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Advanced Calculus, part I Analysis and Topology in Normed Spaces

Andrew Cooper

Section 8.1 Uniform Convergence

First, let’s see why pointwise convergence isn’t as awesome as we might like it to be:
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Checkpoint 8.1.1.

Consider the functions \(p_n:[0,1]\to\mathbb{R}\) given by \(p_n(x)=x^n\text{.}\) What is the limit as \(n\to\infty\text{?}\)
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Make a list of all the nice properties that the \(p_n\) have. (For example, they’re all polynomials.) Which of these properties does the limit have?
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Checkpoint 8.1.2.

Consider the functions \(g_n:[0,1]\to\mathbb{R}\) given by
\begin{equation*} g_n(x)=\begin{cases}0&\text{ if }x\lt \frac{1}{n} \\ n^2\left(x-\frac{1}{n}\right)&\text{ if }\frac{1}{n}\leq x\lt\frac{2}{n}\\ -n^2\left(x-\frac{3}{n}\right)& \text{ if }\frac{2}{n}\leq x\leq \frac{3}{n}\\ 0& \text{ if }x\geq \frac{3}{n}\end{cases} \end{equation*}
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  1. Sketch a graph of \(g_n\text{.}\)
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  2. What is \(\displaystyle \int_{[0,1]}g_n(x)\ dx\text{?}\)
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  3. What is \(\displaystyle\lim_{n\to\infty}g_n\left(0\right)\text{?}\)
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  4. Pick an arbitrary \(0\lt x\lt 1\text{.}\) What is \(\displaystyle\lim_{n\to\infty}g_n\left(x\right)\text{?}\)
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So we’re going to develop another useful notion of convergence for sequences of functions.
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Checkpoint 8.1.3.

Write out the definition of pointwise convergence using quantifiers.
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Definition 8.1.4.

We say that sequence \(f_n:A\to V\) of functions converges uniformly to \(f:A\to V\) if for each \(\epsilon\gt 0\text{,}\) there is \(N\in\mathbb{N}\) so that \(n\gt N\) guarantees \(\forall x,\ d(f_n(x),f(x))\lt \epsilon\text{.}\)
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We write \(f_n\rightrightarrows f\text{.}\)
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Checkpoint 8.1.5.

Why is this "uniform" convergence?
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Uniform convergence behaves quite a bit like convergence of sequences in a normed space. For example:
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Let’s see which other parts of the theory of convergence carry through to uniform convergence.
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Definition 8.1.8.

We call a sequence of functions \(f_n:A\to V\) uniformly Cauchy if for each \(\epsilon\gt 0\) there is \(N\in\mathbb{N}\) so that \(m,n\gt N\) guarantees \(\forall x\in A, d(f_n(x),f_m(x))\lt \epsilon\text{.}\)
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Proof.

Notice that for each \(x\in V\text{,}\) \((f_n(x))_{n\in\mathbb{N}}\) is a Cauchy sequence in \(V\text{.}\) Therefore \(\left(f_n(x)\right)_{n\in\mathbb{N}}\) converges to some element of \(V\text{,}\) which we’ll call \(f(x)\text{.}\)
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Our job now is to prove that the convergence \(f_n\to f\) is in fact uniform.
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Given \(\epsilon\gt 0\text{,}\) there is \(N\in\mathbb{N}\) so that \(m,n\gt N\) guarantees \(d\left(f_n(x),f_m(x)\right)\rVert\lt \frac{\epsilon}{2}\text{.}\) For any \(m,n\gt N\text{,}\) we have
\begin{equation*} d\left(f(x),f_n(x)\right)\leq d\left(f(x),f_m(x)\right)+d\left(f_m(x),f_n(x)\right)\lt \frac{\epsilon}{2}\ \ \ . \end{equation*}
Now fixing \(x\text{,}\) we can take the limit \(m\to\infty\text{;}\) then \(d\left(f(x),f_n(x)\right)\leq \frac{\epsilon}{2}\lt \epsilon\text{.}\)
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Because \(N\) does not depend on \(x\text{,}\) this shows that \(f_n\rightrightarrows f\text{.}\)
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What’s going on here?
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Definition 8.1.10.

The set
\begin{equation*} B(A;V)=\left\{f:A\to V\middle\vert \exists B:\forall x\in A, \lVert f(x)\rVert\leq B\right\} \end{equation*}
is a vector space. If we equip it with the norm
\begin{equation*} \displaystyle\lVert f\rVert_A=\sup_{x\in A}\left\lVert f(x)\right\rVert \end{equation*}
then we obtain a normed space (which we will also call \(B(A;V)\)).
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Checkpoint 8.1.13.

Proof.

Here’s an outline: for arbitrary \(c\in A\text{,}\) we need to show continuity at \(c\text{.}\) Given \(\epsilon\gt 0\text{,}\)
\begin{equation*} d(f(c),f(x))\leq d(f(c),f_n(c))+d(f_n(c),f_n(x))+d(f_n(x),f(x))\ \ \ . \end{equation*}
Now, we need to make all of that less than \(\epsilon\) by controlling \(d(x,c)\text{.}\)
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