Prove that $\int |\nabla u||\nabla v|+|u||v|\leq |u|_{H^1}|v|_{H^1}$

Question

Prove that $\int |\nabla u||\nabla v|+|u||v|\leq |u|_{H^1}|v|_{H^1}$. Here $H^1$ is Sobolev Space. Then my attempt is first (im not sure) apply Inequality'Holder so $ \int |\nabla u||\nabla v|+|u||v|\leq |\nabla u|_{L^2}|\nabla v|_{L^2}+| u|_{L^2}|v|_{L^2}$. My second step (im not sure) i want to apply $ab\leq \frac{a^2+b^2}{2}$ but with this my last inequality $\leq \frac{|u|_{H^1}+ |v|_{H^1}}{2}$. I will apprecciate one hint for finish the statement, thank you!!! Im working with $|U|_{H^1}^{2}=|u|_{L^2}+|\nabla u|_{L^2}$

Answer 1

Applying AM-GM as in your idea, you get $$\int (|\nabla u| |\nabla v| + |u| |v|) \le \frac{1}{2}(\|\nabla u\|_{L^2}^2 + \|u\|_{L^2}^2+ \|\nabla v\|^2_{L^2} + \|v\|^2_{L^2}) =\frac{1}{2} (\|u\|_{H^1}^2 + \|v\|_{H^1}^2). $$ This actually renders your Hölder's inequality step unnecessary.

Now let $c>0$ be an arbitrary constant and apply the above inequality with $u$ replaced by $cu$ and $v$ replaced by $c^{-1} v$. This yields the improvement $$\int (|\nabla u| |\nabla v| + |u| |v|) \le \frac{1}{2}(c^2 \|u\|_{H^1}^2 + c^{-2} \|v\|_{H^1}^2).$$ Then optimize over $c$; the minimum on the right side is attained when $$c = \sqrt{\frac{\|v\|_{H^1}}{\|u\|_{H^1}}}$$ in which case you have as desired $$\int (|\nabla u| |\nabla v| + |u| |v|) \le \|u\|_{H^1} \|v\|_{H^1}.$$ This standard trick is sometimes called a "Peter-Paul inequality", after the saying about robbing Peter to pay Paul: you make one term smaller at the expense of making the other larger.