# Establishing an inequality.

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.

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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.
thanks. How about the part $f\in L^1(\mathbf R)$? Also, why did you split $f$ the way you did? – Kuku Apr 19 '12 at 15:29
The above shows that $f 1_{[-1,1]} \in L^1(\mathbb{R})$, and also that $f 1_{[-1,1]^C} \in L^1(\mathbb{R})$. It follows (using Minkowski's inequality) that the sum in also in $L^1(\mathbb{R})$. – copper.hat Apr 19 '12 at 21:43