I tried to make a proof, where I use a Weierstrass function. I was surprised at how easy it was, and thus a little doubtful as to the correctness of the proof. I've looked it over, and didn't find any errors. So I post it here in case the good people here can verify my proof.

Heres the proof:

Consider the banach space $X=(C_b(\mathbb{R}), ||\cdot||_\infty)$, where $C_b(\mathbb{R})$ is the set of bounded continuous functions from a $\mathbb{R}$
to $\mathbb R $. We wish to prove that the nowhere differentiable continuous functions are dense in this banach space. \ Firstly we recall that K. Weierstrass proved that:\ \begin{equation*} \label{eq:weierstrass} W_{a,b}(x) = \sum_{n=0}^\infty a^n \cos(b^n\pi x) \end{equation*} is nowhere differentiable (and continuous) when $0<a<1$, $b$ positive odd integer and $ab > 1+ \frac{3}{2} \pi$. Specifically we have that $W_{\frac12,25}$ is nowhere differentiable and continuous. Furthermore we have that $|| W_{\frac12,25} ||_\infty \leq 2 < 3$.\

Now lets consider the set $N\subset C_b(\mathbb{R})$ of nowhere differentiable functions. Take some $f \in C_b(\mathbb{R})$. Denote by $D \subseteq \mathbb R$ the set of points where $f$ is differentiable. Now define $g_n$ by: \begin{equation*} g_n(x) = \begin{cases} f(x) + \frac{1}{n} W_{\frac12,25}(x), \quad x \in D\\ f(x) \quad x \in \mathbb R \setminus D. \end{cases} \end{equation*} Note that $g_n \in N$ since we added $W_{\frac12,25}$ to the differentiable parts of $f$ (if $g_n$ where differentiable at a point $x\in D$ then so would $g_n (x) -f(x) =\frac{1}{n}W_{\frac12,25}(x)$). Finally we see that: \begin{align*} ||g_n(x) -f(x)||_\infty \leq ||\frac{1}{n}W_{\frac12,25}||_\infty \leq \frac{3}{n} \to 0, \qquad n \to \infty. \end{align*} Hence $N$ is dense in $C_b(\mathbb R)$.

  • 1
    $\begingroup$ You're right to worry, this is wrong. Your $g_n$ is not continuous. $\endgroup$ – David C. Ullrich Jul 16 '16 at 13:14
  • $\begingroup$ Ofcourse! thanks $\endgroup$ – Martin Jul 16 '16 at 13:15
  • $\begingroup$ Well back to the Baire argument for me =D $\endgroup$ – Martin Jul 16 '16 at 13:17
  • $\begingroup$ As David already pointed out - it isn't in general. $\endgroup$ – Martin Jul 16 '16 at 13:18

Of course the given proof is wrong because $g_n$ is not continuous. But I was hugely amused to realize that we don't need that category argument; the given proof can be easily fixed, and the fix has a deliciously paradoxical flavor:

The fact that the nowhere-differentiable functions are dense is immediate from the fact that the differentiable functions are dense!

Indeed, say $f\in C_b$ and $\epsilon>0$. Choose a differentiable function $g$ with $$||f-g||<\epsilon.$$Then $g+\epsilon W$ is nowhere differentiable, and $$||f-(g+\epsilon W)||<3\epsilon.$$

Detail The fact that the differentiable functions are dense is slightly more problematic than for, say, $C_0$, because $f\in C_b$ need not be uniformly continuous, so the convolution with an approximate identity ("mollifier") need not converge to $f$ uniformly.

But we could use an approximate identity to write $$f=\sum_{n=-\infty}^\infty f_n,$$where $f_n$ is supported in $(n,n+2)$. (That sum converges uniformly on compact sets, but not in $C_b$.) Now convolution with an approximate identity gives us differentiable $g_n$ supported in $(n,n+2)$ such that $||f_n-g_n||<\epsilon$; now if $g=\sum g_n$ we have $||f-g||<2\epsilon$.

  • $\begingroup$ Haha - you where right. I did enjoy this argument. $\endgroup$ – Martin Jul 20 '16 at 14:24

Slightly different proof, using the fact that $\limsup_{y\to x}|W(y) - W(x)/(y-x)| = \infty$ for every $x.$ On each interval $[n,n+1]$ there is a polynomial $P_n$ that agrees with $f$ at the endpoints, and such that $|P_n-f|<\epsilon$ inside. For each $n\in \mathbb Z,$ define $g= P_n +\epsilon W$ on $[n,n+1].$ Then $g$ is well defined by the endpoint arrangements, $g \in C_b,$ and $g$ is not differentiable at each point of $\cup_{n\in \mathbb Z}(n,n+1).$ But also $g$ is not differentiable at any $n,$ because the relevant difference quotients of $P_{n-1}$ and $P_n$ are bounded there, while those of $\epsilon W$ are not. Thus $g$ is nowhere differentiable, and we have $\|g-f\|_\infty \le 3\epsilon.$

  • $\begingroup$ Very nice approach too. Thanks. $\endgroup$ – Martin Jul 20 '16 at 14:25

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.