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We know that a function in $L^1(\mathbb{R}^d)\cap L^2(\mathbb{R}^d)$ and its Fourier transform have the same $L_2$ norm. I wonder if there is any result about the function's Fourier transform's $L_p$ norm, for $p \in [1,\infty)$. We may assume that the function is $L_p$ integrable for any $p$.

In case no exact equality is known, any inequality that bounds the $L_p$ norm from above and below would be great.

Edit: David has given the result for upper bounding the Fourier transform in $p\in(1,2]$, a result known as the Hausdorff-Young inequality. It would be great to have some result for $p>2$ and lower bounds as well.

Thanks in advance.

user2130010
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  • The FT is a bounded operator on $L^{2}$ of norm 1 and it is also a bounded operator from $L^{1}$ to $L^{\infty}$ with norm at most 1. Using the so-called Interpolation Theorems we can prove that for any $p \in (1,2)$ there is a q such that the FT is a bounded operator from $L^{p}$ into $L^{q}$ and we can also get a bound on the operator norm. This gives an inequality of the type $||g||_q \leq C ||f||_p$ for $f \in L^{p}$, g being the FT of f. Here the constant C depends only on p. – Kavi Rama Murthy Feb 05 '18 at 07:20
  • Right. But more is true: What you say is correct for $1/p+1/q=1$, not just for some $q$. And in fact if $p\ge 2$ and $q=p'$ then $||\hat f||_q\le||f||_p$. – David C. Ullrich Feb 05 '18 at 14:10
  • @DavidC.Ullrich Hi David, thanks for the note. You mean $||\hat{f}||_q \leq ||f||_p$, for $1/p+1/q=1$, right? Is there a reference for this result? – user2130010 Feb 05 '18 at 21:04
  • It's Someone's Inequality... goggle fourier transform norm inequalities... right, Hausdorff-Young. – David C. Ullrich Feb 05 '18 at 21:42
  • OOPS: When I said $p\ge2$ I meant $1\le p\le 2$. (The same as $q\ge2$...) – David C. Ullrich Feb 05 '18 at 22:24
  • @DavidC.Ullrich Found it. Many thanks. – user2130010 Feb 05 '18 at 23:31

1 Answers1

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The Hausdorff-Young inequality, namely $$ \|\hat{f}\|_{p'}\le C_p \|f\|_{p}, $$ where $1\le p\le 2$ and $1/p+1/p'=1$, is the most you can get. For any $p>2$ there exists $f\in L^p(\mathbb R^n)$ such that $\hat{f}$ (taken in distributional sense) is not a function: more precisely, $\hat{f}$ is a distribution with strictly positive order; see Hörmander's first volume (of his 4 volume series), Theorem 7.6.6.

In particular, if $p>2$ then there is no function space $X$ such that $\mathcal F\colon L^p(\mathbb R^n)\to X$ is well-defined.

Remark. This can be considered as the starting point of the restriction problem for the Fourier transform.

Giuseppe Negro
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  • @Medo: I wrote "section 7.6.6" instead of "theorem 7.6.6". Corrected now. – Giuseppe Negro Jun 02 '18 at 18:01
  • I might be missing something basic, but do the result about Fourier Transform in $L^p$ mean this "is the most you can get" about the $L^p$-norm of FT in $L^2$? – fqq Jun 10 '20 at 13:08
  • No. I mean that the Hausdorff-Young inequality is the only possible mapping theorem of the Fourier transform in Lp spaces. That means that, if the Fourier transform is a continuous mapping of Lp into Lq, then q is the conjugate of p and $1\le p \le 2$. Recently, Terry Tao explained this in his harmonic analysis course, it is in the very beginning of his notes, if you need more details. – Giuseppe Negro Jun 10 '20 at 14:39