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Let $\Omega$ be a bounded domain in $\mathbb{R}^n$ with smooth boundary. Assume that $f\in L^2(\Omega)$ and $f^{\prime\prime}\in L^2(\Omega)$.

Does one have $f\in H^2(\Omega)$?

Useless comments:

Here I assume that $f^{\prime\prime}$ is a function (the second derivative in the weak sense exists). It seems a pretty stupid question since once you have f in $H^2$ something like Ehrling's lemma will control the norm of the gradient by the $L^2$ norm of $f$ and its second derivative. But my point is to show that $f^\prime$ is $L^2$. Moreover a function in $H^2$ will implies a boundary trace which is $H^{3/2}$ and it seems unclear to me that it can be satisfyed with only $L^2$ conditons which do not see the boundary at all...

user42070
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3 Answers3

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Let $Q$ be a maximal dyadic cube such that $2Q\subset \Omega$. Convolving $f$ with a mollifier $\varphi_\epsilon$, we obtain a smooth function $f_\epsilon$ such that $\int_Q f_\epsilon^2$ and $\int_Q |D^2f_\epsilon |^2$ are controlled by $\int_{2Q} f^2$ and $\int_{2Q} |D^2f |^2$ respectively. Use "something like Ehrling's lemma" to control $\int_Q |Df_\epsilon|^2$. Summing over $Q$, get a bound on $H^2$ norm of $f_\epsilon$ independent of $\epsilon$. By weak compactness, there is a weakly $H^2$ convergent sequence $f_{\epsilon_k}$ as $\epsilon_k\to 0$. Let $g$ be its limit.

At the same time, $f_{\epsilon_k}\to f$ in $L^2$. Since

  • $L^2$ convergence implies convergence in the sense of distributions;
  • weak $H^2$ convergence also implies convergence in the sense of distributions;
  • distributional limit is unique

it follows that $g=f$, and thus $f\in H^2$.

  • Thanks a lot for your nice answer. Is it perfectly clear that the bound we get on the H2 norm of $f_\epsilon$ is independent of $\epsilon$"? – user42070 May 26 '14 at 16:29
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If you take the cantor distribution function $F(t)$ on $[0,1]$ and integrate this to obtain $f$ on $[0,1]$, then $f''$ exists almost everywhere on $[0,1]$, and $f''=0$ a.e.. That definitely doesn't work on $\mathbb{R}$ because $H^{2}[0,1]$ consists of all twice absolutely continuous functions on $[0,1]$, which $\int F \,dt$ is not.

Disintegrating By Parts
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  • I think I get the idea but I believe that your counter example does not fit my hypothesis since $f$ is assume to have a weak derivative which if i am right implies that $f$ is absolutely continuous. – user42070 May 26 '14 at 16:05
  • I didn't read that you were assuming a weak derivative. I thought by "useless comments" you were criticizing something you read. I didn't realize any assumptions in there were part of the problem. – Disintegrating By Parts May 26 '14 at 16:50
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This follows easily from Nirenberg's interpolation inequality $$ \|u'\|_p \leq C \|u\|_{W^{2,r}}^{1/2} \|u\|_q^{1/2} $$ valid for $\frac{1}{p}=\frac{1}{2}\left(\frac{1}{q}+\frac{1}{r} \right)$. See Brezis, Functional analysis, page 234. Or the original paper by L. Nirenberg, On elliptic partial differential equation, 1959.

Siminore
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    Do not you already need $u$ to be in $H^2$ already in the right hand side? – Hui Zhang May 21 '14 at 12:11
  • Thanks for you answer. However as pointed out by Hui Zhang, the things is that I cannot assume f to be in $H^2$ to use this interpolation. – user42070 May 26 '14 at 15:52