Friday, March 22, 2013 CC-BY-NC
Interchange of limits and differentiability

Maintainer: admin

Recall the theorem from the end of last class:

Theorem: Let $J \subseteq \mathbb R$ be a bounded interval and $(f_n)$ be a sequence of functions from $J$ to $\mathbb R$. Suppose there exists some $x_0 \in J$ such that $f_n(x_0)$ converges and suppose that $(f_n')$ exists on $J$ and converges uniformly on $J$ to $g$. Then $(f_n)$ converges uniformly on $J$ to a function $f$ that is differentiable and $f' = g$.

Proof: Let $J = [a,b]$ and let $x \in J$ for $m, n \in \mathbb N$, by MVT applied to $f_m - f_n$ on interval with endpoints $x$ and $x_0$ (i.e. $[x,x_0]$ or $[x_0, x]$) then there exists a $y$ in that interval such that

$$ \frac{f_m(x) - f_n(x) - (f_m(x_0) - f_n(x_0))}{x-x_0} = f_m'(y) - f_n'(y) $$

i.e. $f_m(x) - f_n(x) = f_m(x_0) - f_n(x_0) + (x-x_0)[f_m'(y) - f_n'(y)]$ thus $\| f_m - f_n \|_J \leq |f_m(x_0) - f_n(x_0)| + (b-a)\| f_m' - f_n' \|_J$. Since $f_n(x_0)$ converges and $f_n'$ converges uniformly, then $f_n$ converges uniformly on $J$. Consequently $f_n \rightrightarrows f$ and $f_n$ are continuous as they are differentiable, thus $f$ is continuous on $J$.

Let $c \in J$. We will show that $f$ is differentiable at $c$. Again apply MVT to $f_m - f_n$ on interval with endpoints $c$ and $x \in J$ to get a $z$ in that interval such that

$$ \frac{f_m(c) - f_n(x) - (f_m(c) - f_n(c))}{x-c} = f_m'(z) - f_n'(z) $$

i.e. for $x \neq c$,

$$ \left|\frac{f_m(x) - f_m(c)}{x-c} - \frac{f_n(x) - f_n(x)}{x-c}\right| \leq \| f_m' - f_n' \|_J $$

since $f_n'$ converges uniformly on $J$, thus for $\epsilon > 0$ given there exists some $N \in \mathbb N$ such that if $m, n \geq N$ then $\| f_m' - f_n'\|_J < \frac{\epsilon}{3}$ thus if $n, m \geq N$ and $x \neq c$,

$$ \left| \frac{f_m(x) - f_m(c)}{x-c} - \frac{f_n(x) - f_n(c)}{x-c} \right| < \frac{\epsilon}{3} $$

Now take the limit as $m \to \infty$ to get

$$ \left| \frac{f(x) - f(c)}{x-c} - \frac{f_n(x) - f_n(c)}{x-c} \right| \leq \frac{\epsilon}{3} $$

by squeeze theorem. Also $f_n'(c) \to g(c)$ thus there exists some $N_1 \in \mathbb N$ such that if $n \geq N_1$,

$$ |f_n'(c) - g(c)| < \frac{\epsilon}{3} $$

Let $k = \sup\{N, N_1\}$, $f_k'(c)$ exists so for a given $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x-c| < \delta$ then

$$ \left|\frac{f_k(x) - f_k(c)}{x-c} - f_k'(c)\right| < \frac{\epsilon}{3} $$

Thus if $0 < |x-c| < \delta$,

$$ \left|\frac{f(x) - f(c)}{x-c} - g(c)\right| = \left|\frac{f(x) - f(c)}{x-c} - \frac{f_k(x) - f_k(c)}{x-c} + \frac{f_k(x) - f_k(c)}{x-c} - \frac{f(x) - f(c)}{x-c} + \frac{f(x) - f(c)}{x-c} - g(c)\right| $$

which we can split up into the three obvious parts with the triangle inequality, so that the whole statement is less than $\frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$, i.e. $f$ is differentiable at $c$ and $f'(c) = g(c)$. As $c \in J$ is arbitrary we conclude that $f$ is differentiable on $J$ and $f' = g$, i.e. $\lim f_n' = (\lim f_n)'$

We will now use this theorem to prove the existence of $e^x$, $\sin x$, $\cos x$...

What is so important about $e^x$? Well, $(e^x)' = e^x$ and $(ke^x)' = ke^x$ for all $x \in \mathbb R$.

Theorem: There exists a function $E(x) : \mathbb R \to \mathbb R$ such that

i) $E'(x) = E(x)$ for all $x \in \mathbb R$
ii) $E(0) = 1$

Proof: Well, at least the idea behind the proof. The idea is that we're going to construct a sequence of functions with a nice property, and we're gonna show that the sequence satisfies the properties of the previous theorem. What properties? $E_n' = E_{n-1}$, for instance. That is,

$$ E_n(x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \ldots + \frac{x^n}{n!} $$

from which it is easy to see that $E_n' = E_{n-1}$. We'll so similar things with $\sin x$ and $\cos x$.