Inequality in $L^2$ space $(\|v\| + \|v'\|)(\|w\|+\|w'\|) \le \sqrt{2} (\|v\|^2+\|v'\|^2)^{1/2}\sqrt{2}(\|w\|^2+\|w'\|^2) ^{1/2}$?

Question

As a part of a proof I stuck at an inequality. Suppose $u$, $v$, and $w$ are functions in $L^2$ space, $$ (\|v\| + \|v'\|)(\|w\|+\|w'\|) \le \sqrt{2} (\|v\|^2+\|v'\|^2)^{1/2}\sqrt{2}(\|w\|^2+\|w'\|^2) ^{1/2} $$ Considering the norm is in $L^2$. How did those $\sqrt{2}$ appear?

Answer 1

The scaling of the inequality looks strange. Suppose your inequality is true and then scale every vector by $\lambda >0$. Divide both sides by $\lambda^2$ and take $\lambda$ goes to zero. Then you get $v,v',w,w'$ all equal to zero. Are you sure of this formulation?

Added The corrected inequality is true because $(a+b)^2\leq2(a^2+b^2)$

Answer 2

This is the Cauchy-Schwarz inequality for vectors in $\mathbb R^2$: if $a,b\in\mathbb R$, $$ |a+b| = |\begin{pmatrix}1\\1\end{pmatrix}\cdot\begin{pmatrix}a\\b\end{pmatrix}|\le \sqrt{1^2+1^2}\cdot\sqrt{a^2+b^2}=\sqrt2(a^2+b^2)^{\frac12}. $$