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$\newcommand\norm[1]{\left\|#1\right\|}\newcommand\inner[2]{\langle #1,#2\rangle}$ Let $u\in \dot{H}^1(\mathbb{R}^n)$ for $n\geq 3$ where $\dot{H}^{1}$ denotes the homogeneous Sobolev space that is the closure of smooth functions with compact support $C^{\infty}_c(\mathbb{R}^n)$ with respect to the norm $$\|u\|_{\dot{H^{1}}} = \|{\nabla u}\|_{L^{2}}.$$

We know that in general, we have the following inequality, $$\inner{u}{v} = \int u v \leq \|\nabla u\|_{L^2} \|v\|_{H^{-1}}$$ for $u\in \dot H^{1}$ and $v\in H^{-1}.$ However, I am trying to show an inequality of the form, $$\norm{u}_{\dot{H}^1}\leq C\norm{v}_{H^{-1}}^{\alpha }$$ where $u\in \dot{H}^{1}, v\in H^{-1}$ for some exponent $\alpha > 0.$ By definition of the dual space we know that $$\norm{v}_{H^{-1}} = \sup_{\norm{\nabla u}_{L^2}\leq 1}\inner{u}{v}$$ and so to lower bound $\norm{v}_{H^{-1}}^{\alpha}$ it seems like we have to lower bound the inner product (or the integral as defined above) which does not seem very natural.

So my question is whether there are interpolation estimates involving the dual norm similar to the usual interpolation inequality:

$$\int f^{a} g^{b} \leq \norm{f}_{L^p}^a \norm{g}_{L^q}^{b}$$

where $f\in L^p, g\in L^q$ with $a/p + b/q = 1.$ Here if by some chance the $L^p$ norm for instance was replaced by the dual norm then we could obtain a lower bound of the dual norm raised to a power in terms of the integral (without having to raise it to any power). Of course, this is just wishful thinking but if anyone is aware of results that might look like this, it will be much appreciated.

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  • $\begingroup$ What is the relation between $u$ and $v$ in the inequality you are trying to prove? $\endgroup$
    – fedja
    Commented Mar 6, 2021 at 13:43
  • $\begingroup$ $u$ is a function in $\dot{H}^{1}$ and $v = \Delta u + \lambda u + u^{p}$ for some $\lambda, p >0$ $\endgroup$
    – Student
    Commented Mar 7, 2021 at 17:15

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