Show that there is a continuous real function $h$ on $[0,1]$ such that $\lim \sup |\frac{h(x+t)-h(x)}{t}| = \infty$ for all $x \in [0,1)$

The reason the authors go to $C[0,2]$ is to give you room to the right: For $x\in[0,1],$ we have $x+t\in [0,2]$ for all $t\in [0,1].$ Myself, to make it a litle simpler, I would define $E_m$ to be the subset of $f\in C[0,2]$ such that there exists an $x\in [0,1]$ such that

$$\left | \frac{f(x+t)-f(x)}{t}\right|\le m$$

for all $t\in (0,1].$

There are some problems with your proof that $E_m$ is closed. You fix $x$ and $t$ and then take limits. But there is no reason to think that all $f_n$ behave well at this one $x.$ Note also that you used only pointwise convergence, and not uniform convergence.

Here is a remedy: Suppose $f_n$ is a sequence in $E_m$ and $f_m\to f$ uniformly on $[0,2].$ Then for each $n$ there exists $x_n\in [0,1]$ such that

$$\tag 1\left | \frac{f_n(x_n+t)-f_n(x_n)}{t}\right|\le m,\,\,t\in (0,1].$$

Now we can assume $x_n$ converges to some $x_0,$ for this is true of some subsequence. Fix $t\in (0,1]$ and then use uniform convergence to see

$$\left | \frac{f_n(x_n+t)-f_n(x_n)}{t}\right| \to \left | \frac{f(x_0+t)-f(x_0)}{t}\right|$$

This shows $f\in E_m$ as desired.

Fix $m.$ Here's a sketch to show $E_m$ is nowhere dense in $C[0,2]:$ Define $g(x) = |x|$ on $[-1,1]$ and then extend $g$ to $\mathbb R$ by making $g$ periodic. (The function $g$ is your standard "accordion function".) Note that $g(kx)$ fails the $E_m$ condition at every $x\in [0,1]$ if $k>m.$

Let $p$ be a polynomial. Then $p(x)+g(k^2x)/k, k=1,2,\dots$ converges uniformly to $p.$ But $p$ is smooth and $g(k^2x)/k$ fails the $E_{k-1}$ condition at all $x.$ Letting $k\to \infty$ shows $p$ is the uniform limit of functions not in $E_m.$ Since the polynomials are dense in $C[0,2],$ $E_m$ is nowhere dense.

I think it is easier to consider the Banachspace $ C[0,2] $ with the usual Norm $ \| f \|_{\infty} = \sup_{x \in [0,2]} | f(x) | $. (You don't lose anything because convergence in this Norm is equivalent to uniform convergence.) But now the questions:

(1) As zhw. pointed out. It's wrong.

(2) Because you consider $ x \in [0,1] $ the quotient is only defined for functions on a slightly larger intervall.

(3) Let $ O_m $ be the complement of $ E_m $. Then: $$ O_m = {E_m}^C = \{ f \in C[0,2] | \forall x \in [0,2] \exists 0 < t \le 2-x: |\frac{f(x+t)-f(t)}{t}| > m \} = \{ f \in C[0,2] | \forall x \in [0,2]\sup_{0 < t \le 2-x} {|\frac{f(x+t)-f(t)}{t}|} > m\} $$

Now we show that $ O_m $ is dense. Therefore fix $ f \in C[0,2] $ and $ \epsilon > 0 $. By Weierstrass Approximation Theorem there exists a polynomial p with $ \| p - f \|_{\infty} < \frac{\epsilon}{2} $. Further let $ y_{\alpha} \in C[0,2] $ with:

• $ y_{\alpha} : [0,2] \rightarrow [0,\frac{\epsilon}{2}] $ is continous
• $ y_{\alpha}(0) = 0 $
• $ y_\alpha $ increases from 0 with constant slope $ \alpha $ until the value $ \frac{\epsilon}{2} $ is reached
• then $ y_\alpha $ decreases with constant slope $ - \alpha $ until the value $ 0 $ is reached
• $ y_\alpha $ is continoued periodiclly on $ [0,2] $

In Germany such a function is called "Sägezahnfunktion" but I don't know the English wort. Now let $ g_\alpha = p + y_\alpha $. Cleary $ \| f - g_\alpha \|_\infty< \epsilon $. If we manage to show $ g_\alpha \in O_m $ for one $ \alpha > 0 $ we have finished. Choose $ \alpha > m + \| p \|_\infty $. Then $$ | \frac{g_\alpha(x+t) - g_\alpha(x)}{t} | \ge | \frac{y_\alpha(x+t)-y_\alpha(x)}{t} | - | \frac{p(x+t)-p(x)}{t} |$$ by the reverse triangle inequality. By the mean value theorem there is a $ \xi $ with $ (p(x+t)-p(t)) \cdot t^{-1} = p(\xi) $ and we have $$ | \frac{g_\alpha(x+t) - g_\alpha(x)}{t} | \ge | \frac{y_\alpha(x+t)-y_\alpha(x)}{t} | - \| p \|_\infty $$ Taking the supreme yields $$ \sup_{{0 < t \le 2-x}} {| \frac{g_\alpha(x+t) - g_\alpha(x)}{t} |} \ge \sup_{{0 < t \le 2-x}} {| \frac{y_\alpha(x+t) - y_\alpha(x)}{t} |} - \| p \|_\infty \ge \alpha - \| p \|_\infty $$ where the last inequality comes from a similar mean value argument as above because you can take $ t $ so small that $ y_{\alpha} $ has constant slope on $ (x,x+t) $. We have showed that $ g_\alpha \in O_m $ if $ \alpha > m + \| p \|_\infty $. Therefore $ O_m $ is dense in $ C[0,2] $.

Especially we get $ \forall \epsilon > 0 \forall x \in E_m : B_\epsilon(x) \cap O_m \neq \emptyset $. Because $ O_m = {E_m}^C $ this can be restated as $ E_m = \overline{E_m} $ has no interior point which means that $ E_m $ is nowhere dense.