# L2 norm of a function w.r.t. its gradient

I was wondering whether it is possible to generalise the statement of this post, i.e. I would like to show something like $$\|f\|^2_{L^2(\Omega)}\leq C \|\nabla f\|^2_{L^2(\Omega)},$$ where $\Omega=[0,1]^n$ and $f:\Omega\to\mathbb{R}$ is a suitable function with $f=0$ on the vertices of $\Omega$ (and $C$ does not depend of $f$). Although this seems to be a standard question I was not able to find anything on this for $n\geq2$.

My approach was the following (given that everything is well-defined). By $f(0)=0$ we know that: $$f(x)=\int^1_0\langle\nabla f(hx),x\rangle dh=\int^{|x|}_0\langle\nabla f(uw),w\rangle du,$$ where $w=x/|x|$. If I am not mistaken, using polar coordinates (cf. this post) should give the identity $$\|f\|^2_{L^2(\Omega)}=\int^{\rho_n}_0\int_{A_n}f(rw)^2dwr^{n-1}dr$$ with $A_n=S^{n-1}\cap[0,1]^n$, $S^{n-1}$ sphere and $\rho_n=\sqrt{n}$ being the "diagonal" (or diameter) of $\Omega$. Next I would like to combine the two preceding equalities to obtain the desired inequality. This should give (if I did not mix anything up with the gradient) the equation $$\|f\|^2_{L^2(\Omega)}=\int^{\rho_n}_0\int_{A_n}\Big(\int^r_0\langle\nabla f(uw),w\rangle du\Big)^2dwr^{n-1}dr.$$ Cauchy-Schwarz now yields $$\|f\|^2_{L^2(\Omega)}\leq\int^{\rho_n}_0\int_{A_n}\int^r_0|\nabla f(uw)|^2dudwr^ndr$$ but I do not see how I could end up with $\|\nabla f\|^2_{L^2(\Omega)}$ since I would need the term $u^{n-1}$ to pop-up somewhere.

Any ideas are appreciated especially for the special case $n=2$. In particular, I would like to consider Sobolev functions with $s\in(1,2)$ to obtain $$\|f\|^2_{L^2(\Omega)}\leq C \|\nabla f\|^2_{L^2(\Omega)}\leq C' \sum_{|\alpha|=1}\| f^{(\alpha)}\|^2_{H^{s-1}(\Omega)},$$ such that $C'$ does not depend on $f$ and I think that I already showed that the second inequality holds.

• This is the Poincaré-inequality. – Jan Bohr Apr 3 '18 at 10:28
• Poincaré's inequality requires that $f$ vanishes on the entire boundary of $\Omega$. In the question, it is only required that $f = 0$ on the vertices of $\Omega$. However, the condition "$f = 0$ on the vertices of $\Omega$" does not make sense for $f \in H^1(\Omega)$ in dimensions $n \ge 2$. – gerw Apr 3 '18 at 11:03
• Still, one might use Poincaré in this way: Extend $f$ to the unit ball $B^n$, say to $\tilde f$, such that $\tilde f=0$ on the boundary of $B^n$ and $|\nabla\tilde f|_{H^{s-1}(B^n)}\leq C'\sum_{|\alpha|=1}\|f^{(\alpha)}\|^2_{H^{s-1}(\Omega)}$. Then one might use $\|f\|^2_{L^2(\Omega)}\leq\|\tilde f\|^2_{L^2(B^n)}\leq C\|\nabla\tilde f\|^2_{L^2(B^n)}$ and try to bound the latter one by $|\nabla\tilde f|_{H^{s-1}(B^n)}$. – User123 Apr 3 '18 at 11:12
• @gerw: That is why I wrote "if everything is well-defined". If $s>n/2$ then the condition should make sense and we could also apply Poincaré, right? Since you replied to the last questions I posed: I would like to show the posted inequality for $C$ independent of $f$ and I know that this is true for $f\in H^s(\mathbb{R}^n),\ s\in(0,1)$. – User123 Apr 3 '18 at 11:15
• It should be possible to construct an explicit counterexample: Define $f_\epsilon(x) = \log|\log(\epsilon+|x)| - \log|\log \epsilon|$.Then reflect $f_\epsilon$ such that $f_\epsilon$ is defined and continuous on $[0,1]^2$. This $f_\epsilon$ belongs to $H^1$ and is zero at the corners. Then $\|\nabla f_\epsilon\|_{L^2}$ should be uniformly bounded, while $\|f_\epsilon\|_{L^2}$ blows up for $\epsilon\to0$, – daw Apr 6 '18 at 12:25