Interpolation theorem on$(0,2 \pi)$. Suppose $f \in L^2$ , the distribution derivative $f'' \in L^2$ ,then can we show that $f' \in L^2$?

Question

Interpolation theorem on $(0,2 \pi)$. Suppose $f \in L^2$ , the distribution derivative $f'' \in L^2$ ,then can we show that $f' \in L^2$ ?

Set $f(x)=\sum a_n e^{inx}$ and $ f''(x)=\sum b_n e^{inx}$ , then we have $$f''(x)=\lim_N \sum_{-N}^N -in^2 a_n e^{inx}$$ in the sense of distribution , so it suffice to show $b_n = -in^2 a_n$ . However , $e^{inx} \notin C_c^{\infty}(0,2 \pi)$ so we can not use it as a test function . And I can not imagine what does the Fourier series of such functions look like .

Answer 1

Let $\chi_{\epsilon}$ be mollifiers, it’s easy to see that $\chi_{\epsilon} * f$ (resp. $\chi_{\epsilon} * f’’=(\chi_{\epsilon} * f)’’$) is a smooth function with $L^2$-norm bounded by a constant times that of $f$ (resp. that of $f’’$), and that converges in $L^2$ to $f$ (resp. $f’’$).

But if $g$ is any smooth function on the circle, then $\int_a^b{g’\overline{g’}}=[\overline{g}g’]_a^b-\int_a^b{\overline{g}g’’}$ (as suggested by md2perpe’s comment), thus $\|g’\|_2^2 \leq \|g\|_2 \|g’’\|_2$. It follows that $\chi_{\epsilon} * f’=(\chi_{\epsilon}*f)’$ is Cauchy in $L^2$ and thus converges to a $L^2$ function $g$, but also converges to $f’$ in distribution. Therefore, $g=f’$ is $L^2$.