Let $\Omega\subset \mathbb{R}^N$ and $H_0^1(\Omega)$ the standard Sobolev space. Assume that $1<q<p<2^\star$ and $$\mathcal{S}=\{u\in H_0^1(\Omega):\ \|u\|=1\}.$$

Define $C_q,C_p$ by $$C_q=\inf_{u\in \mathcal{S}}\frac{1}{\|u\|_q},$$

and $$C_p=\inf_{u\in \mathcal{S}}\frac{1}{\|u\|_p}.$$

Once $q<p<2^\star$, we can assume the existence of $u_q,u_p\in \mathcal{S}$ such that $$C_q=\frac{1}{\|u_q\|_q}, C_p=\frac{1}{\|u_p\|}.$$

My question is the following:

Can we have $u_q=u_p$, or equivalently, does the maximum of the function $\|u\|_q\|u\|_p$ over $\mathcal{S}$ is attained for any function, which maximizes both $\|u\|_q$ and $\|u\|_p$ in the same time?

My guess by now is that this is not true, however, I have no idea to how to prove it. Any reference on this matter is appreciated.

I have asked the same question here.


1 Answer 1


Any maximizer $u\in\{ H^1_0(\Omega): \|\nabla u\|_2=1 \}$ of $\|u\|_p$ is a nonconstant, nonnegative function solving $-\Delta u = \lambda u^{p-1}$, with $\lambda=\lambda_p>0$. So if $u$ maximizes both $\|u\|_p$ and $\|u\|_q$, then $\lambda_p u^p=\lambda_q u^q$ a.e., which is only possible if $p=q$.

  • 1
    $\begingroup$ I see. If $u$ is a maximizer, we can use the Lagrange multiplier theorem to infer that $-\Delta u=\lambda |u|^{p-2}u$. Then, an argument with the test function $v=u^-$ yields that $u\ge 0$. The rest is as you said. Thank you. $\endgroup$
    – Tomás
    Aug 7, 2016 at 17:13
  • $\begingroup$ Exact (more precisely, I guess I should have said "u has constant sign", so u can be assumed nonnegative w.l.o.g.) $\endgroup$ Aug 7, 2016 at 22:15

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