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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

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    Are the exponents correct? You have the same term $\left(\int_Mu^2\right)^2$ appearing on both sides of the inequality. – 2017-02-20
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    @MartinR Thank you for your remark. I will correct directly – 2017-02-20
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    @MartinR Now they are correct – 2017-02-20
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    @AnthonyCarapetis – 2017-02-20
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    Why did you remove your question before? Exactly the same question, asked again. – 2017-02-20
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    @mickep I'm trying to show it in a simpler form – 2017-02-20
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    If you don't have any additional constraints, Hölder's inequality is the only relationship between the $L^p$ norms. Scaling the metric shows your first suggestion cannot be true - you definitely need $C$, which will be some power of the volume. Write $u^2 = |u|^a |u|^b 1$, apply Hölder with exponents $p,q,r$ and solve the constraint equations $a+b=2, 1/p + 1/q + 1/r = 1, ap = 4, bq = 1, \rm etc.$ In this case I don't think it works - the power of 2 on the $L^1$ norm is too much. – 2017-02-20

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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.

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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 $0$.

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    please can you explain a liitle bit your answer; whatever you can admit the condition $\int_M u=0$ – 2017-02-20
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    Imagine a function that takes the value 1 on some finite set of your manifold, and -1 on a set with the same measure, then $\int_{M} u=0$, but $\int_M u^2 \neq 0$ – 2017-02-20
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    I'm so sorry for this mistake, the function $u$ must be in absolute value. I correct it – 2017-02-20