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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?
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