# The Unapologetic Mathematician

## The Problem With Pointwise Convergence

I wrote this up yesterday between my sections of college algebra, but forgot to post it afterwards. Oops.

We’ve got a problem with the topology of pointwise convergence. The subspace of continuous functions isn’t closed. What does that mean? It means that if we take a sequence of continuous functions, their pointwise limit may not be continuous.

Here’s an example in the real numbers. Let $f_n(x)=\frac{x^{2n}}{1+x^{2n}}$, which is a sequence of well-defined continuous functions on the entire real line. But if we take the pointwise limit $f(x)=\lim\limits_{n\rightarrow\infty}f_n(x)$ we find that $f(x)=0$ for $|x|<1$, that $f(x)=1$ for $|x|>1$, and that $f(x)=\frac{1}{2}$ for $x=\pm1$. So the functions in the sequence are continuous at $x=\pm1$, but the limiting function isn’t. It would be one thing if the sequence just failed to converge at some points — closedness doesn’t require all sequences to converge — but the pointwise limit clearly exists, and it fails to be continuous.

What we need is a stronger sense of convergence: one in which fewer sequences converge in the first place, and hopefully one in which the continuous functions turn out to be closed. But it should also obey the same definition as that of the pointwise limit when it does exist. And to find it we’ll need to recast the question of continuity in the limit.

Remember that a function is continuous at a point $x_0$ if it agrees with its limit there. That is, if $\lim\limits_{x\rightarrow x_0}f(x)=f(x_0)$. But the function $f$ should be the pointwise limit of the sequence $f_n$: $f(x)=\lim\limits_{n\rightarrow\infty}f_n(x)$. And each of these functions is continuous: $\lim\limits_{x\rightarrow x_0}f_n(x)=f_n(x_0)$. Putting these together, the condition for continuity in the limit is

$\lim\limits_{x\rightarrow x_0}\lim\limits_{n\rightarrow\infty}f_n(x)=\lim\limits_{n\rightarrow\infty}\lim\limits_{x\rightarrow x_0}f_n(x)$.

So our question is really about when we can exchange limits. For which sequences of functions do the dependence on $x$ and that on $n$ play well enough together to allow these limits to be exchanged? We’ll answer that question tomorrow.

September 4, 2008 - Posted by | Analysis, Functional Analysis