Inequality in $L^2$

Question

Let $u,v\in L^2(\mathbb{R}^d)$,

I want to prove the following ineqality

$$\|u\|^2_{L^2(\mathbb{R}^d)}\ge a \|v\|^2_{L^2(\mathbb{R}^d)}-\|u-v\|^2_{L^2(\mathbb{R}^d)}$$

for all $u,v\in L^2(\mathbb{R}^d)$ where $a>0$ is an optimal number to be determined.

Thanks.

Answer 1

It is easy to see that $(a+b)^2 \le 2(a^2 + b^2)$ for any two real numbers $a$ and $b$.

Thus with $a = u$ and $b=v-u$ you get $$\frac 12 v^2 \le u^2 + (u-v)^2.$$ Integrate over $\mathbb R^d$ to discover $$\frac 12 \|v\|_{L^2(\mathbb R^d)}^2 \le \|u\|_{L^2(\mathbb R^d)}^2 + \|u-v\|_{L^2(\mathbb R^d)}^2$$