7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Sep 17 '18 at 5:54
  • $\begingroup$ @KaviRamaMurthy $g$ fails the condition at $x$ but may still satisfy it at some $y.$ $\endgroup$ – zhw. Sep 20 '18 at 19:29
2
+50
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I'm not a native German speaker, but I think you mean a saw tooth function for 'Sägezahnfunktion'. $\endgroup$ – Nigel Overmars Sep 20 '18 at 19:43
  • $\begingroup$ Do you mean the function $ f(t) = t - \lfloor t \rfloor $? This is not meant because it has discontinueties. $\endgroup$ – Christian Sep 20 '18 at 19:57
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.