Let $f: \mathbb{R}^n \rightarrow \mathbb{C}$ be in $L^2\cap L^1,$ then the Fourier transform is in $L^2 \cap L^\infty.$

Does this imply that we can take common norms in the sense that we can estimate

$$\sup_{x_i \in \mathbb{R}} \sqrt{\int_{\mathbb{R}^{n-1}} \left\lvert \hat{f}(x_1,...,x_i ,...,x_n) \right\rvert^2 dx_1...dx_{i-1}dx_{i+1}...dx_{n}}$$ or alternatively $$ \sqrt{\int_{\mathbb{R}^{n-1}}\sup_{x_i \in \mathbb{R}} \left\lvert \hat{f}(x_1,...,x_i ,...,x_n) \right\rvert^2 dx_1...dx_{i-1}dx_{i+1}...dx_{n}}?$$

Maybe this can even be bounded by mixed norms

$$\int_{\mathbb{R}} \sqrt{\int_{\mathbb{R}^{n-1}} \left\lvert f(x_1,...,x_i ,...,x_n) \right\rvert^2 dx_1...dx_{i-1}dx_{i+1}...dx_{n}}dx_i$$ or

$$ \sqrt{\int_{\mathbb{R}^{n-1}}\left(\int_{\mathbb{R}} \left\lvert f(x_1,...,x_i ,...,x_n) \right\rvert dx_i\right)^2 dx_1...dx_{i-1}dx_{i+1}...dx_{n}}?$$

However, I do not see yet whether this is true or not.


1 Answer 1


Yes, such bounds follow by considering partial Fourier transforms and then using the norm bounds you quoted. Let me take $n=2$ for ease of notation. Write $g(x,t)=\int f(x,k) e^{2\pi i kt}\, dk$. Then, for example (and as desired), $$ \sup_y \|\widehat{f}(x,y)\|_{L^2(dx)} =\sup_y \|g(x,y)\|_{L^2(dx)}\le \left\| \int |f(x,k)|\, dk \right\|_{L^2} \le \int\|f(x,k)\|_{L^2(dx)}\,dk , $$ by the integral version of Minkowski's inequality for the final step.


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