# A form of Nash's inequality, $\|f\|_2\le C \|f\|_{1}^{\alpha} \|f'\|_2^\beta$

For $f\in \mathcal{S}(\mathbb{R})$ can anyone help me prove the following Nash inequality,

$$\|f\|_2 \le C \|f\|_{1}^{\alpha} \|f'\|_2^\beta.$$

I believe $\alpha$ and $\beta$ should be $2/3$ and $1/3$ respectively and the following inequality should help

$$\|f\|_2^2\leq 2\lambda \|f\|_1^2+\frac{1}{4\pi^2\lambda^2}\|f'\|_2^2.$$

• If you already have the second inequality, you're almost done. Minimize over $\lambda$. – user147263 Jul 28 '15 at 16:53
• Sorry I know I'm being really dim but I'm trying that and still can't get it, could you expand on it? – Kevin Jul 28 '15 at 17:19
• The inequality should be $\le$ otherwise it's false for zero function. – user147263 Jul 28 '15 at 17:25

To derive Nash's inequality from $$\|f\|_2^2\leq 2\lambda \|f\|_1^2+\frac{1}{4\pi^2\lambda^2}\|f'\|_2^2\tag{1}$$ you should choose the value of $\lambda$ that minimizes the right hand side. To this end, consider the function $$g(\lambda)=2\lambda \|f\|_1^2+\frac{1}{4\pi^2\lambda^2}\|f'\|_2^2$$ of a real variable $\lambda$. Then $$g'(\lambda)=2\|f\|_1^2-\frac{1}{2\pi^2\lambda^3}\|f'\|_2^2$$ The minimizing value of $\lambda$ is obtained by solving $g'(\lambda)=0$. Plug this value into the inequality (1).