This is a practice problem for an applied analysis qualifying exam. Obviously I know how to prove the Poincare Inequality for a general $H_0^1(\Omega)$, but here they want you to use a specialized strategy to prove it in one dimension. I figured the problem out…I think. But I'm suspicious of my strategy because I got $c=1$. Am I missing something obvious/doing something illegal?
Prove the Poincare inequality: for any $u \in H_0^1(0,1)$
$$ \int_0^1 u^2 dx \leq c \int_0^1 (u')^2 dx$$
for some constant $c>0$.
Hint: Write $u(x) = \int_0^x u'(s)ds$, then square this identity.
Proof:
Let $u(x) = \int_0^xu'(s)ds \Rightarrow |u(x)| \leq \int_0^x |u(s)|ds$.
Taking the supremum of both sides wrt $x$ we have
$$||u||_\infty \leq \int_0^1 |u'(s)|ds.$$
We then square both sides (the involved quantities are nonnegative, so the direction of the inequality if preserved) and apply Cauchy Schwarz to obtain
$$||u||_\infty^2 \leq \bigg( \int_0^1 |u'(s)|ds\bigg) ^2 \leq \int_0^1(u'(s))^2ds \int_0^1 1^2 ds = ||u'||_{L^2(0,1)}^2.$$
Finally, we have
$$||u||_{L^2(0,1)}^2 = \int_0^1 |u|^2 dx \leq \int_0^1 ||u||_\infty^2 dx = ||u||_\infty^2 \int_0^1 1 dx = ||u||_\infty^2.$$
This combined with the work above gives the desired result that $||u||_{L^2(0,1)}^2 \leq c||u'||_{L^2(0,1)}^2$ for $c=1$.
Best Answer
You haven't exactly followed the hint, but your proof seems correct.
As pointed out by Chee Han, you could follow the hint by squaring the given identity (using the Cauchy-Schwarz inequality like you did), integrating from $0$ to $1$ and exchanging the order of integration. If I'm not mistaken, this would give a better constant: $c = \frac 12$.
This could be further improved by using $u(x) = \int_0^x u'(s)ds$ for $0 < x \le \frac 12$ and $u(x) = -\int_x^1 u'(s)ds$ for $\frac 12 \le x < 1$, i.e. using the shortest path from $x$ to the boundary of $(0,1)$.
If you're looking for the optimal constant $c = \frac{1}{\pi^2}$, it is attained for $u(x) = \sin(\pi x)$. This can be seen e.g. via the Fourier transform.