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Given $ \Omega$ a compact subset of $\mathbb{R}^n$ and $f\in H^1(\Omega,\mathbb{C})$ with zero average, I wonder if there is an inequality of the form $$   \int_\Omega \phi(|f|^2)\varphi(|f|^2)\ dx   \leq C\left( \int_\Omega \phi(|f|^2)\ dx \right)^{1/a}      \left(\int_\Omega \varphi(|f|^2)\ dx \right)^{1/a} $$ where $a>1$, $C>0$ (might depend on $\Omega$) and $\varphi,\phi$ are real-valued and convex functions. Any idea is welcome, thanks for advance.

I checked this for the simple case of  $\varphi,\phi$ being constants and it works (it is enough to take $C\geq 1$).

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This kind of inequality cannot hold in general.

Indeed, suppose that $\phi(u)=\varphi(u)=u$ for all $u$. Let $f=tg$, where $t\in(0,\infty)$ and $g$ is a function in $H^1(\Omega,\mathbb{C})$ with zero average and such that $\int_\Omega |g|^4\,dx>0$. Then the left-hand side of your inequality is $t^4\int_\Omega |g|^4\,dx$ and the right-hand side is $Ct^{4/a}(\int_\Omega |g|^2\,dx)^{2/a}$, which is less than the left-hand side if (i) $a>1$ and $t$ is large enough or (ii) $0<a<1$ and $t$ is small enough.

Finally, this ineqiality cannot hold in general even for $a=1$. E.g., suppose that $\Omega=[-1,1]^n$, $\phi(u)=\varphi(u)=u$ for all $u$ (as before), and $f(x)=x_1^{2m+1}$ for natural $m$ and $x=(x_1,\dots,x_n)\in\Omega$. Then the left-hand side of your inequality is $2^n/(8m+5)$, and the right-hand side is $C\,2^{2n}/(4m+3)^2$, which is less than the left-hand side if $m$ is large enough.

So, your inequality fails to hold in general for any given real $a>0$.

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  • $\begingroup$ But althoug $t$ is large and K is small, the inequality holds for some small enough $a$, am I wrong? I mean, $$ t^{4-4/a}\leq CK^{2/a-1}$$ holds true for some small $a$ and large enough C, $\endgroup$ Sep 17, 2019 at 22:18
  • $\begingroup$ When I say small $a$, I mean $a$ close to $1$ in the previous comment. $\endgroup$ Sep 17, 2019 at 22:26
  • $\begingroup$ @R.N.Marley : In your question, $a>1$. However, as now shown, your inequality cannot hold in general for any given real $a>0$. $\endgroup$ Sep 18, 2019 at 2:51
  • $\begingroup$ Thank you for your answer. $\endgroup$ Sep 18, 2019 at 10:22

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