Non-standard form of Sobolev's embedding: Does $\Vert f\Vert_{L^2}^2\leq C \Vert f\Vert_{L^\infty}\Vert f\Vert_{H^1}$ hold??

Question

I am wondering exactly the question of the title. Is it true that there exists a constant $C>0$ such that for all $f\in H^1(\mathbb{R})$ it holds $$ \qquad \qquad \Vert f\Vert_{L^2}^2\leq C \Vert f\Vert_{L^\infty}\Vert f\Vert_{H^1} \quad ?? \qquad (*) $$ I know that by Sobolev's embedding we have that $$ \Vert f\Vert_{L^\infty}\leq C \Vert f\Vert_{H^1}, $$ and of course we have the trivial bound $\Vert f\Vert_{L^2}\leq \Vert f\Vert_{H^1}$. However, it is not clear to my if something like the inequality in $(*)$ holds. Does anyone has any thoughts?

Answer 1

Such an inequality can't hold for scaling reasons: Let, for $a>0$, $f_a(x):= a^{-1/2}f(x/a)$. Then $\| f_a\|_{L^2}=\| f\|_{L^2}$, $\| f_a\|_{L^\infty}= a^{-1/2}\| f\|_{L^\infty}$, and $\| f_a\|_{H^1}= \| f_a\|_{L^2} + \| (f_a)'\|_{L^2}= \| f\|_{L^2} + a^{-1}\| f'\|_{L^2}$. Letting $a\to \infty$ we reach a contradiction if such an inequality held.

In fact this argument shows that if you want to replace the $L^\infty$ norm by an $L^q$ norm, then necessarily $q\leq 2$.

Answer 2

No, this inequality cannot hold for any $f\in H^1(\mathbb{R})$.

When you wonder if such a functional inequality can hold, a standard test is the scaling argument. Here is how it goes.

Let $f\in H^1(\mathbb{R})$ be non-identically $0$ and define $f_\lambda(x) = \sqrt{\lambda}f(\lambda x)$ with $\lambda>0$. Then one easily computes $\|f_\lambda\|_{L^2} = \|f\|_{L^2}$, $\|f_\lambda\|_{L^\infty} = \sqrt{\lambda}\|f\|_{L^\infty}$ and $\|f_\lambda'\|_{L^2} = \lambda\|f'\|_{L^2}$. By contradiction assume that $(\ast)$ hold. Then since $f_\lambda\in H^1(\mathbb{R})$, it must also satisfy $(\ast)$. Therefore $$ \|f\|_{L^2}^2 \leq C\,\sqrt{\lambda}\|f\|_{L^\infty}(\|f\|_{L^2}^2 + \lambda^2\|f'\|_{L^2}^2)^{1/2} $$ By letting $\lambda\to 0$ we obtain a contradiction.