I have the norms $$ ||f||_{L^{\infty,n}} = \sup_{x\in X}\left|f(x)\bigg(\frac{1}{1+||x||_2}\bigg)^n\right| \\ ||f||_{F^{n,1}} = \int_{\mathbb{R}^d}(1+||\xi||)^n|\hat{f}(\xi)|d\xi $$ with $X\subseteq \mathbb{R}^d$ for $d\in\mathbb{N}$, $\hat{f}$ the Fourier Transform of $f$ and $n\in\mathbb{N}$.
I would like to know whether
- $||f||_{L^{\infty,n}} < \infty \implies ||f||_{F^{n,1}} < \infty$
- $||f||_{F^{n,1}} < \infty \implies ||f||_{L^{\infty,n}} < \infty$
I have been able to show that 1. does not hold, since for $f(x)=\sin{(1/||x||)}$ it holds that $||f||_{L^{\infty,n}} = 1 < \infty$ but $\hat{f}$ does not exist.
So far I have not been able to show that 2. does or does not hold.
The things I think are useful for proving the statement or providing a counter example are
- $\widehat{f\star g} = \hat{f}\hat{g}$
- $\hat{\hat{f}}(x)=f(-x)$
- If $f$ has absolutely continuous $k$-th partial derivates, then $|\hat{f}(\xi)|\leq \frac{C}{(1+||\xi||)^k}$
- $||\hat{f}||_{L^\infty} = \sup_{\xi\in\mathbb{R}^d} |\hat{f}(\xi)|\leq ||f||_{L^1}$
I would like help with 2.. Could you help me prove the statement or provide a counter example?
I believe I have a counterexample for (2).
One important thing about the Fourier transform is that it trades smoothness for rapid decrease and vice versa. Note that $$\int (1+|\xi|)^n|\hat{f}(\xi))| \ d\xi < +\infty$$ implies that $\hat{f}(\xi)$ decreases more rapidly than $(1+|\xi|)^n$. This implies that $f$ will have some level of regularity, but not necessarily rapid decrease. On the other hand, we will have $$\sup_X\left| f(x)\left( \frac{1}{(1 + |x|)^n} \right) \right| < +\infty$$ if $f$ grows no faster than $(1+|x|)^n$. However, a rapidly decreasing Fourier transform only gives smoothness, so it doesn't seem like (2) should hold. In particular, we just want a smooth function that grows faster than $(1+|x|)^n$.
Let's look at the simplest case: $d=1$. I don't see that you specify what $X$ must be, so I'll assume that you have some freedom in choosing and take $X=\mathbb{R}$. Based on the above, it seems intuitively that $f(x) = x^{n+1}$ will be a good candidate for a counterexample. Then, $$\hat{f}(\xi) = 2\pi i^{n+1}\delta^{(n+1)}(\xi).$$ This tells us that $$\int (1+|\xi|)^n |\hat{f}(\xi)| \ d\xi = 2\pi \int (1+|\xi|)^n\delta^{(n+1)}(\xi) \ d\xi < +\infty.$$ However, $$\sup_\mathbb{R}\left| x^{n+1}\left( \frac{1}{(1+|x|)^n} \right) \right| = +\infty.$$
On the other hand, if $X$ is bounded I suspect that this might be true. Since $\|f\|_{F^{n,1}} < +\infty$ gives $f$ a certain amount of smoothness, it will be bounded on bounded sets.