If you find a mistake, omission, etc., please let me know by e-mail.

The orange balls mark our current location in the course, and the current problem set.

Metric Topology I
Basic definitions and examples: the metric spaces Rⁿ and other product spaces; isometries; boundedness and function spaces

The “sup metric” on X^S is sometimes also called the “uniform metric” because d(f, g)≤r is equivalent to a bound d(f(s), g(s))≤r for all s in S that is “uniform” in the sense that it's independent of the choice of s. Likewise for the sup metric on the space of bounded functions from S to an arbitrary metric space X (see the next paragraph).

If S is an infinite set and X is an unbounded metric space then we can't use our definition of X^S as a metric space because sup_S d_X(f(s), g(s)) might be infinite. But the bounded functions from S to X do constitute a metric space under the same definition of d_X^S. A function is said to be “bounded” if its image is a bounded set. You should check that d_X^S(f, g) is in fact finite for bounded f and g.

Now that metric topology is in 55b, not 55a, the following observation can be made: if X is R or C, the bounded functions in X^S constitute a vector space, and the sup metric comes from a norm on that vector space: d(f, g) = ||f−g|| where the norm ||·|| is defined by ||f || = sup_s |f(s)|. Likewise for the bounded functions from S to any normed vector space. Such spaces will figure in our development of real analysis (and in your further study of analysis beyond Math 55).

The “Proposition” on page 3 of the first topology handout can be extended as follows:

iv) For every point p of X there exists a real number M such that d(p,q) < M for all q of E.

In other words, for every p in X there exists an open ball about p that contains E. Do you see why this is equivalent to (i), (ii), and (iii)?

Metric Topology III
Introduction to functions and continuity
tweaked 27.i.11: the Appendix gives a proof that (C, d) is a metric space, but probably not the [only] proof…
tweaked 30.i.11 to correct a minor typo on page 2: at the end of the proof of the Theorem (Rudin 4.8), the ε-neighborhood is of course N_ε( f (p)), not N_r( f (p)).

Metric Topology IV
Sequences and convergence, etc.
tweaked 29.i.11 for overlines (see 27.i.11 tweak above)

Metric Topology V
Compactness and sequential compactness
tweaked 29.i.11 for overlines (see 27.i.11 tweak above)

Metric Topology VI
Cauchy sequences and related notions (completeness, completions, and a third formulation of compactness)
tweaked 9.ii.11 (reference in p.3 is to Problem 4 of PS1, not Problem 5)

Here is a more direct proof of the theorem that a continuous map f : X → Y between metric spaces is uniformly continuous if X is compact. Assume not. Then there exists ε > 0 such that for all δ > 0 there are some points p, q in X such that d(p, q) < δ but d(f (p), f (q)) ≥ ε. For each n = 1, 2, 3, …, choose p_n, q_n that satisfy those inequalities for δ = 1/n. Since X is assumed (sequentially) compact, we can extract a subsequence {p_ni} of {p_n} that converges to some p in X. But then {q_ni} converges to the same p. Hence both f (p_ni) and f (q_ni) converge to f (p), which contradicts the fact that d(f (p_ni), f (q_ni)) ≥ ε for each i.

You may have already seen “little oh” and “big Oh” notations. For functions f, g on the same space, “f = O(g)” means that  g  is a nonnegative real-valued function, f takes values in a normed vector space, and there exists a real constant M such that |f(x)|≤Mg(x) for all x. The notation “f = o(g)” is used in connection with a limit; for instance, “f(x) = o(g(x)) as x approaches x₀” indicates that f, g are vector- and real-valued functions as above on some neighborhood of x₀, and that for each ε>0 there is a neighborhood of x₀ such that |f(x)|≤εg(x) for all x in the neighborhood. Thus f '(x₀) = a means the same as “f (x) = f (x₀) + a(x−x₀) + o(|x−x₀|) as x approaches x₀”, with no need to exclude the case x = x₀. Rudin in effect uses this approach when proving the Chain Rule (5.5).

Apropos the Chain Rule: as far as I can see we don't need continuity of f  at any point except x (though that hypothesis will usually hold in any application). All that's needed is that x has some relative neighborhood N in [a,b] such that f (N) is contained in I. Also, it is necessary that f  map [a,b] to R, but g can take values in any normed vector space.

The derivative of f /g can be obtained from the product rule, together with the derivative of 1/g — which in turn can be obtained from the Chain Rule together with the the derivative of the single function 1/x. Once we do multivariate differential calculus, we'll see that the derivatives of f +g, f −g, fg, f /g could also be obtained in much the same way that we showed the continuity of those functions, by combining the multivariate Chain Rule with the derivatives of the specific functions x+y, x−y, xy, x/y of two variables x,y.

As Rudin notes at the end of this chapter, differentiation can also be defined for vector-valued functions of one real variable. As Rudin does not note, the vector space can even be infinite-dimensional, provided that it is normed; and the basic algebraic properties of the derivative listed in Thm. 5.3 (p.104) can be adapted to this generality, e.g., the formula (fg)' = f 'g + fg' still holds if f, g take values in normed vector spaces U, V and multiplication is interpreted as a continuous bilinear map from U × V to some other normed vector space W.

Rolle's Theorem is the special case f (b) = f (a) of Rudin's Theorem 5.10; as you can see it is in effect the key step in his proof of Theorem 5.9, and thus of 5.10 as well.

We omit 5.12 (continuity of derivatives) and 5.13 (L'Hôpital's Rule). In 5.12, see p.94 for Rudin's notion of “simple discontinuity” (or “discontinuity of the first kind”) vs. “discontinuity of the second kind”, but please don't use those terms in your problem sets or other mathematical writing, since they're not widely known. In Rudin's proof of L'Hôpital's Rule (5.13), why can he assume that g(x) does not vanish for any x in (a,b), and that the denominator g(x)−g(y) in equation (18) is never zero?

NB The norm does not have to come from an inner product structure. Often this does not matter because we work in finite dimensional vector spaces, where all norms are equivalent, and changing to an equivalent norm does not affect the definition of the derivative. The one exception to this is Thm. 5.19 (p.113) where one needs the norm exactly rather than up to a constant factor. This theorem still holds for a general norm but requires an additional argument. The key ingredient of the proof is this: given a nonzero vector z in a vector space V, we want a continuous functional w on V such that ||w|| = 1 and w(z) = |z|. If V is an inner product space (finite-dimensional or not), the inner product with z / |z| provides such a functional w. But this approach does not work in general. The existence of such w is usually proved as a corollary of the Hahn-Banach theorem. When V is finite dimensional, w can be constructed by induction on the dimension of V. To deal with the general case one must also invoke the Axiom of Choice in its usual guise of Zorn's Lemma.

We next start on univariate integral calculus, largely following Rudin, chapter 6. The following gives some motivation for the definitions there. (And yes, it's the same Riemann who gave number theorists like me the Riemann zeta function and the Riemann Hypothesis.)

The Riemann-sum approach to integration goes back to the “method of exhaustion” of classical Greek geometry, in which the area of a plane figure (or the volume of a region in space) is bounded below and above by finding subsets and supersets that are finite unions of disjoint rectangles (or boxes). The lower and upper Riemann sums adapt this idea to the integrals of functions which may be negative as well as positive (recall that one of the weaknesses of geometric Greek mathematics is that the ancient Greeks had no concept of negative quantities — nor, for that matter, of zero). You may have encountered the quaint technical term “quadrature”, used in some contexts as a synonym for “integration”. This too is an echo of the geometrical origins of integration. “Quadrature” literally means “squaring”, meaning not “multiplying by itself” but “constructing a square of the same size as”; this in turn is equivalent to “finding the area of”, as in the phrase “squaring the circle”. For instance, Greek geometry contains a theorem equivalent to the integration of x²dx, a result called the “quadrature of the parabola”. The proof is tantamount to the evaluation of lower and upper Riemann sums for the integral of x²dx.

An alternative explanation of the upper and lower Riemann sums, and of “partitions” and “refinements” (Definitions 6.1 and 6.3 in Rudin), is that they arise by repeated application of the following two axioms describing the integral (see for instance L.Gillman's expository paper in the American Mathematical Monthly (Vol.100 #1, 16–25)):

• For any a,b,c (with a < b < c), the integral of a function from a to c is the sum of its integrals from a to b and from b to c;
• If a function f on [a,b] takes values in [m,M] then its integral from a to b is in [m(b−a), M(b−a)] (again assuming a < b).

The latter axiom is a consequence of the following two: the integral of a constant function from a to b is that constant times the length b−a of the interval [a,b]; and if f ≤ g on some interval then the integral of f over that interval does not exceed the integral of g. Note that again all these axioms arise naturally from an interpretation of the integral as a “signed area”.

The (Riemann-)Stieltjes integral, with dα in place of dx, is then obtained by replacing each Δx = b−a by Δα = α(b)−α(a).

In Theorem 6.12, property (a) says the integrable functions form a vector space, and the integral is a linear transformation; property (d) says it's a bounded transformation relative to the sup norm, with operator norm at most Δα = α(b)−α(a) (indeed it's not hard to show that the operator norm equals Δα = α(b)−α(a)); and (b) and (c) are the axioms noted above. Property (e) almost says the integral is linear as a function of α — do you see why “almost”?

Recall the “integration by parts” identity: fg is an integral of f dg + g df. The Stieltjes integral is a way of making sense of this identity even when f and/or g is not continuously differentiable. To be sure, some hypotheses on f and g must still be made for the Stieltjes integral of f dg to make sense. Rudin specifies one suitable system of such hypotheses in Theorem 6.22.

Here's a version of Riemann-Stieltjes integrals that works cleanly for integrating bounded functions from [a,b] to any complete normed vector space.
corrected 27.ii.11 to fix a minor typo (“reasonable hypotheses”, not “reasonable hypothesis”).

Riemann-Stieltjes integration by parts: Suppose both f and g are increasing functions on [a,b]. For any partition a = x₀ < … < x_n = b of the interval, write f(b)g(b) − f(a)g(a) as the telescoping sum of f(x_i)g(x_i) − f(x_i−1)g(x_i−1) from i=1 to n. Now rewrite the i-th summand as

f (x_i) (g(x_i)−g(x_i−1)) + g(x_i−1) (f (x_i )−f (x_i−1)). [Naturally it is no accident that this identity resembles the one used in the familiar proof of the formula for the derivative of fg !] Summing this over i yields the upper Riemann-Stieltjes sum for the integral of f dg plus the lower R.-S. sum for the integral of g df. Therefore: if one of these integrals exists, so does the other, and their sum is f (b)g(b) − f (a)g(a). [Cf. Rudin, page 141, Exercise 17.]

We'll cover most of the new parts of Chapter 7:

• One counterexample, 7.4;
• Weierstrass M, 7.10, extended to vector-valued functions;
• uniform convergence and ∫  (7.16, again in vector-valued setting, with the target space V normed and complete);
• uniform convergence and d/dx (7.17, also taking values in a complete normed vector space V, which for us is either finite-dimensional or an inner-product space since we're not proving Hahn-Banach); and
• another counterexample, 7.18.

We also postpone discussion of Euler's Beta and Gamma integrals (also in Chapter 8) so that we can use multivariate integration to give a more direct proof of the formula relating them.

The result concerning the convergence of alternating series is stated and proved on pages 70-71 of Rudin (Theorem 3.42).

The original Weierstrass approximation theorem (7.26 in Rudin) can be reduced to the uniform approximation of the single function |x| on [−1,1]. From this function we can construct an arbirtrary piecewise linear continuous function, and such piecewise linear functions uniformly approximate any continuous function on a closed interval. To get at |x|, we'll rewrite it as [1−(1−x²)]^1/2, and use the power series for (1−X)^1/2. This power series (and more generally the power series for (1−X)^A) is the first part of Exercise 22 for Chapter 8, on p.201; we've already outlined another approach in PS5, Problem 3 (under the assumption of the standard formula for differentiating x^r with respect to x, which as we note there is not too hard for r rational). We need (1−x)^1/2 to be approximated by its power series uniformly on the closed interval [−1,1] (or at least [0,1]); but fortunately this too follows from the proof of Abel's theorem (8.2, pages 174-5). Actually this is a subtler result than we need, since the Xⁿ coefficient of the power series for (1-X)^1/2 is negative for every n>0. If a power series in X has radius of convergence 1 and all but finitely many of its nonzero coefficients have the same sign, then it is easily shown that the sum of the coefficients converges if and only if f(X) has a finite limit as X approachess 1, in which case the sum equals that limit and the power series converges uniformly on [0,1]. That's all we need because clearly (1-X)^1/2 extends to a continuous function on [0,1]. (For an alternative approach to uniformly approximating |x|, see exercise 23 on p.169.)

Rudin's notion of an “algebra” of functions is almost a special case of what we called an “algebra over F” in 55a (with F = R or C as usual), except that Rudin does not require his algebras to have a unit (else he wouldn't have to impose the “vanish on no point” condition). The notion can be usefully abstracted to a “normed algebra over F”, which is an algebra together with a vector space norm ||·|| satisfying ||xy|| ≤ ||x|| ||y|| for all x and y in the algebra. Among other things this leads to the Stone-Čech theorem.

In the first theorem of Chapter 8, Rudin obtains the termwise differentiability of a power series at any |x| < R by applying Theorem 7.17. That's nice, but we'll want to use the same result in other contexts, notably over C, where the mean value theorem does not apply. So we'll instead give an argument that works in any complete field with an absolute value — this includes R, C, and other examples such as the field Q_p of p-adic numbers. If the sum of c_n xⁿ converges for some nonzero x with |x| = R, then any x satisfying |x| < R has a neighborhood that is still contained in {y : |y| < R}. So if f (x) is the sum of that series, then for y ≠ x in that neighborhood we may form the usual quotient (f (x)−f (y)) / (x−y) and expand it termwise, then let y→x and recover the expected power series for f '(x) using the Weierstrass M test (Theorem 7.10).

An alternative derivation of formula (26) on p.179: differentiate the power series (25) termwise (now that we know it works also over C) to show E(z) = dE(z)/dz; then for any fixed w the difference E(w+z) − E(w) E(z) is an analytic function of z that vanishes at z = 0 and is thus zero everywhere.

Error in Rudin: the argument on p.180 that “Since E is strictly increasing and differentiable on [the real numbers], it has an inverse function L which is also strictly increasing and differentiable …” is not quite correct: consider the strictly increasing and differentiable function taking x to x³. What's the correct statement? (Hint: the Chain Rule tells you what the derivative of the inverse function must be.)

An explicit upper bound of 4 on π can be obtained by calculating cos(2) < 1 − 2^2/2! + 2^4/4! = −1/3 < 0. For much better estimates, integrate (x−x²)⁴ dx/(x²+1) from 0 to 1 and note that ½ ≤ 1/(x²+1) ≤ 1. :-)   [That was noted in my recent Math Table talk, but attendance was depressed by (I'm told) midterms season.]

Rudin uses a fact about convex functions that is only presented as an exercise earlier in the book (p.100, #23). Namely: let f  be a convex function on some interval I, and consider the slope s(x, y) := (f (x)−f (y)) / (x−y) as a function on the set of (x, y) in I × I with x > y; then s is an increasing function of both variables. The proof is fortunately not hard. For instance, to prove that if x > y' > y then s(x, y' ) > s(x, y), write y' as px + qy with p + q = 1, and calculate that s(x, y' ) > s(x, y) is equivalent to the usual convexity condition. The case x > x' > y works in exactly the same way.

While we alas suppress most of Fourier analysis, we can easily prove the following: if f :R/2πZ → C is a continuous function whose Fourier coefficients a_n := (2π)⁻¹∫ _R/2πZ exp(−inx) f (x) dx satisfy ∑_n|a_n| < ∞ then f equals its Fourier series. Proof : the difference is a continuous function all of whose Fourier coefficients vanish. By Stone-Weierstrass, it can be uniformly approximated by trigonometric polynomials, etc. (This special case of Stone-Weierstrass is also a theorem of Fejér, who obtained an explicit sequence of trigonometric polynomials converging to the function; see Körner.) A nice example is f (x) = B_k(2πx) for each k≥2, where B_k is the kth Bernoulli polynomial; this yields the values of ζ(k) for k = 2, 4, 6, …, and much else.

The rather obscure integration by parts in Rudin, p.194 is not necessary. A straightfoward choice of “parts” yields

This may seem to go in a useless direction, but the elementary observation that

recovers the recursion (97).

In addition to the trigonometric definite integrals noted by Rudin (formula 98), Beta functions also turn up in the evaluation of the definite integral of u^a du / (1+u^b)^c over (0,∞): let t = u^b / (1+u^b). What is the value of that integral? Can you obtain in particular the formula π / (b sin(a π/b)) for the special case c=1?

The limit formula for Γ(x) readily yields the product formula:

Γ(x) = x⁻¹ e^−Cx Prod(exp(x/k) / (1+x/k), k=1,2,3,...)

We next begin multivariate differential calculus, starting at the middle of Rudin Chapter 9 (since the first part of that chapter is for us a review of linear algebra — but you might want to read through the material on norms of linear maps and related topics in pages 208–9). Again, Rudin works with functions from open subsets of Rⁿ to R^m, but most of the discussion works equally well with the target space R^m replaced by an arbitrary normed vector space V. If we want to allow arbitrary normed vector spaces for the domain of f, we'll usually have to require that the derivative f ' be a continuous linear map, or equivalently that its norm ||f '|| = sup_|v|≤1|f '(v)| be finite.

As in the univariate case, proving the Mean Value Theorem in the multivariate context (Theorem 9.19) requires either that V have an inner-product norm, or the use of the Hahn-Banach theorem to construct suitable functionals on V. Once this is done, the key Theorem 9.21 can also be proved for functions to V, and without first doing the case m=1. To do this, first prove the result in the special case when each D_j f (x) vanishes; then reduce to this case by subtracting from f the linear map from Rⁿ to V indicated by the partial derivatives D_j f (x).

The Inverse function theorem (9.24) is a special case of the Implicit function theorem (9.28), and its proof amounts to specializing the proof of the implicit function theorem. But Rudin proves the Implicit theorem as a special case of the Inverse theorem, so we have to do Inverse first. (NB for these two theorems we will assume that our target space is finite-dimensional; how far can you generalize to infinite-dimensional spaces?) Note that Rudin's statement of the contraction principle (Theorem 9.23 on p.220; cf. the final problem of problem set 5) is missing the crucial hypothesis that X be nonempty! The end of the proof of 9.24 could be simplified if Rudin allowed himself the full use of the hypothesis that f is continuously differentiable on E, not just at a: differentiability of the inverse function g at b = f(a) is easy given Rudin's construction of g; differentiability at any other point f(x) follows, since x might as well be a, and then the derivative is continuous because g and f ' are.

The proof of the second part of the implicit function theorem, which asserts that the implicit function g not only exists but is also continuously differentiable with derivative at b given by formula (58) (p.225), can be done more easily using the chain rule, since g has been constructed as the composition of the following three functions: first, send y to (0, y); then, apply the inverse function F⁻¹; finally, project the resulting vector (x,y) to x. The first and last of these three functions are linear, so certainly C¹; and the continuous differentiability of F⁻¹ comes from the inverse function theorem.

Here's an approach to D_ij=D_ji that works for a C² function to an arbitrary normed space. As in Rudin (see p.235) we reduce to the case of a function of two variables, and define u and Δ. Assume first that D₂₁ f vanishes at (a,b). Then use the Fundamental Theorem of Calculus to write Δ(f,Q) as as the integral of u'(t) dt on [a, a+h], and then write u'(t) as an integral of D₂₁ f (t,s) ds on [b,b+k]. Conclude that u'(t) = o(k) and thus that Δ(f,Q) / hk approaches zero. Now apply this to the function f − xyD₂₁ f (x,y) to see that in general Δ(f,Q) / hk approaches D₂₁ f (x,y). Do the same in reverse order to conclude that D₂₁ f (x,y)=D₁₂ f (x,y). Can you prove D₁₂(f ) = D₂₁(f ) for a function f to an arbitrary inner product space under the hypotheses of Theorem 9.41?

We omit the “rank theorem” (whose lesser importance is noted by Rudin himself), as well as the section on determinants (which we treated at much greater length in Math 55a).

An important application of iterated partial derivatives is the Taylor expansion of an m-times differentiable function of several variables; see Exercise 30 (Rudin, 243-244). As promised at the start of Math 55a and/or Math 55b, this also applies to maxima and minima of real-valued functions f  of several variables, as follows. If f  is differentiable at a local maximum or minimum then its derivative there vanishes, as was the case for a function of one variable. Again we say that a zero of the derivative is a “critical point” of f. Suppose now that f  is C² near a critical point. The second derivative can be regarded as a quadratic form. It must be positive semidefinite at a local minimum, and negative semidefinite at a local maximum. Conversely, if it is strictly positive (negative) definite at a critical point then that point is a strict local minimum (maximum) of f. Compare with Rudin's exercise 31 on page 244 (which however assumes that  f  is C³ — I don't know why Rudin requires third partial derivatives).

Next topic, and last one from Rudin, is multivariate integral calculus (Chapter 10). Most of the chapter is concerned with setting up a higher-dimensional generalization of the Fundamental Theorem of Calculus that comprises the divergence, Stokes, and Green theorems and much else besides. With varying degrees of regret we'll omit this material, as well as the Lebesgue theory of Chapter 11. We will, however, get some sense of multivariate calculus by giving a definition of integrals over Rⁿ and proving the formula for change of variables (Theorem 10.9). this will already hint why in general an integral over an n-dimensional space is often best viewed as an integral not of a function but a “differential n-form”. For instance, in two dimensions an integral of “f (x, y) dx dy” can be thought of as “f (x, y) dx ∧ dy”, and then we recover the formula involving the Jacobian from the rules of exterior algebra. You'll have to read the rest of this chapter of Rudin, and/or take a course on differential geometry or “calculus on manifolds”, to see these ideas developed more fully.

Once we have the change of variables formula, we'll return to the section of Chapter 8 concerning the Euler's Beta and Gamma integrals and give a more natural treatment of the formula relating them (Theorem 8.20).

Math 55b concludes with an introduction to complex analysis (a.k.a. “functions of one complex variable”). We'll start with contour integrals and the fundamental theorems of Cauchy, roughly following the exposition in Ahlfors, chapter III (p.82 ff.). We'll prove:

• Cauchy's theorem for a rectangle (if f  is differentiable on a neighborhood of the rectangle, then the contour integral of f (z) dz on the boundary of the rectangle vanishes), using the subdivision trick that Ahlfors attributes to Goursat.
• The variant when f  is defined on the complement in the rectangle of finitely many points ζ, as long as (z−ζ) f (z) → 0 on each such point
• Same theorems for a circle, using the same subdivision trick.
• Likewise for the annulus between two concentric circles: the integrals over the two circles are equal. (One proof: change variables to a rectangle of height 2π using complex exponential.)
• Cauchy's integral formula for a rectangle or circle: under the same hypotheses, if a is any interior point then f (a) is (1/2πi) times the contour integral of f (z) dz / (z−a).
• Consequences:
  1. If f  is differentiable in a circle of radius R > r > 0 about a then f (a) is the average of f (a + re^iθ) over θ in [0, 2π]; corollary(!): Fundamental Theorem of Algebra; also: |f (·)| has no local maximum unless it is constant (“maximum principle” for analytic functions).
  2. [via power series expansion of 1 / (z−a)]: f  is analytic, and its power-series expansion about any z₀ converges in |z−z₀| < r if that open circle is contained in the interior of our rectangular or circular contour. This again yields the maximum principle above, and also shows that |f (·)| has no nonzero local minimum unless it is constant.
  3. Same argument (or contour integration by parts) also yields an integral formula for the derivative f '(z₀), which in turn gives Liouville's theorem: a bounded analytic function on all of C is constant. [An analytic function on all of C is also known as an entire function.]
  4. Using also the integral formula for higher derivatives of an analytic function: The uniform limit f  of a sequence {f_n} of analytic functions is analytic, and each term in a power series expanion of f  is the limit of the corresponding terms for the f_n. Thus the same is true for a uniformly convergent sum of analytic functions. This is very useful for constructing/defining analytic functions; e.g. the Riemann zeta function ζ(s) is defined for Re(s)>1 as ∑_nn^−s (summed over n=1,2,3,…).
• (“analytic continuation”) An analytic function f on a neighborhood of z that vanishes on distinct points z_n → z is identically zero. (Proof: by induction each coefficient in the power series expansion of f about z vanishes.) Hence if two analytic functions f, g agree at each z_n then they are equal.
• Cauchy's integral formula also works under the weaker hypothesis we've seen above, allowing a finite number of exceptional ζ; the formula shows that in this case f  extends to an analytic function on all of the interior of our rectangular or circular contour. Thus such ζ is called a removable singularity.
• More generally, if f  is analytic in a punctured neighborhood of ζ, and (z−ζ)ⁿ f (z) → 0 for some positive integer n, then (z−ζ)ⁿ⁻¹ f (·) has a removable singularity at ζ. Then f  is said to have a pole at ζ: we can write f (z) as an polynomial of degree <n in 1 / (z−ζ) plus an analytic function on a neighborhood of ζ. We say ζ is a single, double, triple, etc. pole if that polynomial in 1 / (z−ζ) has degree 1, 2, 3, …, or a pole of order d if the polynomial has degree d. (A removable singularity then has d=0 but is rarely called a “pole of order zero”.) If there's no such n then f  is said to have an essential singularity at ζ. The standard example of such a function is f (z) = exp(1/z) with ζ = 0.
• If f  is analytic on an open set E, and has a set P of poles, then f is said to be meromorphic on the union of E and P. In the special case of an analytic function (i.e. with P empty) we also say f  is holomorphic. As the holomorphic functions form a ring, the meromorphic functions on a subset of C form a field; it turns out to be the fraction field of the holomorphic functions: for any meromorphic function f  we can find a holomorphic function h that vanishes at the poles of f  to the necessary multiplicities for g=hf  to be holomorphic, and then f =g/h. [NB: Analytic continuation also works for meromorphic functions; indeed it's easy to see that a function continuous at z cannot have a sequence of poles approaching z, so by restricting to a small enough neighborhood we get a holomorphic function.]
• Example: if f  is analytic on the unit disc and vanishes at the origin then f/z is analytic too; this yields the Schwarz Lemma.

• For an analytic function f  on a punctured neighborhood of ζ, the residue at ζ of the differential f (z) dz is invariant under locally invertible analytic changes of variable; i.e. is the same as the residue of f (g(w)) g'(w) dw at the preimage of ζ under g, assuming g' does not vanish there. The residue is defined as the integral of (1/2πi) f (z) dz on (say) a circle around ζ. If f (z) has a power series expansion ∑_nc_n (z−ζ)ⁿ in a punctured neighborhod of ζ then the residue is the coefficient c₋₁.
• The integral of a differential on a closed contour (at least one of our standard contours: circles, rectangles, and their images under invertible conformal maps) is 2πi times the sum of the residues at the poles or essential singularities the contour encloses, assuming the differential is analytic except at those finitely many points.
• This has numerous applications to the evaluation of (often non-elementary) definite integrals of elementary functions. Some paradigmatic examples:
  • ∫₀^∞ dx / (x² + 1)  =  π / 2
  • ∫₀^∞ dx / (x² + 1)²  =  π / 4
  • ∫₀^∞ cos(cx) dx / (x² + 1)  =  (π / 2) e^−|c|   for all real c
  • ∫₀^∞ sin(cx) dx / x  =  π / 2   for all c > 0 [via the “principal value” of the integral of exp(icx) dx / x over (−∞,∞)]
  • For α in (0,1), the special value B(α, 1−α) of the Beta function is given by
    ∫₀¹ dx / (x^α(x + 1))  =  π / sin(απ)
• An important special case: the logarithmic differential df / f of a meromorphic function f  has only simple poles, at the zeros and poles of f , with residue n at a zero of order n and −n at a pole of order n. Hence the integral of df / f = (f '(z) / f (z)) dz on a closed contour is 2πi times the difference between the number of zeros and poles enclosed by the contour, counted with multiplicity. This assumes that f  is analytic on the interior of the contour and has neither zero nor pole on the contour itself. This formula is often called the “argument principle”, because the integral can be interpreted as i times the change of the “argument” (= imaginary part of the logarithm) of f  around the contour. A nice and occasionally useful corollary is Rouché's theorem: if |g(z)| < |f (z)| for every z on the contour, then f +g has the same change of argument around the contour as f, and thus the same count of zeros minus poles with multiplicity on the interior of the contour.
• The Gamma function extends to a meromorphic function on C, satisfying the same functional equation Γ(z+1) = z Γ(z), and holomorphic except for simple poles at z = 0, −1, −2, −3, … . Given that there exists a meromorphic function Γ on C that agrees with the usual Gamma function on the positive real axis, this extension is unique (analytic continuation again), but it is not immediate that such an extension is possible. We give several approaches:
  1. Use the Euler integral to define Γ(z) for z = x + iy of positive real part x; prove the functional equation for such x (either by analytic continuation or by the usual integration by parts); and then use the functional equation to inductively extend Γ to x > −1, x > −2, x > −3, etc., at each stage finding a simple pole at 0, −1, −2, etc.
  2. Use the formula Γ(z) = lim_n→∞ n! n^z / (z (z+1) (z+2) … (z+n)) . This sequence of analytic functions converges uniformly on bounded subsets of C − {0, −1, −2, −3, …}, whence its limit is analytic.
  3. Let C be a contour that goes from “∞ − iε” horizontally to −ε−iε, then vertically to to −ε+iε, and horizontally to to “∞ + iε”. The integral along C of t^z−1 e^−t dt defines an entire function of z that equals (1−e^2πiz) Γ(z) for Re(z) > 0, so we can solve for Γ(z) for any complex z.
  Each of these can be used to prove the functional equation; the second most easily yields the following additional properties:
  • For each x > 0, the function | Γ(x+iy) | of a real variable y is maximized at y = 0, and is increasing for y ≥ 0 and decreasing for y ≤ 0;
  • In particular, Γ(z) ≠ 0 for all complex z;
  • The Stirling approximation holds for the complex Gamma function, in the following sense: given ε > 0, we have Γ(z) asymptotic to (2π)^1/2 z^z−(1/2) e^−z as |z| →∞  as long as z has argument in [ε−π, π−ε].
  [NB We did not prove this last part in class, and yes, it does reduce to the usual Stirling approximation to n! for z = n+1 even though it looks different.]
• The Beta identity B(x,y) = Γ(x) Γ(y) / Γ(x+y) still holds for x and y of positive real part, and can be used to define B(x,y) for all complex x and y not among the poles of Γ. In particular, the identity B(α, 1−α) =  =  π / sin(απ) holds for all non-integer complex numbers α. We used this, together with the formula Γ(z) = lim_n→∞ n! n^z / (z (z+1) (z+2) … (z+n)) , to give a proof (admittedly more convoluted than necessary or usual) of the product expansion of the sine, and thus (via logarithmic differentiation) of the partial-fraction decomposition of cot(x) and (differentiating once more) sec²(x), etc. Evaluation at special points like π/2 yields classical formulas: Wallis's product, the values of ζ(k) for k = 2, 4, 6, … (again), and others.
• (not covered in the final exam) Introduction to conformal mapping and the Riemann mapping theorem; and 1.5 proofs of Müntz's generalization of the Weierstrass approximation theorem.

Problem sets 3 and 4: Metric topology cont'd
Yes, in problem 10i all vector spaces are over F=R or C. (We already know from last term it's not true over Q…)
Problem 8 is the Lebesgue Covering Lemma (handout VI already let this cat out of the bag); r is a “Lebesgue number” for the open cover. Problem 10(ii) is A-5 on the 1999 Putnam exam (which specified n = 1999 — hah). Problem 13 is Urysohn's Lemma.

Problem set 5: Topology grand finale
Problem 3 is the proof Rudin gives in Chapter 8, pages 184-185. The proof we outlined last term is Exercises 26 and 27 on pages 202.
Problem 8 is the Arzelà-Ascoli theorem. (It turns out to also be largely given in Rudin Chapter 7, see 7.22 ff.) The Wikipedia page gives some typical applications to real and complex analysis.

Problem set 6: Univariate differential calculus
In problem 2 the relevant vector spaces are all finite-dimensional, so all norms are equivalent and differentiability etc. is well defined. Also, “f  o g” is the function taking x to the composition of the linear operators f (x): V → W and g(x): U → V, not composition of functions as in the Chain Rule.
Belated correction 16.iii.11: in problem 2(ii), the map t → f (t)⁻¹ is indeed defined and differentiable in some neighborhood of x, but not necessarily on all of [a, b]. (And its derivative at x is given by the formula −f (x)⁻¹f '(x) f (x)⁻¹, which nicely extends the familiar −f (x)⁻²f '(x) for a nonzero scalar-valued function in a way consistent with the behavior of the derivative and inverse of the matrix transpose.)

Problem set 7: Univariate integral calculus
Corrected 6.iii.11: yes, this problem set is #7, not #2…
The nice solution to Problem 6 is to use a Riemann sum associated to a partition by a geometric series, rather than the usual arithmetic series.

Problem set 8: Univariate integral calculus, cont'd
Corrected 23.iii.11: In problem 1, the series for E(x) starts at n=0, not n=1. (To be sure if you did the n=1 version then the n=0 answer is immediate…)
Belated correction 25.iii.11: In problem 2, x₀ is in the ground field k, not the “function field” K.

Problem set 9: Multivariate differential calculus
Tweaked 26.iii.11: added parentheses in Problem 7 around the “n+1” in “(n+1)-dimensional” (and yes it's a “subspace” in the topological sense but not the vector-space sense…);
Corrected 28.iii.11 to fix a typo (missing “a” in “[Cauchy-Riemann] equations”), to state explicitly in Problem 7 that the root t₀ is real, and to change the eigenvector of M from v₀ to v(M) in Problem 8.

Problem set 10: Multivariate integral calculus, etc.
Δ [Not to be confused with this version of PS10; note the due dates. But if you find a complex number s of real part <1 such that ∑_nμ(n) n^−s converges, please let me know!]
Corrected 7.iv.11 to fix a typo in the generalization of problem 7: the sum of the x_i^s_i should be ≤1, not =1.

Problem set 11: Convexity, multivariate change of variables, etc.
Corrected 10.iv.11: the two Rudin problems from p.290 were already assigned last week :-( so omitted.
Corrected 14.iv.11: in the last problem, the shell has |x| in (r₁, r₂), not (a, b).
and belatedly Corrected 25.iv.11: the numerator on the right-hand side of the displayed equation (in Problem 5) is Γ(ν+1), not Γ(v+1) [with “nu”, not “vee”].
and very belatedly :-( Corrected 7.v.11: in Problem 5 again, it's ν that must exceed |λ|, not just |ν|. (If ν is sufficiently negative the integral doesn't even converge.) And yes, the hypothesis on ν is needlessly strong (it's a relic of an earlier incorrect formula for the integral) but if you did it already for ν > |λ| then you needn't re-do the proof for a wider range of ν.

Problem set 12: Complex analysis I
Belatedly Corrected 22.iv.11: meromorphic functions, plural, in Problem 7; and |1−x_k|, not (1−x_k), in each factor of the numerator displayed in Problem 9

13th and last problem set: Complex analysis II: residues and contour integration
Belatedly Corrected 29.iv.11, but only to fix a transparent grammar error in the preamble