Is a function in $L^2$ which second derivative is in $L^2$ in $H^2$?

Question

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...

Answer 1

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 $f\in H^2$.

Answer 2

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.

Answer 3

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.