I am tasked to find a constant $c>0$ such that for all $C^1$ functions in $(0, 1)$ this variational problem is true:
$$cu(0)^2 \leq \int^1_0 u'^2 + u^2 dt$$
My instinct told me to calculate the minimizer of that funcitonal, and I got that it must be of the form $c_1e^t + c_2e^{-t} = \bar{u}$, where $\bar{u}$ is the minimizer.
Then we can compute the minimum quite easily in the general case and get that $c_1(e^{2} - 1)$ is a lower bound for the functional.
But all this allows me to state is that:
$$c_1(e^{2} - 1) \leq \int^1_0 u'^2 + u^2 dt$$
I am not sure ow to introduce the $u(0)$ term.
Using the fundamental theorem of calculus,
$$u (0) = -\int_0^t u'(s)ds + u(t) ,$$ thus by Holder inequality, \begin{align} u^2(0) &\le 2\left( \int_0^t u'(s)\right)^2 ds + 2u^2(t) \\ & \le 2\left(\int_0^t 1 ^2 ds \int_0^t u'^2(s) ds\right) ds + 2u^2(t) \\ &\le 2\int_0^1 u'^2(s) ds+2 u^2(t) \end{align}
Integrating with respect to $t$ from $0$ to $1$,
$$u^2(0) \le 2\int_0^1 u'^2(s)+2 \int_0^1 u^2(t)dt.$$