could anyone help me to understand this: $$f(x) :=\left\{\begin{array}{ll}{u(x)-u_{x}(x),} & {x<\xi} \\ {u(x)+u_{x}(x),} & {x>\xi}\end{array}\right.$$ where ξ is determined by $$u(\xi)=\max _{x \in \mathbb{R}} u(x)$$ One calculates the integral $$\int_{\mathbb{R}} f^{2}(x) d x=\int_{-\infty}^{\xi}\left(u-u_{x}\right)^{2} d x+\int_{\xi}^{\infty}\left(u+u_{x}\right)^{2} d x$$ $$=\int_{-\infty}^{\xi}\left(u^{2}+u_{x}^{2}\right) d x-u^{2}\left.(x)\right|_{-\infty} ^{\xi}+u^{2}\left.(x)\right|_{\xi}^\infty=\|u\|_{H^{1}}-2 u^{2}(\xi)$$
It implies ( how i can find this )
$$\|u\|_{L^{\infty}(\mathbb{R})} \leq \frac{\sqrt{2}}{2}\|u\|_{H^{1}}$$
You are missing a square: you should have $$\int_{-\infty}^\xi (u^2 + u_x^2) \, dx \le \|u\|_{H^1}^2.$$
Since $\displaystyle 0 \le \int_{\mathbf R} f(x)^2 \, dx$ your inequality leads to $$0 \le \int_{\mathbf R} f(x)^2 \, dx \le \|u\|_{H^1}^2 - 2 u(\xi)^2$$ so that $$2u(\xi)^2 \le \|u\|_{H^1}^2.$$ Now take the square root.