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Let the $f$ be a function in $L^s$ where $s \in [1,\infty) $. For which $r$ Fourier transform $\hat{f}$ belongs to $L^r$?

I'd be grateful for any kind of help including providing a literature or hints.

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up vote 6 down vote accepted

If $s > 2$, this question quickly becomes technical and requires the theory of distributions. It is not hard to prove that any $f \in L^p$, $1 \le p \le \infty$ induces a tempered distribution $T_f$ and that the Fourier transform of $T_f$ is itself a tempered distribution. As a reference for this, I recommend chapter two of the book Partial Differential Equations by Rauch. However, it turns out that for any $s > 2$, there exists an $f \in L^s$ such that $\hat{f}$ is a tempered distribution that is not induced by any proper function. In the case $s \le 2$ we are in luck and the question can be answered via interpolation. The Riesz-Thorin Convexity Theorem states:

Assume $T$ is a bounded operator $L^{p_1} \to L^{p_2}$ and at the same time $L^{q_1} \to L^{q_2}$. Then it is also a bounded operator $L^{r_1} \to L^{r_2}$ where

$$\frac{1}{r_1} = \frac{t}{p_1} + \frac{1-t}{q_1}, \quad \frac{1}{r_2} = \frac{t}{p_2} + \frac{1-t}{q_2}$$

for any $t \in (0,1)$. (Notice that some care must be taken to define exactly how one operator can map between two sets of spaces. What is meant here is that the operator is defined on a common dense subset and can be extended by continuity in both sets of norms. In the case of Fourier transform, you can take the functions of Schwartz class as your dense subset.)

Now we know that the Fourier Transform satisfies the hypothesis with $p_1 = 1, p_2 = \infty$ and $q_1 = q_2 = 2$, the first fact being trivial from the definition, and the second can be found in almost any book (for example Real and Complex Analysis by Rudin.) Thus we get

$$\frac{1}{r_1} = t + \frac{1-t}{2}, \quad \frac{1}{r_2} = \frac{1 - t}{2}$$

which simplifies to

$$r_1 = \frac{2}{1+t}, \quad r_2 = \frac{2}{1-t}$$

So if $r_1$ is given, then the first equation implies that $t = \frac{2}{r_1} - 1$ and substituting this into the second equation yields $r_2 = \frac{2}{1 - (\frac{2}{r_1} - 1)}$ which simplifies to $$r_2 = \frac{1}{1 - \frac{1}{r_1}} = r_1^*$$ Thus the Fourier transform is always bounded from $L^p \to L^q$, where $q$ is the Hölder conjugate of $p$. If you are interested in a more direct proof, I think there is one in the book Functional Analysis by Peter Lax.

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Doesn't $p$ need be lesser or equal than 2? Otherwise $t$ would be negative and so Riesz-Thorin's theorem would fail. – Giuseppe Negro May 17 '12 at 1:34
@GiuseppeNegro True, I have edited the post. – user12014 May 17 '12 at 1:48
Thank you for such a quick and detail answer! I'm not very familiar with functional analysis, so the literature will surely help me fully understand the solution. – Vees May 17 '12 at 17:55

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