Analysis, Part III: Uniform Functions on Metric Spaces

September 27th, 2017

We know that the following holds for sequences in a metric space, if $f$ is continuous:

$$ \lim\limits_{n \rightarrow \infty} f(x_n) = f(\lim\limits_{n \rightarrow \infty} x_n) $$

The weakest form of convergence is pointwise convergence.

Definition: $f_n \rightarrow f$ pointwise if for every $x$ and for every $\epsilon > 0$, there exists $N$ so that if $n \geq N$ then $d(f_n(x), f(x)) < \epsilon$.

But this definition does not allow us to swap the order of limits.

A stronger condition, however, does allow this:

Definition: $f_n \rightarrow f$ pointwise if for every $\epsilon > 0$, there exists $N$ so that for all $x$, if $n \geq N$ then $d(f_n(x), f(x)) < \epsilon$.

The key distinction here is that the value of $N$ works for all $x$. As alluded to earlier, we have the following proposition.

Proposition

A sequence of continuous functions converges uniformly to a continuous function.

Corollary

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

We also have this nice theorem:

Proposition

Let $f_n$ be a sequence of continuous functions which converge uniformly to $f$. Then for any sequence $x_n$ which converges to $x$, $f_n(x_n)$ converges to $f(x)$.

Similarly, uniform limits preserve boundedness in a nice way.

Proposition

A sequence of bounded functions converges uniformly to a bounded function.

Metrics of Uniform Convergence

Definition: Let $X,Y$ be metric spaces. Let $B(X\rightarrow Y)$ denote the space of bounded functions from $X \rightarrow Y$. We let:

$$ d_\infty (f,g) = \sup\limits_{x\in X} d(f(x), g(x)) $$

This is called the $L^\infty$ metric or the “sup norm” metric. Since $f,g$ are assumed bounded in this space, $d(f,g)$ is bounded.

This space with this norm has the nice property that convergence in this space directly correlates with uniform convergence in the regular sense.

Proposition

Let $f_n$ be a sequence of functions in $B(X\rightarrow Y)$; then $f_n$ converges to $f \in B(X\rightarrow Y)$ if and only if $f_n$ converges to $f$ uniformly.

In particular, we define the continuous and bounded functions.

Definition: $C(X\rightarrow Y)$ is the space of bounded and continuous functions from $X$ to $Y$. This a closed subspace of $B(X\rightarrow Y)$

And finally, we have Cauchy sequences:

Proposition

If $Y$ is a complete metric space, then the space $C(X\rightarrow Y)$ is a complete subspace of $B(X \rightarrow Y)$.

The Weierstrass M-Test

Definition: A series of functions $f_n$ converges pointwise to $f(x)$ if the partial sums converge pointwise to $f(x)$

In a similar way we have uniform convergence:

Definition: A series of functions $f_n$ converges uniformly to $f(x)$ if the partial sums converge uniformly to $f(x)$

We can find an easy condition for uniform convergence using the following norm:

Definition: If $f: X \rightarrow \mathbb{R}$ is a bounded function, define the sup norm to be the number:

$$ |f|_\infty = \sup\limits_{x\in X} f(x) $$

And finally we have the immensely useful Weierstrass $M$-Test:

Theorem

Let $f_n$ be a series of bounded real-valued functions such that $\sum\limits_{n=1}^\infty \lvert f_n\rvert_\infty$ is convergent. Then the series $\sum\limits_{n=1}^\infty f_n$ converges uniformly to a continuous function $f$ on $X$.

Uniform Convergence & Integration/Differentation

Theorem

Let $[a,b]$ be an interval and $f_n$ Riemann integrable functions. Suppose $f_n$ converges uniformly on $[a,b]$ to a function $f$. Then $f$ is also Riemann integrable and:

$$ \lim\limits_{n\rightarrow \infty} \int_{[a,b]} f_n = \int_{[a,b]} \lim\limits_{n\rightarrow \infty} f_n = \int_{[a,b]} f $$

An analogy of this theorem exists for series.

Corollary

Let $[a,b]$ be a an interval. Let $f_n$ be a sequence of Riemmann integrable functions on $[a,b]$ such that the series $\sum\limits_{n=1}^\infty f_n$ is uniformly convergent. Then we can say:

$$ \sum\limits_{n=1}^\infty \int_{[a,b]} f_n = \int_{[a,b]}\sum\limits_{n=1}^\infty f_n $$

Derivatives and Uniform Convergence

We ask if the same is true for derivatives. If $f_n$ converges uniformly to $f$, then if we require $f_n$ to be differentiable is $f$ differentiable? And do $f_n’$ converge to $f’$? In general, the answer is no.

As a counterexample consider:

$$ f_n(x) = n^{1/2}\cos nx $$

This sequences converges uniformly to $0$ if we look at the sup norm. However, its derivative at the origin is never $0$. Thus, $f_n’$ need not converge to $f’$.

Similarly, consider the sequence:

$$ f_n(x) = \sqrt{x^2+ \frac{1}{n^2}} $$

We can check that this sequence converges uniformly to $\lvert x \rvert$, which is not differentiable at the origin. This, however, is the theorem we’re looking for – it gives us more or less the converse, so long as we have pointwise convergence at least at one point.

Theorem

Let $[a,b]$ be an interval and $f_n$ be differentiable functions with continuous derivative on the interval. Suppose the derivatives $f_n’$ converge uniformly to a function $g: [a,b] \rightarrow \mathbb{R}$.

Suppose also that there exists a point $x_0$ so that the limit $\lim\limits_{n\rightarrow \infty} f_n(x_0)$ exists (i.e. we have pointwise convergence at some point). Then the functions $f_n$ converge uniformly to a differentiable function $f$ such that the derivative $f’ = g$.

Corollary

Let $[a,b]$ be an interval and $f_n$ be differentiable functions with continuous derivative on the interval. Suppose the series $\sum\limits_{n=1}^\infty \lvert f_n’ \rvert$ converges absolutely.

Suppose also that there exists a point $x_0$ so that $\sum\limits_{n=1}^\infty f_n(x_0)$ converges. Then the series $\sum\limits_{n\rightarrow \infty} f_n$ converges uniformly on $[a,b]$ to a differentiable function $f$ such that:

$$ \frac{d}{dx}\sum\limits_{n\rightarrow \infty} f_n(x)=\sum\limits_{n\rightarrow \infty} \frac{d}{dx} f_n(x) $$

Example

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be the function:

$$f(x)=\sum\limits_{n=1}^\infty 4^{-n} \cos (32^n \pi x) $$

This series is uniformly convergent (Weierstrass M-Test) and continuous, but nowhere differentiable.

Download this Page as a PDF:

Note: Images are replaced by captions.

Link to PDF

Recent Posts