# 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 $x \in [0,1)$

From Gamelin and Greene's Introduction to Topology, 2nd edition, chapter 1 section 6 (Continuity):

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

The book has a Hint: Consider the space $C([0,2])$ of continuous real valued functions on the interval $[0,2]$, with the metric of uniform convergence. Let $E_m$ be the set of $f \in C([0,2])$ for which there is $x \in [0,1]$ satisfying $|f(x+t) - f(x)|/|t| \leq m$ for $t$ positive, $x+t \leq 2$. Show that $E_m$ is a closed nowhere-dense subset of $C([0,2])$.

I am sure that the very last part of the question is hinting at the use of Baire category theorem, so it thus suffices to show $E_m$ is closed and nowhere dense (as any continuous function with compact preimage has compact image, and the set of all bounded functions from a compact subset of the reals are complete). I believe that I got the closed part right by the following argument (correctme if I am wrong): given $\{ f_n \}$ a sequence of functions in $C([0,2])$ converging to some $f$ in the metric of uniform convergence, we know that $f_n(x) \to f(x)$ uniformly. Then, for any given $\epsilon >0$ we can bound the fraction appearing in $E_m$ by the following, by finding sufficiently large $n$: $\frac{|f(x+t) - f(x)|}{t} \leq \frac{|f(x+t) - f_n(x+t)| + |f_n(x+t) - f_n(x)| + |f_n(x) - f(x)|}{t} \leq \frac{2\epsilon}{t} + m$, for any fixed $t$ and $x$ of our concern. By letting $\epsilon \to 0$ we get that $f \in E_m$.

I do not get: (1) if my work so far is correct, (2) where the $2$ comes into play in $C([0,2])$, and (3) how to show the nowhere dense part.

• Partial answer: to show that the set is nowhere dense take any $f$ in it. Then there exists $x$ such that $\frac {|f(x+t_-f(x)|} {|t|} \leq m$ if $x+t \leq 2$. Conider the function $g(x)=f(x)+\epsilon \sqrt {|y-x|}$ where $y$ is chosen such that $\frac {\epsilon} {2\sqrt {|y-x|}} >m$. You can now check that $g$ does not belong to the set. However, $g \to f$ uniformly as $\epsilon \to 0$. This proves that our set has no interior points. Since it is closed, it is nowhere dense. – Kavi Rama Murthy Sep 17 '18 at 5:54
• @KaviRamaMurthy $g$ fails the condition at $x$ but may still satisfy it at some $y.$ – zhw. Sep 20 '18 at 19:29

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.

• I'm not a native German speaker, but I think you mean a saw tooth function for 'Sägezahnfunktion'. – Nigel Overmars Sep 20 '18 at 19:43
• Do you mean the function $f(t) = t - \lfloor t \rfloor$? This is not meant because it has discontinueties. – Christian Sep 20 '18 at 19:57

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.