Uniform Convergence and Differentiation

In general, the uniform convergence of $\{f_n\}$ implies nothing about the sequence $\{f_n^{\prime }\}$. This means require stronger assertions to show that $f_n^{\prime }\to f^{\prime }$ if $f_n\to f$.

Theorem (Term-by-Term Differentiation)

Suppose $\{f_n\}$ is a sequence of functions,

• differentiable on $[a,b]$ and
• such that $\{f_n(x_0)\}$ converges for some point $x_0$ on $[a,b]$.

If $\{f_n^{\prime }\}$ converges uniformly on $[a,b]$, then $\{f_n\}$ converges uniformly on $[a,b]$, to a function $f$, and

$$f^{\prime }(x)=\underset{n\to \infty }{\lim }{f}_n^{\prime }(x)(a\le x\le b)$$

Proof: Let $\epsilon >0$ be given. Choose $N$ such that $n,m\ge N$, implies

$$|f_n(x_0)-f_m(x_0)|<\frac{\epsilon }{2}$$

since $\{f_n\}$ converges uniformly and

$$|f_n^{\prime }(t)-f_m^{\prime }(t)|<\frac{\epsilon }{2(b-a)}(a\le t\le b)$$

since $\{f_n^{\prime }\}$ converges uniformly. Then using MVT, to $f_n-f_m$, we get

$$|[f_n(x)-f_m(x)]-[f_n(t)+f_m(t)]|\le \frac{|x-t|\epsilon }{2(b-a)}\le \frac{\epsilon }{2}$$

for any $x$ and $t$ on $[a,b]$, if $n,m\ge N$. By the triangle inequality, we get

$$\begin{array}{rl}|f_n(x)-f_m(x)|& \le |f_n(x)-f_m(x)-f_n(x_0)+f_m(x_0)|+|f_n(x_0)-f_m(x_0)|\\ & <\epsilon /2+\epsilon /2\\ & =\epsilon \end{array}$$

implies that

$$|f_n(x)-f_m(x)|<\epsilon$$

and so $\{f_n\}$ converges uniformly on $[a,b]$.

This is the same proof technique discussed in Arzela - Ascoli Theorem where you can get to your desired outcome by the triangle inequality and going step by step.

For the second part, let

$$f(x)=\underset{n\to \infty }{\lim }{f}_n(x)(a\le x\le b)$$

Fixing $x\in [a,b]$, define

$${\varphi }_n(t)=\frac{f_n(t)-f_n(x)}{t-x}\varphi (t)=\frac{f(t)-f(x)}{t-x}$$

for $a\le t\le b$ and $t\ne x$. Then

$$\underset{n\to \infty }{\lim }{\varphi }_n(t)=f_n^{\prime }(t)$$

We see from Part 1, using MVT non $f_n-f_m$, we can apply it to $\varphi$ such that

$$|{\varphi }_n(t)-{\varphi }_m(t)|\le \frac{\epsilon }{2(b-a)}$$

so that $\{{\varphi }_n\}$ converges uniformly, for $t\ne x$. Since $f_n\to f$, we conclude that

$$\underset{n\to \infty }{\lim }{\varphi }_n(t)=\varphi (t)$$

uniformly by the Cauchy Criterion. By applying uniform convergence and continuity, we have that

$$f^{\prime }(x)=\underset{t\to x}{\lim }\varphi (t)=\underset{n\to \infty }{\lim }{f}_n^{\prime }(x)$$

which is the definition of a derivative.

If continuity of $f_n^{\prime }$ is also assumed, the proof is shorted by using Fundamental Theorem of Calculus and the Riemann-Stieltjes Integral.

Theorem (Continuity Does Not Imply Differentiable)

There exists a real continuous function on the real line that is nowhere differentiable.

It appears that the main motivation for uniform convergence shows that such an $f$ is uniformly convergent, and thus continuous. But then it means that it can be nowhere differentiable because there is not strong enough assumptions to show so.