# Sobolev Embedding Theorem, why is this not a counterexample.

I don't seem to be able to find my mistake in the following:

Consider Sobolev spaces in $\mathbb{R}^2$.

We know that $W^{1,3}(\mathbb{R}^2)\subset C^{0,\alpha}(\mathbb{R}^2)$ for $\alpha=\frac{1}{3}$ because then we have $k=1,p=3,r=0,n=2$ and hence. $$\frac{k-r-\alpha}{n}=\frac{1}{p}$$ So long story short: Every function in $W^{1,3}(\mathbb{R}^2)$ should have a continuous representative.

Now consider the function $u(x)=|x|$ inside the unit disk and $u(x)=0$ outside the unit disk. It's derivative (which is defined almost everywhere in the classical sense, so we can just take that derivative) fulfills $|\nabla u|=1$ inside the unit disk an $|\nabla u|=0$ outside the unit disk.

And now I get the following contradiction which must be wrong:

• $|u|$ has compact support and is therefore in $L^3$. $|\nabla u|$ has compact support and is therefore in $L^3$. Therefore $u\in W^{1,3}(\mathbb{R}^2)$

• But $u$ can't be made continuous by adjusting it on a set of measure zero. So $u$ can't have a continuous representative.

This would contradict $W^{1,3}(\mathbb{R}^2)\subset C^{0,\alpha}(\mathbb{R}^2)$.

Since it is unlikely that I just disproved Sobolev embedding theory :) I must be making some very stupid mistake in my line of thought...

• I haven't thought much about it yet, but is it possible that the derivative of $u$ is only a distributional derivative and not a weak derivative (in the sense that it is $L^1_{loc}$) due to the jump discontinuity? – Matt Feb 10 '17 at 19:35
• This might actually be it. I'll have to think it through in detail, but that totally makes sense. – Fabian Feb 10 '17 at 19:41
• Yeah, @Matt is right. The derivative $\nabla u$, as a distribution, isn't in $L^3$. It isn't even a function. It's not in general true that if a function is differentiable almost everywhere, then its distributional derivative is the function given a.e. by the derivative. For a simpler example, the distributional derivative of the one-dimensional Heaviside step function is not zero, but rather the Dirac delta. – Nate Eldredge Feb 10 '17 at 21:03

Differentiability almost everywhere does not imply that the function belongs to a Sobolev space. In your example, your function does not belong to $W^{1,3}(\mathbb R^2)$.
To show this, suppose it belongs to $W^{1,3}(\mathbb R^2)$. Let also $\phi$ be a smooth cutoff, supported in $Β(0,3)$, with $\phi(x)=1$ in $B(0,1)$, and $\psi(x)=x_1^3$. Then $\phi\psi\in C_c^{\infty}(\mathbb R^3)$, therefore $$\int_{\mathbb R^3}u\cdot\partial_1(\psi\phi)=-\int_{\mathbb R^3}\partial_1u\cdot\psi\phi.$$ Since $\partial_1|x|=x_1|x|^{-1}$ for $x\neq 0$, and $\psi=\psi\phi$ on the support of $u$, the previous equality implies that $$\int_{\mathbb R^3}|x|\cdot 3x_1^2\,dx=-\int_{\mathbb R^3}x_1|x|^{-1}\cdot x_1^3\,dx.$$ But, this is a contradiction, since the left hand side is positive, while the right hand side is negative.