# Proving inequalities with calculus of variations

I want to prove Friedrichs' inequality: if $$y \in C^1[0, l]$$ and $$y(0)=y(l)=0$$, then $$\int_0^l y(x)^2 dx \leq C \int_0^l y'(x)^2dx.$$

I need to prove it with most optimal constant $$C$$. My approach is as follows:

1. I prove the inequality with some non-optimal constant $$C = C_0$$
2. I rewrite the inequality as $$\frac{1}{C} \leq \frac{\int_0^l y'(x)^2dx}{\int_0^l y(x)^2dx},$$ notice that WLOG $$\int_0^l y(x)^2 dx = 1$$ since $$RHS[\alpha y] = RHS[y]$$ for $$\alpha \in \mathbb R$$. So I get a conditional extrema problem $$J[y] = \int_0^l y'(x)^2dx \to min, \;\;\;\; G[y] = \int_0^l y(x)^2 dx = 1.$$

My questions are:

1. If I find the unique extremal $$\hat y$$ of this variational problem and prove that it is a local minimum, can I conclude that $$\hat y$$ is actually a global minimum?

I'd thought so since the functional $$J[y]$$ has a lower-bound $$1/C_0$$, and the only way it can have a unique extremal is that the extremal is a global minimum (not a saddle or a maximum because of a known lower-bound).

1. Can I conclude the same without proving that $$\hat y$$ is a minimum of any kind (just some extremal) by the same argument?