Let $\phi$ be a function and $\phi \in C^{\infty}(\mathbb{R}_{+},\mathbb{R})$ with compact support and $\mbox{supp }{\phi} \subset [0, \infty)$.
I want to prove that: $$\phi^{2}(0) \leq \|\phi\|^2_{L^{2}}+\|\phi'\|^{2}_{L^{2}}.$$
Someone give an indication that I should begin with:
$$\phi(x)-\phi(0)=\int_{0}^{x}{\phi'(t)}\mbox{dt}$$ and then to prove that:
$$\phi^{2}(0) \leq \int_{0}^{\infty}{|2\phi(x)\phi'(x)|dx} \mbox{ . }$$
Thanks :)
First notice $2\phi \phi' = (\phi^2)'$: $$\phi^2(\infty) - \phi^2(0) = \int^\infty_0 2\phi(x)\phi'(x)\,dx.$$ Compactly supportedness drives $\phi\to 0$ when $x\to \infty$, so: $$ \phi^2(0) = \left|\int^\infty_0 2\phi(x)\phi'(x)\,dx\right| \leq \int_{0}^{\infty}|2\phi(x)\phi'(x)|\,dx $$ Lastly simply using $2ab\leq a^2 + b^2$: $$ \int_{0}^{\infty}|2\phi(x)\phi'(x)|\,dx \leq \int_{0}^{\infty}(|\phi(x)|^2 + |\phi'(x)|^2)\,dx = \|\phi\|^2_{L^{2}}+\|\phi'\|^{2}_{L^{2}}. $$
