{f(k)} ⩾ 0 ⟺ (∃n ∈ ℕ) g(2n) ⩽ k < g(2n + 1);

• ∑_{k=0}^{g(2n+1)−2} a_{f(k)} ⩽ c < ∑_{k=0}^{g(2n+1)−1} a_{f(k)};

• ∑_{k=0}^{g(2n+2)−1} a_{f(k)} < b ⩽ ∑_{k=0}^{g(2n+2)−2} a_{f(k)}.

Since lim_{n→∞} a_{f(n)} = 0, we have

lim inf_{n→∞} ∑_{k=0}^{n} a_{f(k)} = b and lim sup_{n→∞} ∑_{k=0}^{n} a_{f(k)} = c.

If possibly one or both of b and c are infinite, then we modify the construction of f by choosing sequences (b_{n}) and (c_{n}) that have limits b and c respectively and that satisfy b_{n} < c_{n}. Then we require

∑_{k=0}^{g(2n+1)−2} a_{f(k)} ⩽ c_{n} < ∑_{k=0}^{g(2n+1)−1} a_{f(k)}

and ∑_{k=0}^{g(2n+1)−1} a_{f(k)} < b_{n} ⩽ ∑_{k=0}^{g(2n+1)−2} a_{f(k)}.

Suppose (p, q) ↦ a_{p q} is a function from ℕ × ℕ to ℝ (that is, a double sequence). Then we say

lim_{(p,q)→∞} a_{p q} = b

if for all positive ε there is M such that

p, q > M ⟹ |a_{p q} − b| < ε.

Theorem 3.18. Suppose

lim_{(p,q)→∞} a_{p q} = b.

If lim_{q→∞} a_{p q} exists for each p in ℕ, then

lim_{p→∞} lim_{q→∞} a_{p q} = b.

Proof. Let N be such that

p, q > N ⟹ |a_{p q} − b| < \frac{ε}{2}.

Suppose p > N and lim_{q→∞} a_{p q} = c_{p}. There is M such that

q > M ⟹ |a_{p q} − c_{p}| < \frac{ε}{2}.

Hence, if q > max(M, N), then |c_{p} − b| ⩽ |c_{p} − a_{p q}| + |a_{p q} − b| < ε.

Example 3.19. We have

lim_{p→∞} lim_{q→∞} \frac{pq}{p^{2} + q^{2}} = lim_{p→∞} 0 = 0;

but pq/(p^{2} + q^{2}) = 1/2 if p = q; so the double sequence

(p, q) ↦ \frac{pq}{p^{2} + q^{2}}

has no limit.

Suppose (a_{n} : n ∈ ℕ) is a sequence of real numbers. We shall say that ∏ a_{n} converges if, for some p in ℕ, the sequence of products ∏_{k=p}^{n} a_{k} converges to a finite and non-zero limit b. In this case, we write

∏_{k=p}^{∞} a_{n} = b and ∏_{k=0}^{∞} a_{k} = b · ∏_{k=0}^{p−1} a_{k}.

(If p = 0, then ∏_{k=0}^{p−1} a_{k} = 1.) The definition has two immediate consequences:

• If ∏ a_{n} converges, then a_{n} ≠ 0 for all but finitely many n.

• ∏ a_{n} converges if and only if ∏ 1/a_{n} converges.

Examples 3.20. ∏_{k=1}^{n}(1 + 1/k) = ∏_{k=1}^{n}((k + 1)/k) = n + 1, and

∏_{k=2}^{n+1} (1 − \frac{1}{k} ) = ∏_{k=2}^{n+1} \frac{k − 1}{k} = \frac{1}{n + 1};

so ∏(1 ± 1/n) diverges in each case.

Theorem 3.21 (Cauchy condition for products). The product ∏ a_{n} converges if and only if, for every positive ε, there is M such that, for all n and k in ℕ,

n > M ⟹ | ∏_{ℓ=n}^{n+k} a_{ℓ} − 1| < ε.

Proof. Suppose ∏ a_{n} converges. Then for some p there is a positive δ such that

|∏_{ℓ=p}^{p+n} a_{ℓ}| > δ

for all n in ℕ. Also, there is M such that

n > M ⟹ | ∏_{ℓ=p}^{p+n+k+1} a_{ℓ} − ∏_{ℓ=p}^{p+n} a_{ℓ}| < εδ

for all k in ℕ. Division yields

n > M ⟹ | ∏_{ℓ=p+n+1}^{p+n+k+1} a_{ℓ} − 1| < ε.

Suppose conversely that for all positive ε there is M such that

n > M ⟹ | ∏_{ℓ=n}^{n+k} a_{ℓ} − 1| < ε.

for all k in ℕ. Then, in particular, there is p such that

\frac{1}{2} < ∏_{ℓ=p}^{p+k} a_{ℓ} < \frac{3}{2}

for all k in ℕ. Hence if lim_{n→∞} ∏_{ℓ=p}^{p+n} a_{ℓ} exists, then it is not zero, so ∏ a_{n} converges. We can show that this limit exists by the Cauchy criterion. Indeed, we have

| ∏_{ℓ=p}^{p+n+k+1} a_{ℓ} − ∏_{ℓ=p}^{p+n} a_{ℓ}| = | ∏_{ℓ=p+n+1}^{p+n+k+1} a_{ℓ} − 1| · |∏_{ℓ=p}^{p+n} a_{ℓ}| < \frac{3}{2}ε

if n > M.

Theorem 3.22. If (a_{n}) is a sequence of positive terms, then

∏(1 + a_{n}) converges ⟺ ∑ a_{n} converges.

Proof. If 1 < x, then, by the Mean-Value Theorem, 1 ⩽ e^{t} = (e^{x} − 1)/x for some t in (1, x), so 1 + x ⩽ e^{x}. Hence

∑_{k=0}^{n} a_{k} < 1 + ∑_{k=0}^{n} a_{k} ⩽ ∏_{k=0}^{n}(1 + a_{k}) ⩽ ∏_{k=0}^{n} exp a_{k} = exp ∑_{k=0}^{n} a_{k}.

Thus, the increasing sequence of partial sums is bounded if and only if the increasing sequence of partial products is bounded.

The Riemann Zeta-function is defined on the interval (1, ∞) by the formula

ζ(s) = ∑_{n=1}^{∞} \frac{1}{n^{s}} .

(See Examples 3.13.)

Theorem 3.23 (Euler product). Let the set of prime numbers be the range of the increasing sequence (p_{n} : n ∈ ℕ). Then the Riemann Zeta-function is given by

ζ(s) = ∏_{k=0}^{∞} \frac{1}{1 − p_{k}^{−s}} = ∏_{k=0}^{∞} ∑_{n=0}^{∞} \frac{1}{p_{k}^{ns}} .

Proof. Every partial product ∏_{k=0}^{m} _{\frac{1}{1−p_{k}^{−s}}} is the limit of the products

∏_{k=0}^{m} ∑_{ℓ=0}^{n} \frac{1}{p_{k}^{ℓs}} ,

which can be written as the sums ∑_{a} 1/a^{s}, where a ranges over the positive integers whose only prime factors are among the p_{k} such that k ⩽ m, and which are divisible by any one of these p_{k} at most n times. Hence

|ζ(s) − ∏_{k=0}^{m} \frac{1}{1 − p_{k}^{−s}} | ⩽ ∑_{b} \frac{1}{b^{s}} ⩽ ∑_{n=p_{m}}^{∞} \frac{1}{n^{s}} ,

where b ranges over those positive integers divisible by a prime greater than p_{m}. Since lim_{n→∞} ∑_{n=p_{m}}^{∞} 1/n^{s} = 0, the claim follows.

4 Sequences of functions

Suppose I is a sub-interval of ℝ, and (f_{n} : n ∈ ℕ) is a sequence of functions from I to ℝ. If f is another such sequence, then (f_{n}) converges pointwise to f if, for all x in I,

lim_{n→∞} f_{n}(x) = f(x).

Nice properties need not be preserved under the taking of pointwise limits:

Examples 4.1. (0) The sequence of continuous functions x ↦ x^{n} : [0, 1] → ℝ converges pointwise to the non-continuous function

x ↦ \begin{cases}0, & if 0 ⩽ x < 1; \\ 1, & if x = 1. \end{cases}

(1) The sequence of integrable functions x ↦ n^{2}x^{n}(1 − x) : [0, 1] → ℝ converges pointwise to 0, but

∫_{0}^{1} n^{2}x^{n}(1 − x) d x = n^{2} (\frac{x^{n+1}}{n + 1} − \frac{x^{n+2}}{n + 2})|_{0}^{1} = \frac{n^{2}}{(n + 1)(n + 2)},

so the sequence of integrals converges to 1.

(2) The sequence of differentiable functions x ↦ (sin n^{2}x)/n converges to 0, but the sequence of derivatives x ↦ 2n cos nx does not converge.

The sequence (f_{n}) converges uniformly to f if for every positive ε there is M such that for all x in I and all n in ℤ,

n > M ⟹ |f_{n}(x) − f(x)| < ε.

We showed in Math 271 that the uniform limit of continuous functions is continuous.

Let B(I) be the set of bounded real-valued functions on I. This becomes a metric space when we define

d(f, g) = sup{|f(x) − g(x)| : x ∈ I}.

Then a sequence of functions in B(I) converges uniformly if and only if it converges in the metric d.

Lemma 4.2 (Cauchy condition for uniform convergence). A sequence (f_{n}) of functions on I converges uniformly if and only if for all positive ε there is M such that for all x in I, all n in ℤ and all k in ℕ,

n > M ⟹ |f_{n+k}(x) − f_{n}(x)| < ε.

Proof. Suppose (f_{n}) converges uniformly to f. If ε > 0, let M be such that

n > M ⟹ |f_{n}(x) − f(x)| < \frac{ε}{2}.  (iv)

If n > M, then also |f_{n+k}(x) − f(x)| < ε/2, so |f_{n+k}(x) − f(x)| < ε.

Conversely, if (f_{n}) satisfies the Cauchy condition, then each sequence (f_{n}(x)) of real numbers is Cauchy, so it has a limit, say f(x). Hence, for all n, we have

lim_{k→∞} |f_{n}(x) − f_{n+k}(x)| = |f_{n}(x) − f(x)| .

If ε > 0, let M be such that (iv) holds. Then

n > M ⟹ |f_{n}(x) − f(x)| ⩽ \frac{ε}{2} < ε.

So (f_{n}) converges uniformly to f.

A series ∑ f_{n} of functions converges uniformly if the sequence of partial sums ∑_{k=0}^{n} f_{k} converges uniformly.

Lemma 4.3 (Cauchy condition for uniform convergence of series). A series ∑ f_{n} of functions converges uniformly if and only if, for all positive ε, there is M such that for all x in I, for all n in ℤ and all k in ℕ,

n > M ⟹ |∑_{ℓ=n}^{n+k} f_{ℓ}(x)| < ε.

Theorem 4.4 (Weierstraß M-test). Suppose the sequences (f_{n}) of functions and (M_{n}) of numbers are such that, for each x in I,

|f_{n}(x)| ⩽ M_{n}.

If ∑ M_{n} converges, then ∑ f_{n} converges uniformly.

Proof. If ∑ M_{n} converges, then it satisfies the Cauchy criterion for series; so ∑ f_{n} satisfies the Cauchy criterion for series of functions; so ∑ f_{n} converges uniformly.

Theorem 4.5. Suppose each f_{n} is continuous, and ∑ f_{n} converges uniformly to f. Then f is continuous; in particular,

lim_{y→x} ∑_{n=0}^{∞} f_{n}(y) = ∑_{n=0}^{∞} lim_{y→x} f_{n}(y).

Theorem 4.6 (A space-filling curve). There is a continuous surjective function from [0, 1] to [0, 1] × [0, 1].

Proof. Let ϕ be a continuous function on ℝ such that, for all n in ℤ,

ϕ(x) = \begin{cases}0, & if k − 1/6 ⩽ x ⩽ k + 1/6; \\ 1, & if k + 1/3 ⩽ x ⩽ k + 2/3. \end{cases}

Then ϕ is periodic, with period 1; that is, ϕ(x) = ϕ(x − ⌊x⌋). For each e in {0, 1}, let f_{e} be the function given by

f_{e}(x) = ∑_{k=0}^{∞} \frac{ϕ(3^{2k+e}x)}{2^{k+1}}.

The series converge uniformly, by the M-test with M_{k} = 1/2^{k+1}. Since ϕ is continuous, so are the f_{e}. Our function will be

x ↦ (f_{0}(x), f_{1}(x)).

Any point of [0, 1] × [0, 1] can be written, in binary notation, as

(∑_{k=1}^{∞} \frac{a_{k}}{2^{k}} , ∑_{k=1}^{∞} \frac{b_{k}}{2^{k}} ) ,

where a_{k} and b_{k} are in {0, 1}. Now define

c_{n} = \begin{cases}a_{k}, & if 2k − 1 = n; \\ b_{k}, & if 2k = n; \end{cases}

and let

c = ∑_{n=1}^{∞} \frac{c_{n}}{3^{n}} .

Then

3^{k}c − ⌊3^{k}c⌋ = \frac{c_{k+1}}{3} + ∑_{n=1}^{∞} \frac{c_{n+k+1}}{3^{n+1}} ,

so ϕ(3^{k}c) = c_{k+1}.

Theorem 4.7. Let g be of bounded variation on [a, b]. If f_{n} ∈ R(g), and (f_{n}) converges uniformly to f on [a, b], then f ∈ R(g), and (∫_{a}^{x} f_{n} d g) converges uniformly to ∫_{a}^{x} f d g.

Proof. We may assume g is increasing and g(b) ≠ g(a). There is M such that, for all x in [a, b],

n ⩾ M ⟹ |f_{n}(x) − f(x)| < \frac{ε}{3(g(b) − g(a))}.

We use this first to prove f ∈ R(g). For any partition P of [a, b], we have

|U(P, f − f_{M}, g)| , |L(P, f − f_{M}, g)| < \frac{ε}{3}.

Suppose in particular that P is fine enough that

U(P, f_{M}, g) − L(P, f_{M}, g) < \frac{ε}{3}.

Because sup{f(x) : x ∈ [a, b]} ⩽ sup{f(x) − f_{M}(x) : x ∈ [a, b]} + sup{f_{M}(x) : x ∈ [a, b]}, and likewise for the infima, we have

U(P, f, g) − L(P, f, g) ⩽ U(P, f − f_{M}, g) + U(P, f_{M}, g) − (L(P, f − f_{M}, g) + L(P, f_{M}, g)) ⩽ ε.

So Riemann’s condition is satisfied, and f ∈ R(g). We have also

|∫_{a}^{x} f_{n} d g − ∫_{a}^{x} f d g| = |∫_{a}^{x}(f_{n} − f) d g| ⩽

∫_{a}^{x} |f_{n} − f| d g ⩽ ∫_{a}^{b} |f_{n} − f| d g ⩽ \frac{ε}{3},

so the convergence of the sequence of integrals is uniform.

Under the conditions of the theorem, we have

lim_{n→∞} ∫_{a}^{x} f_{n} d g = ∫_{a}^{x}( lim_{n→∞} f_{n}) d g.

If also ∑ f_{n} converges uniformly, then

∑_{n=0}^{∞} ∫_{a}^{x} f_{n} d g = ∫_{a}^{x}(∑_{n=0}^{∞} f_{n}) d g.

A power series is a function

z ↦ ∑_{n=0}^{∞} c_{n}(z − a)^{n}.

At a, this function has the value c_{0}, by definition. The function may, but need not, be well-defined at other points. Nice results are obtained if the coefficients c_{n} and the point a (as well as the argument z) are allowed to be complex numbers.

Theorem 4.8. The vector-space ℝ^{2} becomes a field when equipped with the multiplication given by

(a, b)(c, d) = (ac − bd, ad + bc).

In this field, the multiplicative identity is (1, 0), and

(a, b)^{−1} = \frac{1}{a^{2} + b^{2}} (a, −b).

The field in the theorem is denoted ℂ; its elements are complex numbers. For the complex numbers (1, 0) and (0, 1), we may write

1 and i

respectively; then the complex number (a, b)—that is, a(1, 0) + b(0, 1)—can be written a + bi. Note that i^{2} = −1.

Theorem 4.9. If z, w ∈ ℂ, then |zw| = |z| |w|.

If z ∈ ℂ, then |z| is called the absolute value or modulus of z. We have the triangle inequality, |z + w| ⩽ |z| + |w|. Much of what we have done with sequences of real numbers applies to complex numbers as well. In particular,

• the definition of absolute convergence is meaningful for series of complex numbers;

• the tests for absolute convergence work for such series (since absolute convergence is still convergence of a certain series of real numbers);

• the proof that absolute convergence implies convergence is still valid for such series.

Let ρ = lim sup_{n→∞} \sqrt[n]{|c_{n}|}. Then

lim sup_{n→∞} \sqrt[n]{|c_{n}(z − a)^{n}|} = lim sup_{n→∞} \sqrt[n]{|c_{n}| |(z − a)^{n}|} =

lim sup_{n→∞}(|z − a| \sqrt[n]{|c_{n}|}) = |z − a| ρ.

If ρ > 0, then the series ∑ c_{n}(z − a)^{n}:

• converges absolutely, if |z − a| < 1/ρ;

• diverges, if |z − a| > 1/ρ.

The number 1/ρ (if it exists) is the radius of convergence of the series; the ball B(a; 1/ρ) is the disc of convergence. (If ρ = 0, then the radius of convergence is ∞; if ρ = ∞, then the radius of convergence is 0.)

Theorem 4.10. Every power series is continuous on its disc of convergence.

Proof. Let R be the radius of convergence of ∑ c_{n}(z − a)^{n}, and suppose w ∈ B(a; R). Let δ = (R − |w − a|)/2, and let F = {z ∈ ℂ : |z − w| ⩽ δ}. Then F is a neighborhood of w, so it is enough to show that the series is continuous on F. For this, since z ↦ c_{n}(z − a)^{n} is continous, it is enough to show that the series converges uniformly on F. But F is compact, so some point b of F is furthest from a; that is, if z ∈ F, then

|c_{n}(z − a)^{n}| = |c_{n}| |z − a|^{n} ⩽ |c_{n}| |b − a|^{n} .

Since |b − a| < R, we know ∑ |c_{n}| |b − a|^{n} converges; hence, by the M-test, the series converges uniformly on F.

Since ℂ is a field, the definition of the derivative of a function from ℝ to ℝ is meaningful for functions from ℂ to ℂ. We shall show that any power series is differentiable on its disc of convergence, with derivative the sum of the derivatives of the terms.

Lemma 4.11. ∑_{n=0}^{∞} c_{n}(z − a)^{n} and ∑_{n=1}^{∞} nc_{n}(z − a)^{n−1} have the same radius of convergence.

Proof. The second series is convergent if and only if ∑_{n=1}^{∞} nc_{n}(z − a)^{n} is convergent. Since lim_{n→∞} \sqrt[n]{n} = 1, this series has the same radius of convergence as the first.

Lemma 4.12. |z^{n} − w^{n}| ⩽ n |z − w| (|w| + |z − w|)^{n−1} if n > 0.

Proof. The claim is trivial if w = 0. Assume w ≠ 0, and let t = (z − w)/w. Then we have to prove

|(1 + t)^{n} − 1| ⩽ n |t| (1 + |t|)^{n−1}.

If t is a non-negative real number, then the inequality holds, since then

(1 + t)^{n} − 1 = n ∫_{0}^{t}(1 + x)^{n−1} d x ⩽ nt(1 + t)^{n−1}.

For arbitrary t in ℂ, we have

|(1 + t)^{n} − 1| = |∑_{k=1}^{n} \binom{n}{k}t^{k}| ⩽ ∑_{k=1}^{n} \binom{n}{k} |t|^{k} = (1 + |t|)^{n} − 1,

which, combined with the previous inequality, yields the claim.

Note that, by interchanging z and w, we can write the inequality of the lemma as

|z^{n} − w^{n}| ⩽ n |z − w| (|z| + |z − w|)^{n−1}.

Theorem 4.13. If ∑_{n=0}^{∞} c_{n}(z − a)^{n} has a positive radius of convergence, then the function is differentiable on its disc of convergence, and its derivative is ∑_{n=1}^{∞} nc_{n}(z − a)^{n−1}.

Proof. Write f(z) for ∑_{n=0}^{∞} c_{n}(z − a)^{n}. Let ζ be an element of the disc of convergence; we shall compute f′(ζ). For all elements w of this disc distinct from ζ,

\frac{f(ζ) − f(w)}{ζ − w} = ∑_{n=1}^{∞} c_{n} \frac{(ζ − a)^{n} − (w − a)^{n}}{ζ − w} .

We want to make the difference between this and ∑_{n=1}^{∞} nc_{n}(ζ − a)^{n−1} small. To do this, we break this difference into two pieces, and make each piece small. For any positive integer M,

|\frac{f(ζ) − f(w)}{ζ − w} − ∑_{n=1}^{∞} nc_{n}(ζ − a)^{n−1}| ⩽

∑_{n=1}^{M} |c_{n}| |\frac{(ζ − a)^{n} − (w − a)^{n}}{ζ − w} − n(ζ − a)^{n−1}| +

∑_{n=M+1}^{∞} c_{n} (|\frac{(ζ − a)^{n} − (w − a)^{n}}{ζ − w} | + n |ζ − a|^{n−1}) ⩽

∑_{n=1}^{M} |c_{n}| |\frac{(ζ − a)^{n} − (w − a)^{n}}{ζ − w} − n(ζ − a)^{n−1}| +

∑_{n=M+1}^{∞} n |c_{n}| ((|ζ − a| + |ζ − w|)^{n−1} + |ζ − a|^{n−1})

by Lemma 4.12. Let R be the radius of convergence of f, and let δ_{0} = (R − |ζ − a|)/2. Suppose 0 < |ζ − w| < δ_{0}. Then

|\frac{f(ζ) − f(w)}{ζ − w} − ∑_{n=1}^{∞} nc_{n}(ζ − a)^{n−1}| ⩽

∑_{n=1}^{M} |c_{n}| |\frac{(ζ − a)^{n} − (w − a)^{n}}{ζ − w} − n(ζ − a)^{n−1}| +

2 ∑_{n=M+1}^{∞} nc_{n}(|ζ − a| + δ_{0})^{n−1}.

Say ε > 0. By Lemma 4.11, we can choose M large enough that

2 ∑_{n=M+1}^{∞} nc_{n}(|ζ − a| + δ_{0})^{n−1} < ε/2.

But for each positive n, the derivative of z ↦ (z − a)^{n} is z ↦ n(z − a)^{n−1}; hence, for the chosen M, we can find δ_{1} such that, if 0 < |ζ − w| < δ_{1}, then

∑_{n=1}^{M} |c_{n}| |\frac{(ζ − a)^{n} − (w − a)^{n}}{ζ − w} − n(ζ − a)^{n−1}| < \frac{ε}{2}.

Therefore, if 0 < |ζ − w| < min{δ_{0}, δ_{1}}, then

|\frac{f(ζ) − f(w)}{ζ − w} − ∑_{n=1}^{∞} nc_{n}(ζ − a)^{n−1}| ⩽ ε;

this proves the claim.

Corollary 4.14. If f(z) = ∑_{n=0}^{∞} c_{n}(z −a)^{n}, and R is the radius of convergence of the series, then f is infinitely differentiable on B(a; R), and

c_{n} = \frac{f^{(n)}(a)}{n!}.

Proof. Repeated application of the theorem yields

f^{(k)}(z) = ∑_{n=k}^{∞} n(n − 1) … (n − k + 1)c_{n}(z − a)^{n−k} = ∑_{n=k}^{∞} \frac{n!}{(n − k)!}c_{n}(z − a)^{n−k},

hence f^{(k)}(a) = k!c_{k}.

For which functions f, defined on a neighborhood of a, can we conclude

f(z) = ∑_{n=0}^{∞} \frac{f^{(n)}(a)}{n!}(z − a)^{n}  (v)

on some neighborhood of a? If a ∈ ℂ, then it is enough for f to be oncedifferentiable (but we won’t prove this). If a ∈ ℝ, then it is not even enough to know that f is infinitely differentiable:

Example 4.15. Let f(x) = \begin{cases}exp(−x^{−2}), & if x ≠ 0; \\ 0, & if x = 0. \end{cases} Then f^{(n)}(0) = 0 for all

n; so Equation (v) fails when a = 0.

We can come up with a condition under which a function f satisfies Equation (v):

Lemma 4.16 (Generalized Mean-Value Theorem). Let F and G be realvalued functions continuous on [a, b] and differentiable on (a, b). Then

F′(c)[G(b) − G(a)] = G′(c)[F (b) − F (a)]

for some c in (a, b).

Proof. Apply Rolle’s Theorem to x ↦ F (x)[G(b)−G(a)]−G(x)[F (b)−F (a)].

Theorem 4.17. Suppose f^{(n−1)} and g^{(n−1)} are continuous on [a, b] and differentiable on (a, b). Then for any distinct c and x in [a, b] there is y, strictly between them, such that

f^{(n)}(y) [g(x) − ∑_{k=0}^{n−1} \frac{g^{(k)}(c)}{k!}(x − c)^{k}] = g^{(n)}(y) [f(x) − ∑_{k=0}^{n−1} \frac{f^{(k)}(c)}{k!}(x − c)^{k}] .

Proof. Use the Generalized Mean-Value Theorem, with [c, x] (or [x, c]) in place of [a, b], with F as the function

t ↦ ∑_{k=0}^{n−1} \frac{f^{(k)}(t)}{k!}(x − t)^{k},

and with G as a similar function in terms of g. In particular, G(x) = g(x), so

G(x) − G(c) = g(x) − ∑_{k=0}^{n−1} \frac{g^{(k)}(c)}{k!}(x − c)^{k};

also,

F′(t) = f′(t) + ∑_{k=1}^{n−1} [\frac{f^{(k+1)}(t)}{k!}(x − t)^{k} − \frac{f^{(k)}(t)}{(k − 1)!}(x − t)^{k−1}] ,

a telescoping sum, so F′(t) = f^{(n)}(t)(x − t)^{n−1}/(n − 1)!.

Corollary 4.18 (Taylor’s Theorem). Suppose f^{(n−1)} is continuous on [a, b] and differentiable on (a, b). Then for any distinct c and x in [a, b] there is y, strictly between them, such that

f(x) = ∑_{k=0}^{n−1} \frac{f^{(k)}(c)}{k!}(x − c)^{k} + \frac{f^{(n)}(y)}{k!}(x − c)^{n}.

Proof. In the theorem, let g be t ↦ (t − c)^{n}. We have

g^{(k)}(t) = \frac{n!}{(n − k)!}(t − c)^{n−k},

so the theorem gives

f^{(n)}(y)(x − c)^{n} = n! [f(x) − ∑_{k=0}^{n−1} \frac{f^{(k)}(c)}{k!}(x − c)^{k}] ,

whence the claim.

Lemma 4.19. Suppose f has derivatives of all orders on some neighborhood of c, and for some M, each f^{(n)} is bounded by M^{n} on that neighborhood; then on that neighborhood, Equation (v) holds.

Proof. For any positive a, we have lim_{n→∞}(a^{n}/n!) = 0, since ∑(a^{n}/n!) converges by the ratio test.

5 Differentiation in several dimensions

Suppose in this section that f is a function from ℝ^{n} to ℝ^{m}, and a is a point of ℝ^{n}. We want to investigate the possibilities for differentiating f at a . We can write f as (f_{0}, …, f_{m−1}), where each f_{i} is a function from ℝ^{n} to ℝ; so if n = 1, then we can form (f′_{0}, …, f′_{m−1}).

In the general case, we can take partial derivatives: Define a function δ : ℕ×ℕ → {0, 1} by

δ_{i j} = \begin{cases}1, & if i = j; \\ 0, & if i ≠ j. \end{cases}

Let e_{i} be the element (δ_{i 0}, …, δ_{i n−1}) of ℝ^{n}. Let D_{i}f be the function defined by

D_{i}f(a) = lim_{h→0} \frac{f(a + he_{i}) − f(a)}{h};

this function (where it is well-defined) is the partial derivative of f with respect to the ith coordinate. Note that D_{i}f_{j}(a) is the ordinary derivative at 0 of the function

x ↦ f_{j}(a + xe_{i}).

Example 5.1. Let f : ℝ^{2} → ℝ be given by

f(x, y) = \begin{cases}xy/(x^{2} + y^{2}), & if (x, y) ≠ (0, 0); \\ 0, & if (x, y) = (0, 0). \end{cases}

Then D_{0}f(0, 0) = 0 = D_{1}f(0, 0), but f is not continuous at (0, 0).

For a stronger property than having partial derivatives, we define the generalization called the directional derivative: If u ∈ ℝ^{n}, then

f′(a; u) = lim_{h→0} \frac{f(a + hu) − f(a)}{h}.

Thus f′(a; e_{i}) = D_{i}f(a). In Example 5.1, not all directional derivatives at (0, 0) exist.

Example 5.2. Let f : ℝ^{2} → ℝ be given by

f(x, y) = \begin{cases}xy^{2}/(x^{2} + y^{4}), & if x ≠ 0; \\ 0, & if x = 0. \end{cases}

Then all directional derivatives at 0 exist, since

\frac{f(hu, hv) − f(0, 0)}{h} = \frac{uv^{2}}{u^{2} + h^{2}v^{4}} ,

which has a limit as h goes to 0, whether or not u = 0. However, f(y^{2}, y) = 1/2, unless y = 0; so f is not continuous at (0, 0).

If n = 1, then f is differentiable at a if and only if there is a number, called f′(a), such that

lim_{h→0} \frac{f(a + h) − f(a) − f′(a) · h}{h} = 0.

The function h ↦ f′(a)·h : ℝ → ℝ is linear; call it λ. (A function T : ℝ^{n} → ℝ^{m} is linear if T (au + bv) = aT (u) + bT (v) in all cases.) Then the function x ↦ f(a) + λ(x − a) is an approximation to f near a.

The function f : ℝ^{n} → ℝ^{m} is said to be differentiable at a if there is a linear function λ : ℝ^{n} → ℝ^{m} such that

lim_{h→0} \frac{|f(a + h) − f(a) − λ(h)|}{|h|} = 0.

In this case, λ is the total derivative of f at a and is denoted Df(a). Note well that this is a function; however, we have to check that it is unique:

Theorem 5.3. If Df(a) exists, then Df(a)(u) = f′(a; u) for all u in ℝ^{n}; so the total derivative is uniquely determined by the directional derivatives.

Proof. If Df(a) exists, and u ≠ 0, then

0 = lim_{h→0} \frac{|f(a + hu) − f(a) − Df(a)(hu)|}{|hu|} = 0;

but Df(a)(hu) = hDf(a)(u); hence the claim.

Corollary 5.4. Df(a)(u) = ∑_{i<n} u_{i}D_{i}f(a) for all u.

Proof. It’s true when u = e_{i}; hence it is true in general, by linearity.

Lemma 5.5. For any linear function T : ℝ^{n} → ℝ^{m} there is a number M such that

|T (h)| ⩽ M |h|

for all h in ℝ^{n}. Hence T is continuous at 0.

Proof. |T (x)| = |∑_{i<n} x_{i}T (e_{i})| ⩽ ∑_{i<n} |x_{i}| |T (e_{i})| ⩽ |x| ∑_{i<n} |T (e_{i})|.

Theorem 5.6. Differentiability implies continuity.

Proof. If Df(a) exists, then

lim_{h→0}(f(a + h) − f(a) − Df(a)(h)) = 0;

but also lim_{h→0} Df(a)(h) = 0, by Lemma 5.5; hence

lim_{h→0} f(a + h) = f(a),

whence the claim.

The sum of products in Corollary 5.4 can be written as a matrix product:

Df(a)(u) = [D_{0}f(a) ⋯ D_{n−1}(a)] [ u_{0} ; ⋮ ; u_{n−1}.]

As u, so can each partial derivative D_{i}f(a) be written as a column-vector. Thus Df(a)(u) is the matrix product u · f′(a), where

f′(a) = [\begin{matrix}D_{0}f_{0}(a) & ⋯ & D_{n−1}f_{0}(a) \\ ⋮ & ... & ⋮ \\ D_{0}f_{m−1}(a) & ⋯ & D_{n−1}f_{m−1}(a) \end{matrix}] .

This matrix f′(a) is the Jacobian matrix of f at a.

Theorem 5.7 (Chain-Rule). If the functions f : ℝ^{n} → ℝ^{m} and g : ℝ^{m} → ℝ^{p} are differentiable at a and f(a) respectively, then g ∘f is differentiable at a, and

D(g ∘ f)(a) = Dg(f(a)) ∘ Df(a),

whence (g ∘ f)′(a) = g′(f(a)) · f′(a).

Proof. We have to prove

lim_{h→0} \frac{g(f(a + h)) − g(f(a)) − Dg(f(a))(Df(a)(h))}{|h|} = 0.

The numerator of the fraction can be written as the sum of

g(f(a + h)) − g(f(a)) − Dg(f(a))(f(a + h) − f(a))

and

Dg(f(a))(f(a + h) − f(a) − Df(a)(h)).

Write these summands respectively as

ψ(f(a + h)) and Dg(f(a))(ϕ(h)).

We shall show:

(0) lim_{h→0} \frac{|ψ(f(a + h))|}{|h|} = 0;

(1) lim_{h→0} \frac{|Dg(f(a))(ϕ(h))|}{|h|} = 0.

By Lemma 5.5, there is M such that, for all h,

|Dg(f(a))(ϕ(h))| ⩽ M |ϕ(h)|;

since lim_{h→0} |ϕ(h)| / |h| = 0, this proves (1). For (0), note that

lim_{k→0} \frac{|ψ(f(a) + k)|}{|k|} = 0;

also, if f(a + h) = f(a), then ψ(f(a + h)) = 0. If f(a + h) ≠ f(a), then we have

\frac{|ψ(f(a + h))|}{|h|} = \frac{|ψ(f(a + h))|}{|f(a + h) − f(a)|} · \frac{|f(a + h) − f(a)|}{|h|}.

The second factor here is

\frac{|ϕ(h) + Df(a)(h)|}{h};

hence there is N such that

\frac{|f(a + h) − f(a)|}{|h|} ⩽ \frac{|ϕ(h)|}{h} + N.

Suppose ε > 0. There are positive numbers δ_{0}, δ_{1} and δ_{2} such that

0 < |h| < δ_{0} ⟹ \frac{|ϕ(h)|}{|h|} < 1;
0 < |k| < δ_{1} ⟹ \frac{|ψ(f(a) + k)|}{|k|} < \frac{ε}{N + 1};
0 < |h| < δ_{2} ⟹ |f(a + h) − f(a)| < δ_{2}.

Let δ = min{δ_{0}, δ_{2}}; then

\frac{|ψ(f(a + h))|}{|h|} < ε

whenever 0 < |h| < δ.

Lemma 5.8. If the partial derivatives D_{i}f are defined on a neighborhood of a and are continuous at a, then f is differentiable at a, that is, the total derivative Df(a) exists.

Proof. We may assume m = 1. The D_{i}f must be defined on an open interval I that contains a. (Note then that I is I_{0} × ⋯ × I_{n−1} for some open intervals I_{i} of ℝ such that a_{i} ∈ I_{i}.) We shall show that Df(a) exists by showing that

lim_{h→0} \frac{|f(a + h) − f(a) − ∑_{i<n} D_{i}f(a) · h_{i}|}{|h|} = 0.

Suppose h ∈ I ∖ {a}. We have

f(a + h) − f(a)

= f(a + h_{0}e_{0}) − f(a)
+ f(a + h_{0}e_{0} + h_{1}e_{1}) − f(a + h_{0}e_{0})
+ ⋯
+ f(a + h_{0}e_{0} + ⋯ + h_{n−1}e_{n−1}) − f(a + h_{0}e_{0} + ⋯ + h_{n−2}e_{n−2})

= ∑_{i<n}(f(a + ∑_{j<i+1} h_{j}e_{j}) − f(a + ∑_{j<i} h_{j}e_{j})) = ∑_{i<n} D_{i}f(b_{i}) · h_{i},

where b_{i} = a+∑_{j<i} h_{j}e_{j}+c_{i}e_{i} for some c_{i} between 0 and h_{i}, by the Mean-Value Theorem as applied to the function

x ↦ f(a + ∑_{i<j} h_{j}e_{j} + xe_{i}).

In particular, b_{i} ∈ I. We now have

0 ⩽ \frac{|f(a + h) − f(a) − ∑_{i<n} D_{i}f(a) · h_{i}|}{|h|}

= \frac{|∑_{i<n}(D_{i}f(b_{i}) − D_{i}f(a)) · h_{i}|}{|h|}

⩽ ∑_{i<n} \frac{|D_{i}f(b_{i}) − D_{i}f(a)| · |h_{i}|}{|h|}

⩽ ∑_{i<n} |D_{i}f(b_{i}) − D_{i}f(a)| .

If ε > 0, then, by the continuity of the D_{i}f at a, there is a ball B(a; δ) around a such that, if x is in this ball, then

|D_{i}f(x) − D_{i}f(a)| ⩽ \frac{ε}{n}.

If h is B(a; δ), then so are the b_{i}, which means

\frac{|f(a + h) − f(a) − ∑_{i<n} D_{i}f(a) · h_{i}|}{|h|} < ε.

The claim now follows.

The sufficient condition for differentiability of f at a given by the theorem is that f be continuously differentiable at a.

To the Chain-Rule, we can now state a:

Corollary 5.9. If f is continuously differentiable at a, and the function g : ℝ^{m} → ℝ is differentiable at f(a), then g ∘ f is differentiable at a, and

D_{i}(g ∘ f)(a) = ∑_{j<m} D_{j}g(f(a)) · D_{i}f_{j}(a)

for each i less than n.

Proof. By the Chain-Rule,

[D_{0}(g ∘ f)(a) ⋯ D_{n−1}(g ∘ f)(a)] = (g ∘ f)′(a) = g′(f(a)) · f′(a)

= [D_{0}g(f(a)) ⋯ D_{m−1}g(f(a))] · [\begin{matrix}D_{0}f_{0}(a) & ⋯ & D_{n−1}f_{0}(a) \\ ⋮ & ... & ⋮ \\ D_{0}f_{m−1}(a) & ⋯ & D_{n−1}f_{m−1}(a) \end{matrix}] .

Consideration of the i-th entry of the first and last members of the equation yields the result.

Theorem 5.10 (Inverse Function). Suppose m = n, and f is continuously differentiable on a neighborhood of a, and Df(a) is invertible, that is, det(f′(a)) ≠ 0. Then there are neighborhoods V of a and W of f(a) such that f is a bijection from V to W with a differentiable inverse f^{−1} satisfying

D(f^{−1})(f(x)) = Df(x)^{−1}  (vi)

for all x in V, equivalently,

(f^{−1})′(f(x)) = (f′(x))^{−1}.

Proof. (0) For any function f : ℝ^{n} → ℝ^{n}, let Φ(f) be the statement,

If f is continuously differentiable near a, then a and f(a) have neighborhoods V and W respectively such that f is a bijection from V to W with differentiable inverse.

Now, any linear function on ℝ^{n} is its own derivative everywhere. In particular, let id be the identity on ℝ^{n}. Then

D(id)(x) = id

for all x in ℝ^{n}. If Φ(f), then f^{−1} ∘ f = id on V, so

Df^{−1}(f(x)) ∘ Df(x) = id,

whence (vi). It remains for us to show simply that Φ(f) whenever Df(a) is invertible.

(1) If Φ(f), and T : ℝ^{n} → ℝ^{n} is invertible and linear, then Φ(T ∘ f). In particular, if Df(a) is invertible, and Φ(Df(a)^{−1} ∘ f), then Φ(f). But

D(Df(a)^{−1} ∘ f)(a) = id,

by the Chain-Rule. So it is enough for us to show that Φ(f) whenever

Df(a) = id.  (vii)

(2) Let us now assume that f is continuously differentiable near a, and that (vii) holds. Then

lim_{h→0} \frac{|f(a + h) − f(a) − h|}{|h|} = 0.

But we also have

\frac{|f(a + h) − f(a) − h|}{|h|} ⩾ |\frac{|f(a + h) − f(a)|}{|h|} − 1| .

Therefore

lim_{h→0} \frac{|f(a + h) − f(a)|}{|h|} = 1,

which means in particular that, a has a neighborhood U such that

f(x) = f(a) ⟹ x = a  (viii)

for all x in U.

(3) We can make certain additional assumptions about U. We may assume that U is a closed ball,

U = {x ∈ ℝ^{n} : |x − a| ⩽ r}.

Since f is continuously differentiable near a, the function x ↦ det(f′(x)) is continuous near a, and therefore we may assume that Df(x) is invertible for all x in U; we may also assume

|D_{i}f_{j}(x) − D_{i}f_{j}(a)| ⩽ \frac{1}{2n^{2}}  (ix)

for each i and j.

By (vii), D_{i}f_{j}(a) = δ_{i j}. Let g be the function x ↦ f(x) − x; then

D_{i}g_{j}(a) = D_{i}f_{j}(x) − δ_{i j} = D_{i}f_{j}(x) − D_{i}f_{j}(a).  (x)

For all x and u in U, we have

|u − x| − |f(u) − f(x)| ⩽ |g(u) − g(x)| ⩽ ∑_{i<n} |g_{i}(u) − g_{i}(x)|

⩽ ∑_{i<n} ∑_{j<n} |g(∑_{k=0}^{j} u_{k}e_{k} + ∑_{k=j+1}^{n−1} x_{k}e_{k}) − g(∑_{k=0}^{j−1} u_{k}e_{k} + ∑_{k=j}^{n−1} x_{k}e_{k})|

⩽ ∑_{i<n} ∑_{j<n} \frac{1}{2n^{2}} · |u_{j} − x_{j}| ⩽ ∑_{i,j<n} \frac{1}{2n^{2}} · |u − x| ⩽ \frac{1}{2} · |u − x|

by the Mean-Value Theorem, (ix) and (x), and therefore

|u − x| ⩽ 2 · |f(u) − f(x)| .  (xi)

Hence f is injective on U, and f^{−1} is well-defined and continuous on f(U).

It remains to prove that f(U) is in fact a neighborhood of f(a) and that f^{−1} is differentiable on some neighborhood of f(a).

(4) The boundary of U can be denoted ∂U and is the compact set

{x ∈ U : |x − a| = r}.

Note then that

U ∖ ∂U = B(a; r).

The continuous function x → |f(x) − f(a)| : ∂U → ℝ must attain a minimum value d, which must be positive, by (viii). Now define

W = B(f(a); d/2),

the open ball with center f(a) and radius d/2. If y ∈ W, and u ∈ ∂U, then

d ⩽ |f(u) − f(a)| ⩽ |f(u) − y| + |y − f(a)|;

since also |y − f(a)| < d/2, we have

|y − f(a)| < |f(u) − y| .  (xii)

(5) We shall now show that

W ⊆ f(U ∖ ∂U).

Then we can define

V = (U ∖ ∂U) ∩ f^{−1}(W),

so that V is an open neighborhood of a, and f is a continuous bijection between this and the open neighborhood W of f(a).

Let y ∈ W; we shall find x in U ∖ ∂U such that f(x) = y. Let h be the continuously differentiable function

u ↦ |y − f(u)|^{2} : U → ℝ.

By (xii), we have

u ∈ ∂U ⟹ h(a) < h(u).

Since U is compact, h attains a minimum at some x, which must therefore be in U ∖ ∂U. Then Dh(x) is the zero-function. Now, h is the composition of u ↦ y − f(u) and z ↦ ∑_{i<n} z_{i}^{2}. By the Chain-Rule, we find

2 [y_{0} − f_{0}(x) ⋯ y_{n−1} − f_{n−1}(x)] · f′(x) = [0 ⋯ 0] .

The invertibility of f′(x) yields f(x) = y, as desired.

(6) It remains to show that if x ∈ V, then D(f^{−1})(f(x)) exists. It will be enough to show

lim_{k→0} \frac{|f^{−1}(f(x) + k) − x − Df(x)^{−1}(k)|}{|k|} = 0.

By Lemma 5.5, it will therefore be enough to show

lim_{k→0} \frac{|Df(x)(f^{−1}(f(x) + k) − x − Df(x)^{−1}(k))|}{|k|} = 0.  (xiii)

We re-write this, first defining

h(k) = f^{−1}(f(x) + k) − x,

so that

k = f(x + h(k)) − f(x).

Then

\frac{|Df(x)(f^{−1}(f(x) + k) − x − Df(x)^{−1}(k))|}{|k|}

= \frac{|Df(x)(h(k) − Df(x)^{−1}(k))|}{|k|} = \frac{|k − Df(x)(h(k))|}{|k|}

= \frac{|f(x + h(k)) − f(x) − Df(x)(h(k))|}{|k|}

= \frac{|f(x + h(k)) − f(x) − Df(x)(h(k))|}{|h(k)|} · \frac{|h(k)|}{|k|}.

By (ix), the second factor here is bounded by 2; and the first factor has limit 0 at 0, since lim_{k→0} h(k) = 0. Therefore (xiii), which completes the proof.

Corollary 5.11 (Implicit Function Theorem). Suppose a ∈ ℝ^{n} and b ∈ ℝ^{m} and f : ℝ^{n}×ℝ^{m} → ℝ^{m} is continuously differentiable in a neighborhood of (a, b). Let

f′(a, b) = [A B] ,

where B is a square matrix. If f(a, b) = 0, and det B ≠ 0, then there is a function g : ℝ^{n} → ℝ^{m} such that g(a) = b, and f(x, g(x)) = 0 for all x in a neighborhood of a, and g is differentiable on this neighborhood.

Proof. Let F be the function

(x, y) ↦ (x, f(x, y)) : ℝ^{n} × ℝ^{m} → ℝ^{n} × ℝ^{m}.

Then

F′(a, b) = [\begin{matrix}I & 0 \\ A & B \end{matrix}] ,

so det(F′(a, b)) = det B ≠ 0. Hence (a, b) and (a, 0) have neighborhoods U × V and W respectively such that F is a bijection from U × V to W with a differentiable inverse. We can write this inverse as

(x, y) ↦ (x, h(x, y)),

where h is the appropriate function from W to V. Then

(x, y) = F (x, h(x, y)) = (x, f(x, h(x, y))),

which means

y = f(x, h(x, y)),

when (x, y) ∈ W. Hence, in particular,

0 = f(x, h(x, 0)).

So we may let g be x ↦ h(x, 0).

6 Integration in several dimensions

A compact interval of ℝ^{n} is a subset

I_{0} × … I_{n−1},

where each I_{j} is a closed, bounded interval of ℝ. Every compact interval I has a measure, denoted μ(I) and defined recursively by:

• μ([a, b]) = b − a;

• μ(I × I′) = μ(I) · μ(I′).

Let I be the compact interval I_{0} × … I_{n−1}. A partition of I is a product

P_{0} × ⋯ × P_{n−1},

where each P_{j} is a partition of I_{j}. Let the partition of I be P. If each P_{j} has m_{j} +1 elements, then P determines m_{0} ⋯ m_{n−1} sub-intervals of I in an obvious way. Let J range over the set of these sub-intervals. If t_{J} ∈ J for each J, and f is a function from I to ℝ, then f has a Riemann sum

∑_{J} f(t_{J}) · μ(J)

for the partition P. Then f is Riemann-integrable on I if there is a number A such that, for all positive numbers ε, there is a partition of I such that, for all finer partitions of I into sub-intervals J, for all choices of t_{J} in J,

|∑_{J} f(t_{J}) · μ(J) − A| < ε.

In this case, A is unique and can be denoted

∫_{I} f.

As in the special case where I ⊂ ℝ, so in the general case, integrability is equivalent to a condition known as Riemann’s Condition.

Suppose S is a bounded subset of ℝ^{n}, and f is a real-valued function on S. Define the characteristic function of S to be χ_{S} : ℝ^{n} → ℝ, where

χ_{S}(x) = \begin{cases}1, & if x ∈ S; \\ 0, & if x ∉ S. \end{cases}

Then f · χ_{S} is a well-defined function on ℝ^{n}, with value 0 on ℝ^{n} ∖ S. If I and I′ are two compact intervals including S, and ∫_{I} f · χ_{S} exists, then so does ∫_{I′} f ·χ_{S}, and the two intervals are equal; we can then define this common value to be the integral

∫_{S} f.

We may ask two questions:

• How can we tell whether integrals ∫_{S} f exist?

• If they exist, how can they be computed?

The set S has measure zero if for all positive ε there is a countable set {I_{j} : j ∈ ℕ} of intervals such that

S ⊆ ⋃_{j∈ℕ} I_{j} and ∑_{j=0}^{∞} μ(I_{j}) ⩽ ε.

Let us say that x is a discontinuity of f if f is not continuous at x.

Theorem 6.1 (Lebesgue Criterion). Let f be a bounded real-valued function on a compact interval I of ℝ. Then ∫_{I} f exists if and only if the set of discontinuities of f has measure zero.

Example 6.2. ℚ has measure zero. Indeed, ℚ can be written as {x_{j} : j ∈ ℕ}. If ε > 0, let

I_{j} = [x_{j} − \frac{ε}{2^{j+2}} , x_{j} + \frac{ε}{2^{j+2}} ] .

Then

∑_{j=0}^{∞} μ(I_{j}) = ∑_{j=0}^{∞} \frac{ε}{2^{j+1}} = ε,

and x_{j} ∈ I_{j}. Let f : ℝ → ℝ be given by

f(x) = \begin{cases}0, & if x ∉ ℚ; \\ 1/ |n| , & if x = m/n in lowest terms. \end{cases}

Then the set of discontinuities of f is precisely ℚ, so f is integrable on every compact interval, and each of the integrals is 0.

The boundary of a subset S of ℝ^{n} is the set ∂S of points x of ℝ^{n} such that every ball with center x contains points of both S and ℝ^{n} ∖ S. If S is bounded, then S is said to be Jordan-measurable if ∂S has measure zero.

Theorem 6.3. Let S be a bounded, Jordan-measurable subset of ℝ^{n}, and let f be a real-valued function on S. Then ∫_{S} f exists if and only if the set of discontinuities of f on S has measure zero.