A Hölder continuous function which does not belong to any Sobolev space - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T04:33:23Z http://mathoverflow.net/feeds/question/38751 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38751/a-holder-continuous-function-which-does-not-belong-to-any-sobolev-space A Hölder continuous function which does not belong to any Sobolev space Dorian 2010-09-15T00:24:19Z 2012-12-05T21:47:04Z <p>I'm seeking a function which is Hölder continuous but does not belong to any Sobolev space.</p> <p><strong>Question:</strong> More precisely, I'm searching for a function $u$ which is in $C^{0,\gamma}(\Omega)$ for $\gamma \in (0,1)$ and $\Omega$ a bounded set such that $u \notin W_{loc}^{1,p}(\Omega)$ for any $1 \leq p \leq \infty$. Take $\Omega$ to be bounded, open.</p> <p>My first guess is to do a construction with a Weierstrass function. I know this is differentiable 'nowhere' but that doesn't convince me it isn't weakly differentiable in some bizarre way. Hopefully someone knows of an explicit example.</p> http://mathoverflow.net/questions/38751/a-holder-continuous-function-which-does-not-belong-to-any-sobolev-space/38791#38791 Answer by Pietro Majer for A Hölder continuous function which does not belong to any Sobolev space Pietro Majer 2010-09-15T08:06:44Z 2012-12-05T21:47:04Z <p>Your guess is indeed right. Following a similar idea gives you the Takagi or <a href="http://en.wikipedia.org/wiki/Blancmange_curve" rel="nofollow">blancmange function</a>. It is even quasi-Lipschitz (it has a modulus of continuity $\omega(t)=ct(|\log(t)|+1)$ for a suitable constant $c>0$), thus it's Hoelder of any positive exponent less than 1. It is not even BV in any open interval, thus $W^{1,p} _ {loc}$ for no $p\geq1$. </p> <p><em>Rmk 1.</em> The above example is for dimension 1: but of course it holds in any dimension a fortiori. </p> <p><em>Rmk 2.</em> To get an example with a more classical flavor, actually a Weierstrass function, replace $s(x)$ with $\cos(x)$. I'd say that the resulting Fourier series defines a function with the same features, by the same reasons (the function $\cos(x)$ works better than $\sin(x),$ in view of point 2 below.)</p> <p><em>Rmk 3.</em> Once you know that the Weierstrass function $f(x):=\sum_{k=0}^\infty 2^{-k}\cos(2^k x)$ is nowhere differentiable, you also have that it is BV on no open interval, for BV on an interval would imply differentiability a.e. there. However, for your needs it seems more direct just showing it has infinite variation on any interval. </p> <p><em>Details.</em> </p> <ol> <li><p>To prove that the Takagi function $f(x)$ admits the above modulus of continuity, recall that that $f$ is characterized as the fixed point of the affine contraction $T:C_b(\mathbb{R})\to C_b(\mathbb{R})$ such that $(Tf)(x)=\frac{1}{2}f(2x)+s(x),$ for all $x\in\mathbb{R}$, where $s(x)$ is the distance function from the integers (a zig-zag piecewise 1-periodic function). Just find a $c$ such that the subset of $C_b(\mathbb{R})$ of functions that admit $\omega$ as modulus of continuity is a $T$-invariant set. The latter subset is obviously closed and non-empty, so the fixed point is there. (The above illustrated a standard general technique to prove properties of objects found by means of the contraction principle). </p></li> <li><p>Proving that $f$ is not of bounded variation on $[0,1]$ (hence in no open interval, due to the self-similarity encoded in the fixed point equation), requires a small computation on the partial sum $f_n$ of the series defining $f$. Let $$f_n(x):=\sum_{k=0}^{n-1}\ 2^{-k} s(2^k x).$$ First note that the derivative of $f_n$ only takes integer values, which of course come as a result of the sum of $n$ terms $\pm 1$ (with all the $2^n$ possible signs). In particular, for any $n\in\mathbb{N}$ the function $f_{2n}$ has ${2n \choose n}$ flat intervals of lenght $2^{-2n}$ within the unit interval $I$, and has derivative larger than $2$ in absolute value elsewhere in $I$. Thus, for the subsequent odd index $2n+1,$ the function $f_{2n+1}$ has ${2n \choose n}$ local maxima in $I$ (located in the mid-points of the above intervals). Moreover, passing to $f_{2n+1}$ each maximum point contributes to the increment of the total variation with $2^{-2n}$, while the total variation remains unchanged passing from $2n+1$ to the next even index $2n+2$. The conclusion is that, for any $n$, the total variation of $f_n$ on $I$ is $$V(f_n;I)=\sum_{0\leq k &lt; n/2}{2k\choose k}2^{-2k} =O\big(\sqrt{n}\big),$$<br> since by the classical asymptotics for the central binomial coefficient, ${2k \choose k}=\frac{4^k}{\sqrt{\pi k}}(1+o(1)),\ k\to\infty.$ So actually $V(f_n;I)$ diverges. Yet this would not be sufficient to conclude that $V(f,I)=\infty,$ as the total variation is only lower semicontinuous with respect to the uniform convergence. However, the discrete variation on a given subdivision $P:=\{t_0 &lt; \dots &lt; t_r \}$ $$V(f_n; P\ )=\sum_{i=0}^{r-1}\ \big|f_n(t_{i+1})-f(t_i)\big|$$ does of course pass to the limit under even pointwise convergence. Now the point is that, for the binary subdivision $P_m:=\{ k2^{-m} \ : \ 0 \le k \le 2^m \},$ we have $V(f_n;I)=V(f_n;P_m)$ as soon as $n \geq m$. So for all $m$ letting $n\to\infty$ $$V(f;P_m)=\lim_{n\to\infty }V(f_n;P_m)=V(f_m;P_m)$$ and $$V(f;I)=\sup_{m\in\mathbb{N}}V(f;P_m)=\infty,$$ as we wished to show.</p></li> </ol> http://mathoverflow.net/questions/38751/a-holder-continuous-function-which-does-not-belong-to-any-sobolev-space/38814#38814 Answer by fedja for A Hölder continuous function which does not belong to any Sobolev space fedja 2010-09-15T12:06:14Z 2010-09-15T14:52:27Z <p>Another simple approach is to take the lacunary series $f(x)=\sum_k a_k e^{i(m_k,x)}$ where $m_k\in \mathbb Z^d$ and $|m_{k+1}\gg |m_k|$. For any modulus of continuity $\omega$ such that $\Omega(t)=\omega(t)/t\to \infty$ as $t\to 0$, the condition that $f$ has this modulus of continuity is equivalent to the condition that $|a_k|\le C\omega(|m_k|^{-1})$ if the spectrum is sparce enough to ensure that $\Omega(|m_{k+1}|^{-1})\ge 2\Omega(|m_k|^{-1})$ and $\omega(|m_k|^{-1})\ge\sum_{\ell>k}\omega(|m_\ell|^{-1})$. If $f\in W^{1.p}$, then <code>$\left|\int f\nabla\psi\right|\le C\|\psi\|_{L^q}$</code> for smooth $\psi$. Plugging $e^{-i(m_k,x)}$, we see that unless $a_k=O(|m_k|^{-1})$, we have no chance. Thus, nothing short of Lipschitzness will force $f$ to be in $f\in W^{1.p}$.</p> <p>This formally works only on the torus but you can take any smooth partition of unity $g_j$ on the torus and notice that one of the functions $g_j f$ is also bad. But any of them can be replanted to $\mathbb R^d$ if the supports are small enough. </p> http://mathoverflow.net/questions/38751/a-holder-continuous-function-which-does-not-belong-to-any-sobolev-space/48593#48593 Answer by Dan Lee for A Hölder continuous function which does not belong to any Sobolev space Dan Lee 2010-12-07T22:24:31Z 2010-12-07T22:24:31Z <p>I believe that the Cantor ternary function (aka devil's staircase) is a simple example.</p>