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.

  • $\begingroup$ First of all, thank you for your answer. I assume you mean the Slobodeckij (or fractional Sobolev) semi-norm. I thought of that, too, but I have no clue how to bring it into play here. Could you explain what you mean by "work as usual"? I have to admit that I am not very familiar with this semi-norm. $\endgroup$ – User123 Mar 29 '18 at 11:43
  • $\begingroup$ Btw: I verified the transformation argument and indeed it holds that $s=t$ so the only thing missing would be the first inequality. Can you explain why I would consider functions with $|f|_{H^s(\Omega)}=0$? $\endgroup$ – User123 Mar 29 '18 at 12:31
  • $\begingroup$ @User123: I have added some details concerning the proof of the inequality. I just transformed a proof of Poincaré's inequality. The precise form of the seminorm is not really important. $\endgroup$ – gerw Mar 29 '18 at 17:04
  • $\begingroup$ Thanks again, that looks quite convincing! Is $f$ vanishing at the vertices crucial? I am asking since the transformation argument only holds for regularities in $(0,1)$. To be more precise: $|f|^2_{H^s(\Omega)}$ equals for $s\in(1,2)$ (or is dominated by) $\sum_{|\alpha|=1}\int\int|f^{\alpha}(x)-f^{\alpha}(y)|^2|x-y|^{-2\theta-2}dxdy$ for $\theta=s-1$. Then one has with $g(u,v):=f(x+hu,y+hv)$ that $\|f\|^2_{L^2(\Omega)}=h^2\|g\|^2_{L^2([0,1]^2)}\leq h^2C|g|^2_{H^s([0,1]^2)}=h^{2\theta}C|f|^2_{H^s(\Omega)}$. $\endgroup$ – User123 Mar 29 '18 at 20:34
  • $\begingroup$ To remedy this I was thinking of applying your 1st inequality to the first derivatives of $f$ (which do not vanish at the vertices) to obtain something like $\|f\|^2_{L^2(\Omega)}\leq C_1h^2\sum_{|\alpha|=1}\|f^{\alpha}\|^2_{L^2(\Omega)}\leq C_2h^4\sum_{|\alpha|=1}|g^{\alpha}|^2_{H^{\theta}([0,1]^2)}\leq\ldots\leq C_2 h^{2+2\theta}|f|^2_{H^s(\Omega)}=C_2 h^{2s}|f|^2_{H^s(\Omega)}$ but here we would need the 2nd inequality to holds for functions not vanishing on the vertices. $\endgroup$ – User123 Mar 29 '18 at 20:34

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