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I need to show $|f|_{L^\infty}\leq c|f|_{H^2} = c(\int_{\mathbb R^n} (1+|\xi|^2)^2|\hat f(\xi)|^2 d\xi )^{1/2}$, assume $f\in H^2(\mathbb R^2)$

I think I can trasnfer $f\ = \int \hat f(\xi)e^{2\pi i \xi\cdot x}d\xi$, and $|f(x)|=\int|\hat f(\xi)|d\xi$ and use Cauchy-Schwarz to get the $H^2$ norm. But I encounter a troubles:

  1. the infinity norm is the esssup $|f|$, but this representation kind of far away from this inequality. Is there any other way to represent the infinity norm?
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I don't think this is true for arbitrary $n$. For instance, consider a function that behaves like $1/|x|$ near $x=0$. For large enough $n$, you can ensure that this function and several of its derivatives are all in $L^2$, but it isn't bounded. – Nate Eldredge Dec 14 '12 at 21:38
fixed, is that correct? – Zhixia Zhang Dec 14 '12 at 21:43
Isn't that $|f|_{H^s}^s = \int (1+|\xi|^2)^s|\hat f(\xi)|^2d\xi$ – Zhixia Zhang Dec 14 '12 at 21:49
Sorry, yes, it looks correct now. – Nate Eldredge Dec 14 '12 at 21:52
Sorry, my bad, dimension =2, i did not notice this line – Zhixia Zhang Dec 14 '12 at 21:55
up vote 0 down vote accepted

Take a look in the page 272, Proposition 1.5 from this book

In the demonstration, the author shows that $H^{1+\alpha}(\mathbb{R}^2)$ is continuously immersed in $C^\alpha(\mathbb{R}^2)$ for $\alpha\in (0,1)$. On the other hand, $H^2(\mathbb{R}^2)$ is continuously immersed in $H^{1+\alpha}(\mathbb{R}^2)$ and $C^\alpha(\mathbb{R}^2)$ is continuously immersed in $C^0(\mathbb{R}^2)$, hence you can conclude.

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Sorry, what is the relation here with $L^\infty$? – Zhixia Zhang Dec 15 '12 at 1:45
In this case $|f|_{\infty}=\sup_{x\in\mathbb{R}^2}|f(x)|$ – Tomás Dec 15 '12 at 10:33
But $|f(x)|$ can go to infinity on a measure 0, while the infinity norm is finite. – Zhixia Zhang Dec 15 '12 at 21:37
@TianyiXia, Im using a abusive notation but here $C^{\alpha}(\mathbb{R}^2)$ and $C^0(\mathbb{R}^2)$ mean the space of functions $f$ such that $\|f\|_\alpha<\infty$ and $\|f\|_0<\infty$ respectivley, where $\|f\|_\alpha<\infty$ is the Holder norm and $\|f\|_0<\infty$ is the sup norm. – Tomás Dec 18 '12 at 14:42

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