Given $g$, a continuous real-valued function defined in $\mathbb R^n$,
$K^g:=\{u:e^g \hat{u}\in L^2(\mathbb R^n,(1/2 \pi)^nd \xi)\}$.
Thus $K^g$ can be viewed as a Hilbert space with natural norm $||u||_{K^g}=||e^g\hat{u}||_{L^2(\mathbb R^n,(1/2 \pi)^nd \xi)}$
Prove that if
$g(\xi)/\log(1+|\xi|)$ tends to $\infty \quad$ as $|\xi|\rightarrow \infty$,
then $K^g$ is continously embedded in $C^{\infty}(\mathbb R^n)$
Here is my attempt:
If suffices to show $\forall$ compact $F$ and nonnegative integer $m$,
$\quad \Sigma_{|\alpha|\le m}\max_{x \in F}|D^{\alpha} u(x)| \le C_{m,F}||u||_{K^g}$, where $C_{m,F}$ is a nonnegative constant depending only on F and m. $\quad $With the help of Fourier inverse formula, what I have in mind is
$\quad \max_{x \in \mathbb R^n}|D^{\alpha} u(x)| \le ||\space|\xi|^{\alpha}|\hat{u}(\xi)|\space||_{L^1(\mathbb R^n, (1/2 \pi)^nd \xi)}$.
I also realized that $e^{g(\xi)}=(1+|\xi|)^{A(\xi)}$, where $A(\xi)=g(\xi)/\log(1+|\xi|)\rightarrow \infty$ as $\xi \rightarrow \infty$.
My problem is that I cannot deal with the case $|\xi|$ is not large enough.
I'm not sure whether I am in the right direction. Any comment or answer would be greatly appreciated.
Only large frequencies matter for smoothness. For every $M$, the part of Fourier transform with $\{\xi:|\xi|\le M\}$ contributes a real-analytic term to the function.
You know that integrability of $|\xi|^\alpha \hat u(\xi)$ implies certain smoothness of $u$. So you want to show that this product is integrable for every $\alpha$. On every ball $\{|\xi|\le M\}$ it is integrable simply because $$ \int_{|\xi|\le M}|\xi|^\alpha |\hat u(\xi)| \le M^\alpha \sqrt{\int_{|\xi|\le M}|\hat u(\xi)|^2 \int_{|\xi|\le M}|\xi|^{2\alpha} } <\infty $$ On the set $|\xi|\ge M$ you have your comparison, where $|\xi|^\alpha$ gets dominated by $e^g$.