Proving $\int_{-\pi}^{\pi} |f''(x)|^2dx\leq \left(\int_{-\pi}^{\pi} |f(x)|^2dx\right)^{1/2}\cdot\left(\int_{-\pi}^{\pi} |f''''(x)|^2dx\right)^{1/2}$

Question

Let $f$ be 2$\pi$ periodic. I want to show that $$\int_{-\pi}^{\pi} |f''(x)|^2dx\leq \left(\int_{-\pi}^{\pi} |f(x)|^2dx\right)^{1/2}\cdot\left(\int_{-\pi}^{\pi} |f''''(x)|^2dx\right)^{1/2}$$ where we take $f$ to be $C^4$.

My initial thought was to use Fourier series expansions and apply the Cauchy-Schwarz inequality, but I am not sure how exactly to proceed. Any help would be appreciated.

Answer 1

Hint: Using integration by parts, we get \begin{align} \int^\pi_{-\pi} f''(x)f''(x)\ dx = f''(x)f'(x)\big|_{-\pi}^\pi-\int^\pi_{-\pi}f'''(x)f'(x)\ dx = -\int^\pi_{-\pi}f'''(x)f'(x)\ dx. \end{align} Perform integration by parts one more time, then ... I will let you finish the rest.

Answer 2

The idea is like \begin{align*} \int|f''(x)|^{2}dx&=\sum|\widehat{f''}(m)|^{2}\\ &=\sum|m|^{4}|\widehat{f}(m)|^{2}\\ &\leq\left(\sum|m|^{8}|\widehat{f}(m)|^{2}\right)^{1/2}\left(\sum|\widehat{f}(m)|^{2}\right)^{1/2}\\ &=\left(\sum|\widehat{f}(m)|^{2}\right)^{1/2}\left(\sum|m^{4}\widehat{f}(m)|^{2}\right)^{1/2}\\ &=\left(\int|f(x)|^{2}dx\right)^{1/2}\left(\sum|\widehat{f^{(4)}}(m)|^{2}\right)^{1/2}\\ &=\left(\int|f(x)|^{2}dx\right)^{1/2}\left(\int|f^{(4)}(x)|^{2}dx\right)^{1/2}. \end{align*}