I cannot prove that $f \notin L^\infty(\mathbb{R}^n)$. $f$ is defined as follows:

Question

I cannot prove that $f \notin L^\infty(\mathbb{R}^n)$. $f$ is defined as follows: $$ \widehat{f}(\xi) := \frac{1}{(1+|\xi|)^n \log(2+|\xi|)}. $$

It is easy to check that $\widehat{f}(\xi) \in L^2(\mathbb{R}^n)$, and $\widehat{f}$ is $L^2$-Fourier transform of $f$. I found this $f$ in p.49, "Introduction to Nonlinear Dispersive Equations", F.Linares and G.Ponce.

Answer 1

In fact,

Prop. If $g\in L^2(\Bbb R^d)$, $g\ge0$ and $\int g=\infty$ then $\hat g\notin L^\infty$.

Hint: It follows from Plancherel that $$\int\widehat g(\xi)\frac {y}{(|\xi|^2+y^2)^{(d+1)/2}}\,d\xi=\int g(x)e^{-y|x|}\,dx.$$(There may be some constants missing there that I leave to you to get straight.) Let $y\to0$.

(If you're not familiar with the Poisson kernel and/or its Fourier transform you could see here or you could use something else like a gaussian or a Schwarz function...)

Answer 2

Thank you very much. I got it.

I don't think it is easy to show that$$(e^{-y|x|})\hat{}(\xi) = C_d \frac{y}{(|\xi|^2+y^2)^{(d+1)/2}}.$$ However, we can prove the Proposition without calculating above Fourier transform. (Of course, it is essentially same.)

We know that there exists $\psi \in \mathcal{S}$ such that $\psi(0)=1, 0 \leq \psi \leq 1, \psi$ is radial and monotone decreasing with radius. (E.g. $\psi(x)=e^{-|x|^2}.$) Set $\varphi = \check{\psi}$ and $\varphi_\varepsilon(x) = \varepsilon^{-d}\varphi(x/\varepsilon).$

Assume that $\widehat{g} \in L^\infty(\mathbb{R}^d).$ Then, for any $\varepsilon >0,$ $$ \int g(x)\widehat{\varphi_\varepsilon}(x) dx = \left| \int \widehat{g}(x) \varphi_\varepsilon(x) dx \right| \leq \|\widehat{g}\|_\infty \|\varphi\|_1. $$

On the other hand, by monotone convergence theorem,

$$ \int g(x)\widehat{\varphi_\varepsilon}(x) dx = \int g(x) \psi(\varepsilon x) dx \to \infty (\varepsilon \downarrow 0). $$

This is a contradiction.