Let $f,x^2f \in L^2(\mathbf R)$. How can I show that $f\in L^1(\mathbf R)$ and
$ \|f\|_1 \leq \sqrt 2 \|f\|_2 + \frac{\sqrt 6}{3} \|x^2 f\|_2?$ I have no idea where to begin.
Let $f,x^2f \in L^2(\mathbf R)$. How can I show that $f\in L^1(\mathbf R)$ and
$ \|f\|_1 \leq \sqrt 2 \|f\|_2 + \frac{\sqrt 6}{3} \|x^2 f\|_2?$ I have no idea where to begin.
Write $f = f 1_{[-1,1]} + f 1_{[-1,1]^C}$. Then using Hölder's inequality, bound the two components of $f$ separately. (Some abuse of notation below, but you get the idea.)
$||f 1_{[-1,1]}||_1 \leq ||f||_2 ||1_{[-1,1]}||_2= \sqrt{2} ||f||_2$
$||f 1_{[-1,1]^C}||_1 =|| x^2 f \frac{1}{x^2}1_{[-1,1]^C}||_1 \leq ||x^2f||_2 ||\frac{1}{x^2}1_{[-1,1]^C}||_2 $
Since $\int_1^{\infty}(\frac{1}{x^2})^2 dx = \frac{1}{3}$, we have $||\frac{1}{x^2}1_{[-1,1]^C}||_2 = \sqrt{\frac{2}{3}}$, hence the result.