For functions $u,v \in L^2(\Omega)$, with $\Omega \subset \mathbb{R}^n$, the Cauchy-Schwarz inequality as a special case of the Hölder inequality $(p=q=2)$ states that
(1) $\Vert uv\Vert_{L^1} = \int_\Omega \vert uv \vert \le \sqrt{\int_\Omega \vert u\vert^2}\sqrt{\int_\Omega \vert v\vert^2}= \Vert u \Vert_{L^2} \Vert v\Vert_{L^2}$.
I have come across inequalities of the following form using references to C.S. stating that for $u,v\in H^1(\Omega)$
(2) $ \int_\Omega \vert \nabla u \cdot \nabla v \vert \le \sqrt{\int_\Omega \vert \nabla u\vert^2}\sqrt{\int_\Omega \vert \nabla v \vert^2}= \Vert \nabla u \Vert_{L^2} \Vert \nabla v\Vert_{L^2}$.
Is this formula correct?
Or is there a constant missing on the right side:
The scalar product on the left side of (2) can be written as a sum $\sum_{i=1}^n \partial_iu \partial_iv$ and the triangle inequality can be applied. Then, (1) can be used on each subterm of the form $\int_\Omega \vert \partial_i u \partial_i v \vert$, leading to an inequality of the form of (2), but with an additional constant c=n on the right side.
Best Answer
The formula is correct; one applies first the Cauchy-Schwarz inequality in $\mathbb{R}^n$ to write $$ |\nabla u\cdot \nabla v|\leq |\nabla u||\nabla v|, $$ and then we apply the integral version of C-S to the right hand side.