$G(f) \le \|f\|_{H^s(\mathbb R)},\; s>2 \Rightarrow G(f) \le \|f\|_{H^2(\mathbb R)}$?

Question

If a quantity of a function $f$, call $G(f)$ satisfies $$ G(f) \le \|f\|_{H^s(\mathbb R)} $$ for all $s>2$, then can I conclude that this holds for the limiting case $s\to 2$: $$ G(f) \le \|f\|_{H^2(\mathbb R)} $$ as well? Here $H^s$ denotes the usual Sobolev space.

Answer 1

The short answer is no.

It is possible that $$ \|f\|_{H^s}=\infty $$ for all $s>2$ and $$ \|f\|_{H^2}<\infty. $$ Take for example the function $f$, with $$ \hat f(\xi)=\frac{1}{(1+\xi^2)^{3/2}(1+\log(\xi^2+1))}. $$

Note. If $\|f\|_{H^{s_0}}<\infty$, for some $s_0>2$, then $$ \lim_{s\searrow 2}\|f\|_{H^s}=\|f\|_{H^2}. $$

Answer 2

Okay, I think this works.

Since the function $a^x$ is convex as a function of $x$ we have, putting $a=(1+|\xi|^2)$ that

$$ (1+|\xi|^2)^r\leq (1-\theta)(1+|\xi|^2)^t +\theta (1+|\xi|^2)^s, \qquad r=(1-\theta)t+\theta s. $$ Multiplying both sides by $|\hat{f}|^2$ and integrating over $\mathbb{R}$ we get that the function $h(t)=\| f\|_{H^t}^2$ is convex on an interval where it's finite. In particular, it's continuous there. Your result follows from this as long as you know that $\| f\|_{H^s}$ is finite for some $s>2$ at least.