Good morning,
I want to prove that: $$\left(\int_Mu^2\right)^2\leq C\left(\int_Mu^4\right)^{1/2}\left(\int_M|u|\right)^2$$ where $u\in H_1^2(M):=\{f\in L^2(M):|\nabla f|\in L^2(M)\}$ and $M$ is a compact smooth Riemannian manifold of dimension $n$;
I have tried Holder's inequality, but I didn't got it. any idea please
Any help will be appreciated. Thank you
On
The exponents here just don't add up - as I mentioned in my comment, you can't make Hölder's inequality give you this, and thus it can't be true. To show this rigorously, let $u$ be an indicator function $1_E$ where ${\rm vol}(E) = V$. Plugging this in, we get $V^2$ on the LHS but $V^{5/2}$ on the RHS; so letting $V \to 0$ shows your inequality cannot be true for any finite $C$.
These $u$ are not smooth, of course; but if the inequality was true for all $u \in C^\infty$ it would also be true for all $u \in L^4$ by smooth approximation.
I'm not sure this is true, take any function such that $\int_M u =0$ (except $u$ the zero function), then you're getting some positive number is bounded above by $0$.