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Given $V$ an inner product space with norm $(‖v‖_V)^2$=$∫_Ω(v^2 (x)+|∇v|^2 )dx$. Prove that

$$(∫_Ω(|v||w|+|∇v||∇w|)dx)^2 ≤ ∫_Ω(|v|^2+|∇v|^2 )dx ∫_Ω(|w|^2+|∇w|^2 )dx=(‖v‖_V)^2(‖w‖_V)^2.$$

Any idea to prove this inequality??

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1 Answer 1

This is the Cauchy-Schwarz inequality applied to $|v|$ and $|w|$ in your specific inner-product space. It can be proved generically. Another standard way is to observe that the quadratic $\|x+ty\|^2$ in $t\in\mathbb{R}$ must have nonpositive discriminant, since it is nonnegative on $\mathbb{R}$. This is for the real case. Do the same trick in the complex case, but consider $\|x+te^{i\theta}y\|^2$ for any, or the appropriate, $\theta$.

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