1

I have $s\in\mathbb{R}$ and I'm looking at the space $H^s$ of functions $f$, such that $$\lVert f\rVert_{H^s}^2:=\int_\mathbb{R} (1+\lvert x\rvert^2)^s\lvert f(x)\rvert^2\,dx<\infty.$$ I want to know if this norm is equivalent to $$\lVert f\rVert_{s,2}^2:=\int_\mathbb{R} (1+\lvert x\rvert)^{2s}\lvert f(x)\rvert^2\,dx,$$ or at least $\lVert f\rVert_{H^s}\le\lVert f\rVert_{s,2}$.

I would especially like to know if this also holds for negative values of $s$.

My intuition is that this is easy, since we can look at $$\frac{(1+\lvert x\rvert^2)^s}{(1+\lvert x\rvert)^{2s}}$$ which is $1$ for $x=0$, converges to $1$ for $s>0$ if $\lvert x\rvert\to\infty$ and therefore also for $s<0$.

PTO2
  • 11

1 Answers1

1

Yes, the norms are equivalent. All you have to show is that there exist $C_1$, $C_2 > 0$ such that $$ C_1 (1+|x|)^2 \leq 1+|x|^2 \leq C_2 (1+|x|)^2 \quad \forall x \in \mathbb R. $$ This relation holds for e.g. $C_1 = \frac{1}{2}$ and $C_2$ = 1 (*). Then $$C_1^s \| f \|_{H^s}^2 \leq \| f \|_{s,2}^2 \leq C_2^s \| f \|_{H^s}^2 \quad \text{for $s > 0$}, $$ $$C_2^s \| f \|_{H^s}^2 \leq \| f \|_{s,2}^2 \leq C_1^s \| f \|_{H^s}^2 \quad \text{for $s < 0$}. $$

(*) To see $C_1 = \frac 1 2$ use that for $a, b \in \mathbb R$ there holds $$(a+b)^2 = a^2 + b^2 + 2 ab \leq a^2 + b^2 + 2 \left(\frac 1 2 a^2 + \frac 1 2 b^2\right) = 2(a^2 + b^2)$$ by Young's inequality.

Three.OneFour
  • 1,772