Uniform convex space Please I want to know if this space 
$$H^1_{0,p}([0,+\infty))=\lbrace u, u\in AC([0,+\infty)), u(0)=u(+\infty)=0,\sqrt{p}u'\in L^2\rbrace$$
where $p>0$, $p\in L^1((0,+\infty))$ 
$$||u||^2=\int_0^{+\infty} p(t) (u'(t))^2 dt $$
is uniformly convex ?
Please 
Thank you.
 A: Consider $\newcommand{\inner}[2]{\langle #1 \mspace{-3mu}\mid\mspace{-3mu} #2\rangle}$
$$\inner{u}{v} = \int_0^\infty u'(t)v'(t)p(t)\,dt.\tag{1}$$
$\inner{\cdot}{\cdot}$ is a bilinear form, and $\inner{u}{u} = \lVert u\rVert^2$ is zero only for $u \equiv 0$. Hence $\inner{\cdot}{\cdot}$ is an inner product inducing the norm.
Every space whose norm is induced by an inner product is uniformly convex, since the parallelogram identity
$$\lVert u+v\rVert^2 + \lVert u-v\rVert^2 = 2\lVert u\rVert^2 + 2\lVert v\rVert^2\tag{2}$$
holds. If we take $\lVert u\rVert = \lVert v\rVert = 1$ in $(2)$, we find
$$\left\lVert \tfrac{1}{2}(u+v)\right\rVert = \sqrt{1 - \left\lVert\tfrac{1}{2}(u-v)\right\rVert^2},$$
which shows the uniform convexity.
For the (equivalent) definition of uniform convexity where $\lVert u\rVert,\lVert v\rVert \leqslant 1$ is the criterion, note that then
$$\left\lVert\tfrac{1}{2}(u+v)\right\rVert = \sqrt{\tfrac{1}{2}(\lVert u\rVert^2 + \lVert v\rVert^2) - \left\lVert\tfrac{1}{2}(u-v)\right\rVert^2} \leqslant \sqrt{1 - \left\lVert\tfrac{1}{2}(u-v)\right\rVert^2},$$
so $\lVert u\rVert,\lVert v\rVert \leqslant 1$ and $\lVert u-v\rVert \geqslant \varepsilon$ implies
$$\left\lVert\tfrac{1}{2}(u+v)\right\rVert \leqslant \sqrt{1-\frac{\varepsilon^2}{4}}.$$
