Why $\|u\|_{L^{2}}\|v\|_{L^{2}}+\|v'\|_{L^{2}}+\|u'\|_{L^{2}}≤\|v\|_{H^{1}}\|u\|_{H^{1}}$

Question

I see this steps from a book !

$I=]0,1[$ defined on $H^{1}$

$$a(u,v)=\int_{I}u'v'dx+\int_{I}u\cdot\int_{I}v$$

So in solution I find this steps :

\begin{align*} \|u\|_{L^{2}}\|v\|_{L^{2}}+\|v'\|_{L^{2}}+\|u'\|_{L^{2}}&\le(\|v\|_{L^{2}}+\|v'\|_{L^{2}})(\|u\|_{L^{2}}+\|u'\|_{L^{2}})\\ &≤\|v\|_{H^{1}}\|u\|_{H^{1}}\end{align*}

Also why is

$$|v(0)|≤\|v\|_{L^{\infty}}\le\|v\|_{L^{2}}?$$

I know that:

$$\|u\|_{H^{1}(\Omega)}=\sqrt{\int_{\Omega}(|u|^{2}+|\nabla {u}|^{2})\,dx}$$

Also:

$$\|u\|_{L^{2}(\Omega)}=\sqrt{\int_{\Omega}|u|^{2}\,dx}.$$

$$\Omega\subset{\mathbb{R^{n}}$$ But I can't see this inequality!?

Answer 1

$|v(0)| \leq ||v||_{L^{\infty}}$ because $||v||_{L^{\infty}}$ is the maximum value $|v(x)|$ can take.

I think in fact also $||v||_{L^2} \leq ||v||_{L^{\infty}}$ because $||v||_{L^2} = (\int_0^1 v(x)^2 )^{\frac{1}{2}} \leq (\int_0^1 ||v||_{L^{\infty}}^2)^{\frac{1}{2}} = ||v||_{L^{\infty}}$