Let $f \in H^1(0,\pi)$, show that $\int^{\pi}_{0} f^2(x)dx \le \int^{\pi}_{0} (f')^2(x)dx + (\int^{\pi}_{0} f(x)dx)^2$.
$H^1(0,\pi) = W^{1,2}(0,\pi)$, Sobolev space. The question is from PDE course, but I guess it's true for all functions as long as integrals in the inequality are finite.
I tried IBP, Holder inequality, but cannot get the answer. Would appreciate any suggestions where to start. Thanks!
I haven't worked out your problem but it reminds me of Poincare type inequalities. Below is from Partial Differential Equations in Action by Salsa, and I have seen this called Poincare-Wirtinger inequality elsewhere:
Let $\Omega$ be a bounded Lipschitz domain and $ E \subset \Omega$, with $|E|\gt 0$. If $f \in H^1(\Omega)$ let $$ f_E = \frac{1}{|E|} \int_E f \,dx $$ and $ w = f-f_E $. Then $ \int_\Omega w =0$ and we get ( where $C\gt0$ )
$$ \| f - f_E\|_{L^2(\Omega)} \leq C \|\nabla f\|_{L^2(\Omega)}. $$
Square both sides and expand the LHS and you should have all the terms you need. Hope this helps.