Suppose $f$ is smooth and both $f$ and $f^{\prime\prime}$ are in $L^2(\mathbb{R})$. Prove $f^{\prime}$ is again in $L^2(\mathbb{R})$. Someone said I can use the method of Fourier transform. Because since $f$ and $f^{\prime\prime}$ are in $L^2(\mathbb{R})$, then we must have $$ \int |\hat f(\xi)|^2 <\infty \quad \text{and} \int |\xi|^4 |\hat f(\xi)|^2 <\infty, $$ and hence, by Holder's inequality, $$ \int |\xi|^2 |\hat f(\xi)|^2 <\infty-(*). $$ It says in a text book that only if $f^{(k)}$ is in $L^2(\mathbb{R})$, then $\widehat{(f^{(k)})}=(i\xi)^k \hat f$. But in my problem, we do not know $f^\prime$ is in $L^2(\mathbb{R})$. So, my first question is, can we deduce the result by $(*)$?
My second question: I tried to use approximation theorem to figure this out. I hope to find a sequence of functions in $C_c^{\infty}(\mathbb{R})$ to approximate $f$. I know $C_c^{\infty}(\mathbb{R})$ is dense in $L^2(\mathbb{R})$. But can we guarantee that the sequence also converges to the derivatives of $f$ at the same time? I know I can use mollifier to achieve this, but the approximation functions would not have compact support then.
Can someone give me an answer? Thank you.
For any $U\subset \mathbb{R}$ open and bounded, choose $\psi$ s.t. $\psi \in C_c^\infty (\mathbb{R})$ and $\psi \vert_{U^\prime} =1$, where $U^\prime$ satisfying $U \subset\subset U^\prime$. Consider $g=f\psi$. Let $\{\eta^\epsilon\}$ be a family of mollifiers. Then, $$ g^\epsilon\to f \text{ in } L^2(U) \text{$\quad and \quad $} (g^\epsilon)^{\prime\prime} \to f^{\prime\prime} \text{ in } L^2(U) $$ Since we know $f$, $f^{\prime\prime}\in L^2(\mathbb{R})$ and $g\vert_U=f$.
Therefore, $$ ||\widehat{g^\epsilon}||_{L^2(U)}=||g^\epsilon||_{L^2(U)}\to ||f||_{L^2(U)}\leq ||f||_{L^2(\mathbb{R})}, $$ and $$ ||\widehat{(g^\epsilon)^{\prime\prime}}||_{L^2(U)}=||(g^\epsilon)^{\prime\prime}||_{L^2(U)}\to ||f^{\prime\prime}||_{L^2(U)}\leq ||f||_{L^2(\mathbb{R})}. $$ Now, since $g^{\epsilon}$ has compact support, then $\widehat{(g^\epsilon)^\prime}=i\xi \hat {g^\epsilon}$. It then follows from H$\ddot{\text{o}}$lder's inequality and the above two that, $$ ||(g^\epsilon)^\prime||_{L^2(U)}=||\widehat{(g^\epsilon)^\prime}||_{L^2(U)}=||\xi \hat {g^\epsilon}||_{L^2(U)} \leq C (||f||_{L^2(\mathbb{R})},||f^{\prime\prime}||_{L^2(\mathbb{R})})=:M. $$ Notice that $M$ does not depend on $U$. Hence, there exists $h\in L^2(U)$ s.t. $$ (g^\epsilon)^\prime \to h \text{ in } L^2(U) \text{ weakly up to a subsequence.} $$ So, for any $\phi \in C_c^\infty(U)$, $$ -\int g^\epsilon \phi^\prime=\int(g^\epsilon)^\prime \phi \to \int h\phi. $$ But $\int g^\epsilon \phi \to \int f\phi$. So, $$ \int f\phi^\prime=-\int h \phi, $$ which implies that $h=f^\prime$ on $U$. Now, since $ (g^\epsilon)^\prime \to f^\prime \text{ in } L^2(U) \text{ weakly up to a subsequence,} $ then we have $$ ||f^\prime||_{L^2(U)} \leq \liminf ||(g^\epsilon)^\prime||_{L^2(U)} \leq M. $$ Since $U$ can be arbitratry, then $f^\prime \in L^2(\mathbb{R})$.