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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 $C_b(\mathbb R)$.

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    $\begingroup$ 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
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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 $||f-g||<2\epsilon$.

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  • $\begingroup$ Haha - you where right. I did enjoy this argument. $\endgroup$ – Martin Jul 20 '16 at 14:24
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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 $\|g-f\|_\infty \le 3\epsilon.$

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

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