I'm preparing for my prelims and here's a practice question related to Schwartz functions.
Let $f \in \mathcal S(\mathbb R)$. Prove that for some $C > 0$:
$$||f'||_{L^2(\mathbb R)} \leq C||f||_{L^2(\mathbb R)}^{1/2}||f''||_{L^2(\mathbb R)}^{1/2}.$$
I do not know how to approach questions of this sort where they ask for an upper bound on the derivative. Could someone help me with this?
I think $C=1$: Since $f \in \cal{S}(\mathbb{R})$ we have $f^{(k)}(x) \to 0$ $(|x| \to \infty)$ for each $k$. Now $$ \int_{- \infty}^\infty |f'(x)|^2 dx = \int_{- \infty}^\infty f'(x) \overline{f'(x)} dx = [f(x)\overline{f'(x)}]_{-\infty}^\infty - \int_{- \infty}^\infty f(x) \overline{f''(x)} dx $$ $$ =0- \int_{- \infty}^\infty f(x) \overline{f''(x)} dx \le \int_{- \infty}^\infty |f(x)| |f''(x)| dx \le (\int_{- \infty}^\infty |f(x)|^2 dx)^{1/2}(\int_{- \infty}^\infty |f''(x)|^2 dx)^{1/2}. $$ Thus $\|f'\|_2^2 \le \|f\|_2\|f''\|_2$, which gives $\|f'\|_2 \le \|f\|_2^{1/2}\|f''\|_2^{1/2}$.