Reading Notes, Theorems

Convergence of power series: Cauchy-Hadamard

The domain of convergence of a general holomorphic function can be quite arbitrary—in one (complex) dimension any domain is the domain of convergence for some holomorphic function, and in higher dimensions domains of holomorphy may be characterized using certain notions of convexity: holomorphic convexity, which involves the propagation of maximal-principle-type estimates, or pseudoconvexity, which is similar but uses the larger class of plurisubharmonic functions instead of holomorphic functions

The domain of convergence of functions defined by power series, however, are much more restricted and exhibit more symmetry: in one dimension, the domain of convergence of a power series f(z) = \sum_{n=0}^\infty c_n z^n is a disc of radius R satisfying \dfrac 1R = \limsup_{n \to \infty} \left( |c_n|^{1/n} \right). This we may prove by noting that, for |z| > R with R as above, the terms of the series do not converge to zero, and hence the series cannot converge; for |z| < R, the series will converge by comparison with a geometric series. This is the Cauchy-Hadamard theorem. (On the boundary of the disc of convergence the behavior of the power series is more subtle—we can extract some information using results such as Abel’s theorem.)

There is an analogue in higher dimensions, which states that (using multi-index notation) a power series f(z) = \sum_\alpha c_\alpha z^\alpha converges in a polydisc with polyradius \rho iff \limsup_{|\alpha| \to \infty} \sqrt[|\alpha|]{|c_\alpha|\rho^\alpha} \leq 1; the proof of this is exactly analogous to the one given above. Note that the union of all such polydiscs may not itself be a polydisc–e.g. the domain of convergence of the power series \sum_{k=0}^\infty z_1^k z_2^k is precisely the set \{|z_1z_2|| < 1\}, which is not a polydisc.

Nevertheless these domains of convergence still have a certain sort of radial symmetry—they belong to the more general sort of domains known as a complete (a.k.a. logarithmically-convex) Reinhardt domains. These are domains U for which (z_1, \dots, z_n) \in U implies (e^{i\theta_1} z_1, \dots, e^{i\theta_n} z_n) \in U for all \theta_1, \dots, \theta_n \in \mathbb{R} (this is the “Reinhardt” part) and (w_1, \dots, w_n) \in U whenever |w_i| \leq |z_i| (this is the complete, or log-convex part.) Note that one-dimensional complete Reinhardt domains are precisely discs centered at the origin.

We can think of the Cauchy-Hadamard condition in several variables as giving a constraint on the polyradius \rho, so that the domain of convergence \Omega is the union of all polydiscs with polyradius satisfying the given constraint or, equivalently, the logarithmic image \mathrm{Log}(\Omega) of the domain of convergence is the set of all polyradii satisfying the given constraint. \mathrm{Log}(\Omega) is convex (and closed to the right) but \Omega need not be an orthant, as would be the case if \Omega were a polydisc.



Version 1 (215)

Let K be a compact subset of \mathbb{R} and let Y \subset C(K) be an equicontinuous family of functions. Assume that f(x) is bounded for each fixed x \in K as f varies over Y. Then Y is pre-compact, i.e. every sequence in Y has a subsequence converging to some f \in C(K) (equivalently, the closure of Y in C(K) is compact.)

In other words

If \mathcal{F} is an equicontinuous and uniformly bounded family of continuous functions on \mathbb{R}, then any sequence drawn from \mathcal{F} has a subsequence which converges (uniformly on any compact set) to some continuous function.

In complex analytic terms

Any equicontinuous and uniformly bounded collection of continuous functions forms a normal family: this is the second half of Montel’s theorem (the first half states that a uniformly bounded family of continuous functions is equicontinuous.)

Version 2 (Tao)

Let Y be a metric space, be a compact metric space, and let (f_\alpha)_{\alpha \in A} {(f_\alpha)_{\alpha \in A}} be a family of bounded continuous functions from X to Y. The following are equivalent:

  • (i) \{f_\alpha \mid \alpha \in A\} is a precompact subset of the bounded continuous functions from X to Y.
  • (ii) (f_\alpha)_{\alpha \in A} is pointwise precompact (i.e. for every x \in X, the set \{f_\alpha(x) \mid \alpha \in A\} is precompact in Y) and equicontinuous.
  • (iii) (f_\alpha)_{\alpha \in A} is pointwise precompact and uniformly equicontinuous (= equicontinuous + uniform bound on size of neighborhoods (radius of balls))

Proof (Arzelà–Ascoli diagonalisation)

Now a standard technique in analysis, as first shown to you by Sarnak

Given \epsilon > 0, by equicontinuity of \mathcal{Y} there is a \delta > 0 such that \left| f(x) - f(y) \right| < \epsilon whenever $d_X(x,y) < \delta$.

Key observation: since K \subset \cup_{k \in K} B(k, \delta) is compact, there exists a finite subcover \cup_{j=1}^n B(p_j, \delta) \supset K.

Let M_j = \sup_{f \in \mathcal{Y}} \left| f(p_j) \right| < \infty and set M = \max \{M_j \:|\: j=1, \cdots, N\} < \infty.

Given x \in K, $x \in B(p_j, \delta)$ for some $j = 1, \cdots, N$; then $\left| f(x) – f(p_j) \right| < \epsilon$. Then

\[ \left|f(x)\right| < M_j + \epsilon \leq M + \epsilon \]

for all $x \in K$; hence $f$ is uniformly bounded as $f$ varies over $\mathcal{Y}$ and $x$ varies over $K$.

Next, let $E \subset K$ be a countable dense subset of $K$. We claim that any sequence $\{f_n\}\stdseq \subset \mathcal{Y}$ has a subsequence $\{f_{n_j}\}_{j=1}^{\infty}$ for which $\{f_{n_j}(e)\}_{j=1}^{\infty}$ converges for each $e \in E$.

This we prove using a diagonal argument (not quite Cantor’s.) Enumerate $E = \{e_1, e_2, \cdots \}$. Write out a table

\[ \begin{array}{l l l l l}

& f_1 & f_2 & f_3 & \cdots \\

e_1 & f_1(e_1) & f_2(e_1) & f_3(e_1) & \cdots \\

e_2 & f_1(e_2) & f_2(e_2) & f_3(e_2) & \cdots \\

e_3 & f_1(e_3) & f_2(e_3) & f_3(e_3) & \cdots \\

\vdots & \vdots & \vdots & \vdots & \ddots

\end{array} \]

First construct a family of sequences as follows: let $A_1$ be a convergent subsequence of the elements of the first row.

For $n > 1$, let $A’_n$ be the subsequence of the $n\thc$ row of elements picked out by the indices of $A_{n-1}$, and let $A_n$ be a convergent subsequence of $A’_n$.

At each step a convergent subsequence exists because $f \in \mathcal{Y}$ is uniformly bounded, and hence $f_i(e_j) \in S$ for all $i,j \in \nats$, where $S$ is some closed, bounded (hence compact) interval of reals.

Now define the sequence $\{f_{n_j}\}_{j=1}^{\infty}$ as follows: for each $j$, look at the $j\thc$ element of $A_j$, which is of the form $f_k(x_l)$ for some natural numbers $k,l$. Then $n_j := k$, i.e. $f_{n_j} = f_k$. By recursion, this subsequence converges pointwise for every $e_i \in E$.

Finally, we will show that the subsequence we have just constructed is convergent in $C(K)$, by showing that it is uniformly Cauchy, i.e. given any $\epsilon > 0$, there exists a $N \in \nats$ such that $\left| f_n(x) – f_m(x) \right| < \epsilon$ for all $x\in K$ whenever $n,m \geq N$.

We have the finite open cover $\cup_{j=1}^n B(p_j, \delta) \supset K$. For each $j = 1, \cdots, N$, $B(p_j, \delta)$ contains some element of $E$—call it $e’_j$—since $E$ is dense.

Note $\{f_n(e’_j)\}\stdseq$ is a Cauchy sequence for each fixed $j$, since we have already proven that $\{f_n(e)\}\stdseq$ converges for any $e \in E$. Hence there exists a natural number $R_j$ such that whenever $n,m \geq R_j$, we have $\left| f_n(e_j) – f_m(e_j) \right| < \epsilon$.

Pick $R = \max\{R_1, \cdots, R_N\}$. Then if $n,m \geq R$, we have $\left| f_n(e_j) – f_m(e_j) \right| < \epsilon$ for all $j=1, \cdots, N$.

Now, given any $x \in K$, $X \in B(p_j, \delta)$ for some natural number $j$. Then

\[ \left| f_n(x) – f_m(x) \right| \leq \left| f_n(x) – f_n(e’_j) \right| + \left| f_n(e’_j) – f_m(e’_j) \right| + \left| f_m(e’_j) – f_m(x) \right| < \epsilon \]

for all $n,m \geq R$.

There are also other proofs (e.g. using functional analysis, which I shall perhaps explore one day.)

In short

A way of obtaining convergence / compactness in function space from stronger continuity / boundedness assumptions.