Uniform Convergence of Difference Quotient to Derivative using Compactness

Question

I have a question about a proof to a question in another discussion. This is regarding the uniform convergence of the difference quotient to the derivative of a function.

Here is a link to the discussion. The question in the discussion was:

Let $f:[0,1] \to \mathbb{R}$ be continuously differentiable. Prove that, for every $\epsilon > 0$, there exists $\delta > 0$ such that $0 < |h| < \delta$ implies $\left| \frac{f(x+h) - f(x)}{h} - f'(x) \right| < \epsilon$ for all appropriate $x$.

And the answer provided by Hagen von Eitzen was:

Let $\epsilon>0$ be given. For $\delta>0$ let $$U_\delta = \left\{a\in [0,1]\colon 0<|h|<\delta\Rightarrow \left|\frac{f(a+h)-f(a)}h-f'(a)\right|<\epsilon\right\}$$ Clearly, $\delta<\delta'$ implies $U_{\delta'}\subseteq U_\delta$. By continuity of $f$ and $f'$, $U_\delta$ is open and by definition of $f'$, $$[0,1]=\bigcup _{\delta>0}U_\delta.$$ Since $[0,1]$ is compact, there is a finite subcover, i.e. there is a single $\delta>0$ such that $[0,1]=U_\delta$.

My question is: why is $U_{\delta}$ open? I tried creating some examples and this does appear to hold for those examples, but I couldn't see how to extend it to a general proof. I do understand how all of the other parts in the proof work though.

It would also be appreciated if the use of the mean value theorem can be avoided as this was requested in the discussion when the question was asked. However, if no other reasonable approach is known to show that $U_{\delta}$ is open, then the use of the mean value theorem or other similar theorems is welcomed.

Answer 1

It turns out that in general, $U_{\delta}$ is not open.

For a counterexample, consider $f:[0,1] \to \mathbb{R}$ defined by $f(x) := x^{3}$. Set $\varepsilon := 1/2$ and $\delta := 1/\sqrt{2}$. Then

$$U_{\delta} = \left\{ x\in [0,1]: 0 < |h| < \frac{1}{\sqrt{2}} \;\text{and}\; x+h \in [0,1] \;\text{implies}\; \left| \frac{f(x+h) - f(x)}{h} - f'(x) \right| < \frac{1}{2} \right\}.$$

It is first shown that $0\in U_{\delta}$. Note that if $0 < |h| < 1/\sqrt{2}$ and $0+h\in [0,1]$, then $h\in (0,1/\sqrt{2})$. For each $h\in (0,1/\sqrt{2})$,

$$\left| \frac{f(0+h) - f(0)}{h} - f'(0) \right| = \left| \frac{h^{3} - 0}{h} - 0 \right| = |h^{2}| = h^{2} < \frac{1}{2}.$$

This shows that $0\in U_{\delta}$. It will now be shown that $0$ is not an interior point of $U_{\delta}$, which will show that $U_{\delta}$ is not open.

Fix $x\in (0,1-1/\sqrt{2})$. If $h\in (0,1/\sqrt{2})$, then

$$\begin{align*} \left| \frac{f(x+h) - f(x)}{h} - f'(x) \right| &= \left| \frac{(x+h)^{3} - x^{3}}{h} - 3x^{2} \right| \\ &= \left| \frac{x^{3} + 3x^{2}h + 3xh^{2} + h^{3} - x^{3}}{h} - 3x^{2} \right| \\ &= \left| \frac{3x^{2}h + 3xh^{2} + h^{3}}{h} - 3x^{2} \right| \\ &= \left| 3x^{2} + 3xh + h^{2} - 3x^{2} \right| \\ &= \left| 3xh + h^{2} \right| \\ &= 3xh + h^{2}. \\ \end{align*}$$

Consider the function $\phi : [0,1] \to \mathbb{R}$ defined by $\phi (t) := 3xt + t^{2}$. $\phi$ is continuous and as $x>0$,

$$\phi (1/\sqrt{2}) = \frac{3x}{\sqrt{2}} + \frac{1}{2} > \frac{1}{2}.$$

By the continuity of $\phi$, there exists some $h\in (0,1/\sqrt{2})$ such that $\phi (h) > 1/2$. But as

$$\left| \frac{f(x+h) - f(x)}{h} - f'(x) \right| = \phi (h),$$

it follows that

$$\left| \frac{f(x+h) - f(x)}{h} - f'(x) \right| > 1/2$$

and $x \not\in U_{\delta}$. As this holds for all $x\in (0,1/\sqrt{2})$, there does not exist an $\varepsilon >0$ such that $[0,\varepsilon ) \subseteq U_{\delta}$. This shows that $0$ is not an interior point of $U_{\delta}$, and $U_{\delta}$ is not an open subset of $[0,1]$.