I have a rather basic question about Sobolev functions. I would need a reference or proof for the following inequality which seems to be well-known in approximation theory.
Question: Let $\Omega=[x,x+h]\times[y,y+h]$ be a square of side-length $h$ and let $f\in H^s(\Omega)$ be a Sobolev funtion of regularity $s\in(1,2)$ such that $f=0$ on the vertices of $\Omega$. Does it hold that $$\|f\|^2_{L^2(\Omega)}\leq C h^{2s}\|f\|^2_{H^s(\Omega)}?$$
I would like to use such a bound to get the rate of the approximation error of a function on $[0,1]^2$ by its piecewise linearly interpolated counterpart on a grid of size $h$, which explains the assumptions of roots on the vertices of the grid.
Here is my argument for the univariate case: Let $I=[x,x+h]$ and $g:I\to R$ such that $g(x)=g(x+h)=0$.
Assume first that $g\in H^1(I)$. Then weak differentiability, $g(x)=0$ and Cauchy-Schwarz imply \begin{align}\|g\|^2_{L^2(I)}&=\int^{x+h}_xg(t)^2dt=\int^{x+h}_x\Big(g(x)+\int^t_xg'(s)ds\Big)^2dt\\ &\leq\int^{x+h}_x(t-x)\int^t_xg'(s)^2dsdt\leq\int^{x+h}_x(t-x)dt\int^{x+h}_xg'(s)^2ds\\ &=\frac{1}{2}h^2\|g'\|^2_{L^2(I)}\leq\frac{1}{2}h^2\|g\|^2_{H^1(I)}. \end{align}
Assume now that $g\in H^2(I)$. Then additionally using that there is some $x_0\in I$ with $g'(x_0)=0$ (since $g$ has to have an extremum on $I$, by $g(x)=g(x+h)=0$) and applying Cauchy-Schwarz twice yields \begin{align}\|g\|^2_{L^2(I)}&=\int^{x+h}_xg(t)^2dt=\int^{x+h}_x\Big(g(x)+\int^t_xg'(s)ds\Big)^2dt\\ &=\int^{x+h}_x\Big(g(x)+\int^t_x\Big(g'(x_0)+\int^s_{x_0}g''(u)du\Big)ds\Big)^2dt\\ &=\int^{x+h}_x\Big(\int^t_x\Big(\int^s_{x_0}g''(u)du\Big)ds\Big)^2dt\\ &\leq\int^{x+h}_x(t-x)\int^t_x(s-x_0)\int^s_{x_0}g''(u)^2dudsdt\\ &\leq\int^{x+h}_x(t-x)\int^t_x(s-x)dsdt\int^{x+h}_xg''(u)^2du\\ &=\frac{1}{8}h^4\|g''\|^2_{L^2(I)}\leq\frac{1}{8}h^4\|g\|^2_{H^2(I)}. \end{align} Now an interpolation argument gives for $g\in H^s(I),s\in(1,2)$, the inequality $$\|f\|^2_{L^2(I)}\leq Ch^{2s}\|f\|^2_{H^s(I)}.$$ For the bivariate case I have a few problems. For instance if $f\in H^1(\Omega)$ the point evaluations are not necessarily well-defined since the approximated function might not be continuous. But even if I would assume that I do not have to worry about that the same approach would give me (using $f(x)=0$) $$\|f\|^2_{L^2(\Omega)}=\int_{\Omega}\Big(\int^1_0\langle\nabla f(t+u(t-x)),t-x\rangle\Big)^2dudt$$ and I am stuck at this point. For $f\in H^2(\Omega)$ I found a bound in this paper (by the proof of Lemma 1). Has anyone an idea or knows some helpful literature? Thank you!
This is merely a collection of ideas rather than a full answer.
First, we note that point evaluation is continuous on $H^s(\Omega)$, since $H^s(\Omega) \hookrightarrow C(\bar\Omega)$ for $s > 1$ in dimension $2$.
Let us first consider the case $x = y = 0$ and $h = 1$, i.e., $\Omega$ is the unit square.
First, I would try to show $$ \|f\|_{L^2(\Omega)} \le C \, | f |_{H^s(\Omega)},$$ where $| f |_{H^s(\Omega)}$ is the seminorm in $H^s(\Omega)$. This should work as usual since a function $f \in H^s(\Omega)$ with $| f|_{H^s(\Omega)} = 0$ which vanishes at the vertices should be zero.
Now, you could use a transformation argument to transform $\Omega$ to an arbitrary square. Since the $L^2(\Omega)$-norm and the $H^s(\Omega)$-seminorm scale differently, this should produce an $h$-dependent constant: $$ \|f \|_{L^2(\Omega)} \le C \, h^t \, |f|_{H^s(\Omega)}.$$ ($t = s$ looks reasonable, but I am not entirely sure) Finally, you bound the seminorm by the full norm to obtain $$ \|f \|_{L^2(\Omega)} \le C \, h^t \, \|f\|_{H^s(\Omega)}.$$
Edit: Here are some ideas to prove the first inequality (this is just an application of a usual proof of Poincaré's inequality). Let us assume that it does not hold. Then, there exists a sequence $\{f_n\}$ (vanishing at the vertices) with $$1 = \|f_n\|_{L^2(\Omega)} > n \, | f_n |_{H^s(\Omega)}.$$ Then it should hold (maybe by some interpolation arguments) that even the full norm $\|f_n\|_{H^s(\Omega)}$ is bounded. Hence, there exists a weakly convergent subsequence (without relabeling), i.e., $f_n \rightharpoonup f$ in $H^s(\Omega)$. Since the embedding from $H^s(\Omega)$ to $L^2(\Omega)$ is compact, we get $f_n \to f$ in $L^2(\Omega)$. Moreover, we can show $\|f\|_{L^2(\Omega)} = 1$, $|f|_{H^s(\Omega)} = 0$ and $f$ vanishes at the vertices. This is a contradiction.
Edit 2: Let us prove $$\|g\|_{H^s(\Omega)} \le C \, \Big( \|g\|_{L^2(\Omega)} + |g|_{H^s(\Omega)} \Big).$$ We proceed by contradiction and assume that the inequality is false. This gives a sequence $\{g_n\}$ with $$1 = \|g_n\|_{H^s(\Omega)} > n \, \Big( \|g_n\|_{L^2(\Omega)} + |g_n|_{H^s(\Omega)} \Big).$$ Again (up to a subsequence), $g_n \rightharpoonup g$ in $H^s(\Omega)$, and $\|g\|_{L^2(\Omega)} = 0$ and $|g|_{H^s(\Omega)} = 0$. The latter means that the first derivatives of $g$ vanish (since $s \in (1,2)$). Thus, $g$ is constant and, therefore, $g = 0$. The embedding from $H^s(\Omega)$ into $H^1(\Omega)$ is compact. Therefore, $| g_n |_{H^1(\Omega)} \to 0$. Putting everything together, we have $$ 1 = \|g_n\|_{H^s(\Omega)}^2 = \|g_n\|_{L^2(\Omega)}^2 + |g_n|_{H^1(\Omega)}^2 + |g_n|_{H^s(\Omega)}^2 \to 0 + 0 +0$$ which is a contradiction.