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.

Math Night will again happen Monday evenings/nights, usually 8-10, in Leverett Dining Hall, starting January 22. The course CA’s will again hold office hours there.

Rohil Prasad: Thursday 3-4 PM, Science Center room 309A.

Vikram Sundar: Wendesday 4-5 PM, Science Center room 304.

Vikram’s notes for 55b will be here.

Our first topic is the

Metric Topology I

Basic definitions and examples:
the metric spaces **R**^{n}
and other product spaces; isometries; boundedness and function spaces

The “sup metric” on $X^S$ is sometimes also called the “uniform metric” because $d(\,f,g) \leq r$ is equivalent to a bound, $d(\,f(s),g(s)) \leq r$ for all $s \in S$, that is “uniform” in the sense that it’s independent of the choiceof $s$. Likewise for the sup metric on the space of bounded functionsfrom $S$ to an arbitrary metricspace $X$ (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 thebounded functionsfrom $S$ to $X$doconstitute 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 boundedfand g.

Now that metric topology is in 55b, not 55a, the following observation can be made: if $X$ is $\bf R$or $\bf C$, 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 $\left\| \cdot \right\|$ is defined by $\| \, f \| = \sup_{s \in S} | \, f(s)|.$ 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:Metric Topology IIiv) ForIn 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)?every$p \in X$ there exists a real number $M$ such that $d(p,q) \lt M$ for all $q \in E.$

Open and closed sets, and related notions

Metric Topology III

Introduction to functions and continuity

Metric Topology IV

Sequences and convergence, etc.

The proof of “uniform limit of continuous is continuous” also shows that “uniform limit of uniformly continuous is uniformly continuous”: if each $f_n$ is uniformly continuous then a single $\delta$ will work forall $x$.

A typical application is the continuity of power series such as $\sum_{k=0}^\infty x^k/k!$ (which by definition is the pointwise limit of the partial sums $f_n(x) = \sum_{k=0}^n x^k/k!$). The limit isnotuniform as $x$ ranges over all of $\bf R$(or $\bf C$), but itisuniform for $x$ in every ball $B_R(0)$, and since every $x$ is in some such ball the limit function is continuous. Likewise for any power series $\sum_{k=0}^\infty a_k x^k$ with $\{a_k\}$ bounded, as long as $x$ is within the open circle of convergence $|x| \lt 1$: the limit is not in general uniform, but itisuniform in the ball $B_r(0)$ for every $r \lt 1$ (why?), which is good enough because $|x|\lt 1$ means that $x$ is contained in such a ball. In either case, each $f_n$ is easily seen to be uniformly continuous on bounded sets, so the argument noted in the previous paragraph shows that the limit function $f$ is also uniformly continuous in every ball on which we showed that it is continuous (that is, arbitrary $B_R(0)$ for $\sum_{k=0}^\infty x^k/k!$, or $B_r(0)$ with $r \lt 1$ for $\sum_{k=0}^\infty a_k x^k$ with $\{a_k\}$ bounded). [NB we need to use open balls $B$ so that we can test continuity at $x \in X$ on a neighborhoodfor small enough $\epsilon$, the in $B$:$\epsilon$-neighborhood of $x$in $X$ is containedin $B$.]

Metric Topology V

Compactness and sequential compactness

Metric Topology VI

Cauchy sequences and related notions
(completeness, completions, and a third formulation of compactness)

Here is a more direct proof (using sequential compactness) of the theorem that a continuous map $f: X \to Y$ between metric spaces is uniformly continuous if $X$ is compact. Assume not. Then there exists $\epsilon \gt 0$ such that for all $\delta \gt 0$ there are some points $p,q \in X$ such that $d(p,q) \lt \delta$ but $d(\,f(p),f(q)) \geq \epsilon.$ For each $n = 1, 2, 3, \ldots,$ choose $p_n, q_n$ that satisfy those inequalities for $\delta = 1/n.$ Since $X$ is assumed compact, we can extract a subsequence $\{ p_{n_i} \! \}$ of $\{p_n\!\}$ that converges to some $p \in X.$ But then $\{q_{n_i}\!\}$ converges to thesame $p$. Hence both $f(p_{n_i})$ and and $f(q_{n_i})$ converge to $f(p),$ which contradicts the fact that $d(\,f(p_{n_i}), f(q_{n_i})) \geq \epsilon$ foreach $i$.

Our next topic is

You may have already seen “little oh” and “big Oh” notations. For functions $f,g$ on the same space,We next start on“$f = O(g)$” 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 $\left|\,f(x)\right| \le M g(x)$ forall $x$. The notation“$f = o(g)$” is used in connection with a limit; for instance,“$f(x) = o(g(x))$ as $x$approaches $x_0$” indicates that $f,g$ are vector- and real-valued functions as above on some neighborhoodof $x_0$, and that for each $\epsilon \gt 0$ there is a neighborhoodof $x_0$ such that $\left|\,f(x)\right| \le \epsilon g(x)$ for all $x$ in the neighborhood. (Given $g$ and the targetof $f\!$, functions $f=O(g)$ form a vector space, which contains functions $o(g)$ as a subspace.) Thus $f'(x_0) = a$ means the same as“$f(x) = f(x_0) + a (x-x_0) + o(\left|x-x_0\right|)$ as $x$ approaches $x_0\!$”, with no need to exclude the case $x = x_0.$ 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 pointexcept $x$ (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 containedin $I$. Also, it is necessary that $f$ map $[a,b]$to $\bf R$, 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 $1/x$. [Also, if you forget the quotient-rule formula, you can also reconstruct it from the product rule by differentiating both sides of $f = g \cdot (\,f/g)$ and solving for $(\,f/g)';$ but this is not a proof unless you have some other argument to show that the derivative exists in the first place.] 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

notnote, 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 $U \times V$ to some other normed vectorspace $W$. “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. [As you can see from the Wikipedia page, the attribution of this result to Michelle Rolle (1652–1719) is problematic in several ways, and seems to be a good example of “Stigler’s law of eponymy”.]

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. One exception to this is Theorem 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 $\left\| w \right \| = 1$ and $w(z) = \left| z \right|.$ If $V$ is an inner product space (finite-dimensional or not), the inner product with $z \left/ \left| z \right| \right.$ provides such afunctional $w$. 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$V\!$. To deal with the general case one must also invoke the Axiom of Choice in its usual guise of Zorn’s Lemma.

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 $\int x^2 \, dx,$ a result called the “quadrature of the parabola”. The proof is tantamount to the evaluation of lower and upper Riemann sums for the integralSome ofof $x^2 \, dx$. 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)):The latter axiom is a consequence of the following two: the integral $\int_a^b K \, dx$ of a constant function $K$ is $K(b-a);$ and if $f(x) \le g(x)$ for all $x$ in the interval $[a,b]$ then $\int_a^b f(x) \, dx \le \int_a^b g(x) \, dx.$ Note that again all these axioms arise naturally from an interpretation of the integral as a “signed area”.

- For any $a,b,c$ (with $a \lt b \lt c$) we have $\int_a^c f(x)\, dx = \int_a^b f(x)\, dx + \int_b^c f(x)\, dx;$
- If a function $f: [a,b] \to {\bf R}$ takes values in $[m,M]$ then $\int_a^b f(x) \, dx \in [m(b-a),M(b-a)]$ (again assuming
$a \lt b$). The (Riemann-)Stieltjes integral, with $d\alpha$ in place

of $dx$, is then obtained by replacing each $\Delta x = b - a$ by $\Delta\alpha = \alpha(b) - \alpha(a).$Here’s a version of Riemann-Stieltjes integrals that works cleanly for integrating bounded functions from $[a,b]$ to any complete normed vector space.

In Theorem 6.11 (integrable functions are preserved under continuous maps), we readily generalize to the integrability over $[a,b]$ of $h = \phi \circ f$ when $f:[a,b] \to [m,M]$ is integrable and $\phi$ is a continuous map from $[m,M]$ to a complete normed

vector space $V$. If we want to generalize further by letting $\,f$ itself be vector-valued, then we must explicitly assume that $\phi$ isuniformlycontinuous, which Rudin doesn’t have to do in 6.11 because $[m,M]$ is compact.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 $\Delta\alpha = \alpha(b)-\alpha(a)$ (indeed it’s not hard to show that the operator norm equals $\Delta\alpha = \alpha(b)-\alpha(a);$ and (b) and (c) are the axioms noted above. Property (e)

almostsays the integral is linear as a functionof $\alpha$ — 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.

Riemann-Stieltjes integration by parts: Suppose both $\,f$ and $g$ are increasing functions on $[a,b].$ For any partition $a = x_0 \lt \cdots \lt x_n = b$ of the interval, write $\,f(b) g(b) - f(a) g(a)$ as the telescoping sum $\sum_{i=1}^n \left(f(x_i) g(x_i) - f(x_{i-1}) g(x_{i-1})\right).$ Now rewrite the$i$-th summand as $$ f(x_i) (g(x_i) - g(x_{i-1})) + g(x_i) (f(x_i) - f(x_{i-1})). $$ [Naturally it is no accident that this identity resembles the one used in the familiar proof of the formula for the derivativeof $fg$!] Summing thisover $i$ 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 some of the new parts of Chapter 7:
Weierstrass M, 7.10, extended to vector-valued functions;
uniform convergence and $\int$ (7.16, again in vector-valued setting,
with the target space $V$ normed and complete); and the
**Stone-Weierstrass theorem**,
which is the one major result of Chapter 7 we haven’t seen yet.
We then proceed to **power series** and the exponential
and logarithmic functions in **Chapter 8**.
We omit most of the discussion of Fourier series (185–192),
an important topic (which used to be the concluding topic of Math 55b),
but one that alas cannot be accommodated given the mandates of
the curricular review. We’ll encounter a significant special case
in the guise of Laurent expansions of an analytic function on a disc.
See these notes (part 1,
part 2) from 2002-3 on
**Hilbert space** for a fundamental context for Fourier
series and much else (notably much of quantum mechanics),
which is also what we’ll use to give one proof of the
Müntz-Szász theorem
on uniform approximation on $[0,1]$ by linear combinations of arbitrary powers.
[Yes, if I were to rewrite these notes now I would not have to define
separability, because we already did that in the course of developing
the general notion of compactness.]

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

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 arbitrary piecewise linear continuous function, and such
piecewise linear functions uniformly approximate any continuous function
on a closed interval.(*) [They also give yet another example of a natural
vector space with an uncountably infinite algebraic basis.]
To get at $|x|,$ we’ll rewrite it as
$[1 - (1-x^2)]^{1/2},$ and use the power series for $(1-X)^{1/2}.$
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

(*) Let $\,f$ be any continuous function on $[0,1]$.
It is uniformly continuous because $[0,1]$ is compact.
So, given $\epsilon \ge 0$ there exists $\delta \ge 0$ such that
$|x-x'| \lt \delta \Rightarrow |\,f(x)-f(x')| \lt \epsilon$.
Now let $g$ be the piecewise linear function such that $g(x) = f(x)$
at $x=0,\delta,2\delta,3\delta,\ldots,N\delta$ (with
$N = \lfloor 1/\delta \rfloor$) and at $x=1$,
and is (affine) linear on $[N\delta,1]$ and on each
$[(i-1)\delta,i\delta]$ ($1 \leq 1 \leq N$). Exercise:
$|\,f(x)-g(x)| \lt \epsilon$ for all $x \in [0,1]$.
So we have uniformly approximated $\,f$ to within $\epsilon$
by the piecewise-linear continuous

Rudin’s notion of an “algebra” of functions is *almost*
a special case of what we called an
“algebra

In the first theorem of **Chapter 8**,
Rudin obtains the termwise differentiability of a power series at any
$x$ with $x \lt R$ by applying Theorem 7.17.
That’s nice, but we’ll want to use the same result
in other contexts, notably

An alternative derivation of formula (26) on p.179:
differentiate the power series (25) termwise (now that we know
it works also *w* the difference $E(w+z)-E(w)E(z)$
is an *analytic* function

In algebraic terms, identities (26) and (27) say that $E$
(that is, $\exp$) gives *group homomorphisms* from
$({\bf R}, +)$

Small **error in Rudin**: the argument on p.180 that
“Since $E$ [a.k.a. $\exp$] 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
$x \mapsto x^3.$ What’s the
correct statement? (Hint: the Chain Rule tells you what the
derivative of the inverse function must be.)

In any case, we have deliberately omitted the univariate
Inverse Function Theorem in anticipation of the multivariate setting where
the Inverse and Implicit Function Theorems are equivalent. However,
*if* there is a differentiable inverse function then
we known its derivative from the Chain Rule.
So if $L'(y)$ exists then it equals $1/y;$
this together with $L(1) = 0$ gives us the integral formula
$L(y) = \int_1^y dx/x$
(via the Fundamental Theorem of Calculus), and then we can *define*
$L(y)$ by this formula, and differentiate to prove that it is in fact
the inverse function

The same approach identifies $\tan^{-1}(y)$ with $\int_0^y dx/(x^2+1)$
once we have constructed the sine and cosine functions
(Rudin’s

- You can also check that $1 / (x^2+1)$ is $\left( 1/(x-i) - 1/(x+i) \right) / 2i,$ and that the corresponding linear combination of $\log (x \pm i)$ seems to agree with $\tan^{-1}(x)$ — though I don’t think we are quite in position yet to make rigorous sense of this route to $\int dx/(x^2+1)$.
- The power series for $\sin$, $\cos$, and $\tan^{-1},$
and the alternating series for $\pi/4$, long predate 19th-century calculus.
They are often named for Leibniz (1646–1716) or
James Gregory (1638–1675),
but were already known centuries earlier
— together with some computational applications, including
the evaluation of $\pi$ as $4 \cos^{-1}(0)$ —
to the mathematicians of the
Kerala school,
and “are believed to have been discovered by Madhava of Sangamagrama
(c. 1350 – c. 1425) ” according to the “Madhava series” page on Wikipedia.

As far as I can tell the final inequality “$\le 2$” in
Rudin’s (50) can just as easily be “$\le 1$”,
because if we have found a choice

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 ${\bf R}^n$
to ${\bf R}^m$,
but most of the discussion works equally well with the target space
${\bf R}^m$ replaced by an arbitrary normed vector *domain*
*continuous* linear map, or equivalently that its norm
$\| \, f' \| = \sum_{\|v\|=1} \left|\,f'(v)\right|$
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
*V*,
and without first doing the case $m=1.$ To do this,
first prove the result in the special case when each $D_j({\bf x})$
vanishes; then reduce to this case by subtracting from $f$
the linear map from ${\bf R}^n$

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)
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 $\bf f$
is continuously differentiable

[We have seen that even in dimension $1$ there can be a function
$\,f$ that is differentiable everywhere, and has $f'(0) \neq 0$,
but is not locally injective *and is nonzero* for all $x$
in a neighborhood __is__
injective by Rolle’s theorem, and then it does have an inverse
function with the expected derivative. Remarkably this generalizes
to higher dimensions (replacing

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 $\bf b$ given by formula (58) (p.225), can be done
more easily using the chain rule, since $\bf g$ has been
constructed as the composition of the following three functions:
first, send $\bf y$ to $({\bf 0}, {\bf y})$;
then, apply the inverse function ${\bf F}^{-1}$;
finally, project the resulting vector $({\bf x}, {\bf y})$

Here’s an approach to $D_{ij} = D{ji}$
that works for a ${\cal C}^2$ 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

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

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 ${\bf R}^n$
and proving the formula for change of variables (Theorem 10.9).
this will already hint why in general an integral over an

By induction on $n$, the map taking a continuous function $\,f$ on
a box $B = \{ x \in {\bf R}^n: \forall i, x_i \in [a_i,b_i] \}$
to its integral on $B$ is bounded linear map of norm equal to
the volume $\prod_{i=1}^n (b_i-a_i)$ of the box.
The application of Stone-Weierstrass that Rudin uses to derive
Fubini’s theorem (for continuous integrands on a box)
suggests the following generalization:
for *any* compact metric spaces $X_1,\ldots,X_n$, any continuous
$\, f: X_1 \times X_2 \times \cdots \times X_n \to {\bf R}$
can be uniformly approximated by linear combinations of functions
of the form $(x_1,x_2,\ldots,x_n) \mapsto
\, f_1(x_1) \; f_2(x_2) \cdots \, f_n(x_n)$
for continuous $\, f_i: X_i \to {\bf R}.$ The proof is much the same
as in the case of intervals $X_i = [a_i,b_i]$ that Rudin uses.

Rudin’s use of “compact support”
(top of page 247) doesn’t quite match the definition
(10.3, bottom of page 246): as defined there, the only continuous
function of compact support is zero! But all that is needed is that
the support is *contained* in a compact set (which is what
“compact support” actually means in practice), which by Heine-Borel
is equivalent to the assumption that the function has *bounded* support.

The “partition of unity” constructed in Theorem 10.8
works for compact subsets of any metric space, not just ${\bf R}^n$.
In ${\bf R}^n$, it can be done also with differentiable $\psi_i$,
or even ${\cal C}^\infty$ functions (but not analytic ones…),
by choosing differentiable or ${\cal C}^\infty$

Complex Analysis 1:

Outline of solutions and extensions for the complex analysis problems from
the 8th and 9th problem set

Having defined line integrals, we can deduce that if $f$ is analytic in
an open *any* closed path *convex*
region (so that we can consistently *not* required to be simple
(i.e., they may self-intersect). In fact this is true on any
*simply-connected* __not__ required to be complex analytic!(*) The point is that
since $\omega = f(z) \, dz$ is closed, so is its “pullback”
$T^*\omega$ (obtained by substituting for $dx$ and $dy$ the
total derivatives of the

(*) In fact any two connected and simply-connected
regions in $\bf C$ are related by an *analytic* 1:1 map,
unless exactly one of them is all of $\bf C$. But that is
a considerably harder theorem.

Now that we recognize the rectangular $\oint_{\partial R}$ as a special case of a contour integral, we can also recognize $\int_0^{\theta_0} f(Re^{i\theta}) \, d\theta$ as $\int_\gamma f(z) \, \frac{dz}{iz}$ where $\gamma$ is the circular arc from $R$ to $Re^{i\theta_0}$. In particular, the formula $f(a) = (2\pi)^{-1} \int_0^{2\pi} f(a + Re^{i\theta}) \, d\theta$ is tantamount to Cauchy’s integral formula $f(a) = (2\pi i)^{-1} \oint_\gamma f(z) \, \frac{dz}{z-a}$ for a circular contour $\gamma$ centered at $a$ (in each case $\,f$ must be analytic in a neighborhood of the corresponding circular disc). Likewise for our generalization where $a$ can be any point in the open disc, not necessarily its center.

An important application is the
Laurent series
of a function analytic in a neighborhood of an annulus
$\{ z \in {\bf C} : r \leq |z-a| \leq R \}$,
generalizing the power series expansion of an analytic function
in a disc. This time we find that if $r \lt |z_0-a| \lt R$ then
$$
f(z_0) = \frac1{2\pi i} \oint_{|z|=R} f(z) \, \frac{dz}{z-z_0}
- \frac1{2\pi i} \oint_{|z|=r} f(z) \, \frac{dz}{z-z_0}.
$$
The first integral is still $\sum_{n=0}^\infty c_n (z_0-a)^n$
where $c_n = (2\pi i)^{-1} \oint_{|z|=R} f(z) \, dz/(z-a)^{n+1},$
using the geometric series
$$
\frac 1{z-z_0} = \frac 1{(z-a)-(z_0-a)}
= \sum_{n=0}^\infty \frac{(z_0-a)^n}{(z-a)^{n+1}}
$$
uniformly convergent in compact subsets of the open annulus
(and indeed of the circle

Liouville’s theorem
soon follows: *every bounded entire function is constant*.
(An “entire function” is an analytic function
$\,f: {\bf C} \to {\bf C}$.) Write $f(z) = \sum_{n=0}^\infty a_n z^n.$
Since the domain is the entire complex plane,
we can apply the integral formula for $a_n$ with $R$ arbitrarily large.
This shows that $a_n = O(1/R^n),$ and thus $a_n = 0$ for each $n > 0.$
The hypothesis may seem very restrictive, but note that the
Fundamental Theorem of Algebra follows immediately on setting
$\,f(z) = 1/P(z)$ for a polynomial $P \in {\bf C}[z]$ with no complex roots:
$1/P,$ and thus $P,$ must be constant!

The same argument shows more generally that if an entire function grows
no faster than a polynomial then it *is* a polynomial;
more precisely, if for some $d$ we have constants $C,R_0$ such that
$|\,f(z)| \leq C |z|^d$ for all $z\in\bf C$ with $|z| \geq R_0,$
then $\,f$ is a polynomial of degree at

An important example is a *logarithmic derivative*
$(\log \, f)' = \,f'/f$ (better, a logarithmic differential

Now, since $2\pi i {\bf Z}$ is a discrete subset of ${\bf C},$
continuous changes

Problem sets 1 and 2: Metric topology basics

Problem set 3: Metric topology cont’d

Problem set 4: Topology finale; differential-calculus prelude

Problem set 5: More univariate differential calculus; introducing univariate integral calculus

Problem set 6: Riemann(-Stieltjes) integration cont’d

Problem set 7: Fourier series via Stone-Weierstrass;
power series; manipulating and estimating definite integrals to prove
some classical product and sum formulas

Typos in problems 1 (F.Flesher), 2 (C.J.Dowd), and 4,5 (T.Piazza)
**corrected**

Problem set 8:
Introduction to multivariate differentiation —
and to contour integration and complex analysis

Typo in problem 7 (D.Chiu) **corrected** 27.iii.2018

Problem set 9:
More complex analysis, and (counter)examples of multivariate real analysis

Typos in problem 5 and 9 (A.Sun) **corrected** 5.iv.2018

Problem set 10:
Integration in ${\bf R}^k$; more analysis in $\bf C$

Small error in problem 2 (J.Ahn) **corrected** 11.iv.2018;
typo in problem 3, and missing hypothesis a the end of problem 7
(both D.Xiang), **corrected** 15.iv.2018

Problem set 11:
Complex analysis cont’d:

definite integrals and other uses of residues; product formulas;
rational functions; variation on a theme of Jensen

Problem 11 **corrected** (S. Hu)