Define $\varphi(u) = \displaystyle\int_{0}^{1} \displaystyle\frac{{|u'| }^2}{2} - \displaystyle\frac{{u }^2}{4} - hu \ dt$, $u \in H^{1}_{0}(0,1)$ where $h: [0,1] \rightarrow R$ is a continuous functions. Show that $\varphi$ is not bounded below.
I tried to use basic inequalities like the AM - GM , Cauchy Schawrz and the imersion $H^{1}_{0}(0,1) \subset L^4(0,1)$. But these thing not worked. Someone can give me a hintto prove the affirmation?
Thanks in advance.
This is false; the functional is bounded from below. Begin with $h\equiv 0$ for simplicity: the functional
$$ \varphi(u) = \int_{0}^{1} \left(\frac{(u')^2}{2} - \displaystyle\frac{{u }^2}{4}\right)\,dt $$ is never negative. Indeed, the 2nd form of Wirtinger's inequality says that $$\int_0^1 u^2 \le \frac{1}{\pi^2}\int_0^1 (u')^2$$ hence $$ \varphi(u) \ge \left( \frac12 - \frac{1}{4\pi^2}\right) \int_0^{1} \frac{(u')^2}{2} \tag{1}$$
With general $h$, we can bound the term with $hu$ by $-\|h\|_2\|u\|_2 $, which, being linear in $u$, is still dominated by the right side of (1).