I am trying to prove the following:
Let $u \in W^{2,2}(\mathbb R)$. Then $\| u^\prime \|_{L^2}^2 \leq \| u \|_{L^2} \| u^{\prime \prime} \|_{L^2}$ holds (these are meant to be weak derivatives).
As a first step to tackle the problem I want to prove it for testfunctions $\psi \in C_0^\infty(\mathbb R)$.
Using partial integration it holds: $\displaystyle \| \psi\,^\prime \|_{L^2}^2 = \int_{-\infty}^{+\infty} | \psi\,^\prime |^2 dx = \int_{-\infty}^{+\infty} \psi\,^\prime \psi\,^\prime dx = [\psi\,^\prime \psi]_{-\infty}^{+\infty} - \int_{-\infty}^{+\infty} \psi \psi\,^{\prime \prime} dx$.
As $\psi$ and $\psi\,^\prime$ both have compact support, the the term $[\psi\,^\prime \psi]_{-\infty}^{+\infty}$ vanishes and we are left with: $\displaystyle \| \psi\,^\prime \|_{L^2}^2 = - \int_{-\infty}^{+\infty} \psi \psi\,^{\prime \prime} dx$.
(1) Does this inequality have a name?
(2) Isn't it weird, that there is a minus sign? On the left hand side we have a norm!
(3) I have read that this proof continues using Hölder's inequality. But to use that we would require $\displaystyle \int_{-\infty}^{+\infty} |\psi \psi\,^{\prime \prime}| dx$ (with the absolute value!).
Can anyone help me out with these to problems? Thanks.
Once you have: $$ \left\| \psi' \right\|_{L^2}^2 = -\int_\mathbb{R}\psi\psi''\,dx \tag{1}$$ the inequality: $$ \left\| \psi' \right\|_{L^2}^2 \leq \left\|\psi\right\|_{L^2}\cdot\left\|\psi''\right\|_{L^2}\tag{2}$$ just follows from the Cauchy-Schwarz inequality. Nothing strange in the minus sign in the RHS of $(1)$: just think to any non-negative test function that is concave ($\psi''<0$) over its support.
The general version of this inequality is known as the Gagliardo-Niremberg interpolation inequality.