# Integral inequality issue

Given that the bilinear form on the set $$V:= \{v\in C^2[0,1] v(0)=0=v(1)\}$$ is defined as $$[u,v]=\int_0^1 [pu'v'+quv]dx,$$

where $$p\in C^1[0,1], p(x)\ge p_0>0, q(x)\ge 0, q\in C[0,1]$$,

I want to show that $$[u,u]\ge p_0\|u\|_\infty$$.

Here's how I proceed:

$$[u,u]=\int_0^1 pu'^2dx+\int_0^1 qu^2dx\ge \int_0^1 pu'^2dx>p_0\int_0^1 u'^2dx\ge p_0\left(\int_0^1 u'dx \right)^2\\ = p_0 (u(1)-u(0))=0$$

I don't see how to show that $$\|u\|_\infty\le |u'|$$. How does one show this? I definitely would appreciate some help.

Update: I think one can proceed as follows:

$$p_0\|u\|^2_\infty=p_0\left\|\int_0^x u'dx\right\|^2_\infty\le p_0\left\|\int_0^x u'^2dx\right\|_\infty\le p_0\left\|\int_0^x u'^2dx\right\|_\infty\\ \le p_0\int_0^1 u'^2dx<\int_0^1 p(x)u'^2dx\le \int_0^1 p(x)u'^2dx+\int_0^1 q(x)u^2dx=[u,u]$$. Please let me know if you think is correct.

The statement is false. Take $$u(x)=x^2(x-1)$$, $$p=1$$, and $$q=0$$. Then $$u\in V$$ but $$||u||_\infty=4/27>2/15=\int_0^1 u'^2=[u,u]$$.