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.

  • $\begingroup$ 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. $\endgroup$ Sep 17, 2018 at 5:54
  • $\begingroup$ @KaviRamaMurthy $g$ fails the condition at $x$ but may still satisfy it at some $y.$ $\endgroup$
    – zhw.
    Sep 20, 2018 at 19:29
  • $\begingroup$ Baire's theorem shows that the union of Em's nowhere dense functions is also nowhere dense, leaving the rest of C([0,2]) functions, which are not bounded. That C is a complete space comes from a Cauchy Sequence of continuous functions on C comes from a sequence of functions that is Cauchy at every point of [0,2] and is therefore uniformly convergent on that real, closed interval. $\endgroup$
    – goedelite
    May 28, 2021 at 3:04

2 Answers 2


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.

  • $\begingroup$ Thank you. This was really helpful. $\endgroup$
    – Henry Choi
    Sep 8, 2020 at 14:36

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.

  • $\begingroup$ I'm not a native German speaker, but I think you mean a saw tooth function for 'Sägezahnfunktion'. $\endgroup$ Sep 20, 2018 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, 2018 at 19:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .