It is known that the Fourier transform $\mathcal F$ maps $L^2 \to L^2$ as an (isometric) isomorphism and $L^1 \to L^\infty$ as bounded operator. Via Riesz-Thorin this result can be extended to give that $\mathcal F$ also maps $L^p \to L^{p'}$ where $1 = \frac{1}{p} + \frac{1}{p'}$ as a bounded operator, i.e. the Hausdorff-Young equality holds: $$ \|\mathcal F(u) \|_{p'} \leq \|u\|_p $$ for all $u \in L^p(\mathbb{R}^n)$, $p \in (1,2)$.

How do we know that the Fourier transform is only an isomorphism from $L^2 \to L^2$? One could argue that no $L^p$ space is isomorphic to an $L^q$ space for $p \neq q$ using the invariants type and cotype as suggested here but, considering that I only want to show that $\mathcal F$ is not an isomorphism from $L^p \to L^{p'}$, I suppose there is a more simple approach directly related to the Fourier transform. I am thinking of showing the fourier transform from $L^p \to L^{p'}$ is simply not surjective.

Can you think of an easier approach to this problem?

  • $\begingroup$ I'd start with the same approach; the non-surjectivity of $\mathcal F:L_1\to L_\infty$ follows from an easy observation that constant non-zero functions do not belong to $\mathcal F(L_1)$. Maybe in your case you can find a very slowly decreasing function in $L_{p'}$ that can not be obtained as an image of a $L_p$ function. $\endgroup$ Aug 29, 2017 at 14:34
  • $\begingroup$ I want to point out that also injectivity may fail, the continuous extension of an injective operator need not be injective anymore. Concerning your suggestion, some remarks/possible issues: Since the Fourier transform is an isomorphism on the Schwartz functions, its range is always dense on $L^p$ despite in the case $p=1$, so the situation there is quite different. I would rather try to show that its range is not closed. But all of this is not so easy because the Fourier transform on $L^p$ for $p\in(1,2)$ is only given by interpolation and no representation is available. $\endgroup$ Aug 31, 2017 at 7:54
  • $\begingroup$ @SebastianBechtel: In general, the extension of an injective operator need not be injective. But since we have $L^p \subset \mathcal{S}'$ (tempered distributions), and since the Fourier transform is injective on $\mathcal{S}'$, the extension is also injective in this case. Nevertheless, I think your idea is the only sensible one, i.e., one needs to show that we do not have $\mathcal{F} f \|_{L^q} \gtrsim \| f \|_{L^p}$. $\endgroup$
    – PhoemueX
    Aug 31, 2017 at 8:15
  • $\begingroup$ @PhoemueX Good point with the distributional Fourier transform! $\endgroup$ Aug 31, 2017 at 8:35

2 Answers 2


Here is a (modified) argument which I used repeatedly in a paper of mine. Possibly, this is some form of type and cotype argument (I am not too familiar with these notions), but maybe it is elementary enough for you:

I will make extensive use of the Khintchine inequality, which states that if $\varepsilon_1, \dots, \varepsilon_N$ are independent and uniformly distributed in $\{1, -1\}$, then we have $$ \mathbb{E} \left|\sum_{n=1}^N \varepsilon_n x_n\right|^p \asymp \left(\sum_{n=1}^N |x_n|^2\right)^p , $$ for arbitrary $x_1,\dots, x_N \in \Bbb{C}$. Here, $\Bbb{E}$ denotes the expected value with respect to $\varepsilon = (\varepsilon_1,\dots, \varepsilon_N)$. It is crucial that the implied constants above are independent of $x_1,\dots, x_N$ and of $N$; they only depend on the choice of $p \in (0,\infty)$.

Now, assume that $\mathcal{F} : L^p \to L^q$ is an isomorphism. Let us first rule out the case $q = \infty$: If we had $q=\infty$, then there would be some $f \in L^p$ with $\widehat{f} \equiv 1$, where I write $\widehat{f} = \mathcal{F}f$. But we also have $\widehat{\delta_0} \equiv 1$, where we are using the formalism of tempered distributions, and where $\delta_0$ is the Dirac measure at the origin. But this implies $f = \delta_0$ as tempered distributions, which is absurd.

Hence $q \in (0,\infty)$. Now, let $\varepsilon_1,\dots, \varepsilon_N$ as above. For $n \in \Bbb{N}$, let $y_n := (n,0,\dots,0) \in \Bbb{R}^d$, and set $f_n := T_{y_n} 1_{(0,1)^d}$, where $(T_y f)(x) = f(x-y)$. Note that $\widehat{f_n} = M_{-y_n} \widehat{f}$ for $f := 1_{(0,1)}$, where $M_\xi f(x) = e^{2\pi i \langle x, \xi \rangle} f(x)$ denotes the modulation of $f$. Finally, observe that the supports of $f_n, f_m$ are disjoint for $n \neq m$. All in all, these properties imply \begin{align*} N^{q/p} = \Bbb{E} \, \left\| \sum_{n=1}^N \varepsilon_n \cdot 1_{(n,n+1)} \right\|_{L^p}^q & \asymp \Bbb{E} \, \left\| \sum_{n=1}^N \varepsilon_n \cdot \mathcal{F}[ 1_{(n,n+1)} ] \right\|_{L^q}^q \\ & = \Bbb{E} \int \left| \sum_{n=1}^N \varepsilon_n e^{2\pi i \langle y_n, \xi\rangle} \right|^q \cdot |\widehat{f}(\xi)|^q \, d\xi \\ (\text{Fubini's theorem}) & = \int \Bbb{E} \left| \sum_{n=1}^N \varepsilon_n e^{2\pi i \langle y_n, \xi\rangle} \right|^q \cdot |\widehat{f}(\xi)|^q \, d\xi \\ (\text{Khintchine inequality}) & \asymp N^{q/2} \cdot \| \widehat{f} \|_{L^q}^q \\ & \asymp N^{q/2} \cdot \|f\|_{L^p}^q = N^{q/2}, \end{align*} where the implied constants are independent of $N \in \Bbb{N}$. This can only hold for $p = 2$.

Now, let us show that also $q = 2$. To see this, note for arbitrary $f \in L^p = L^2$ by Plancherel's theorem that $$ \|\widehat{f} \|_{L^2} = \| f \|_{L^2} = \| f \|_{L^p} \asymp \|\widehat{f}\|_{L^q} . $$ Thus, we get for arbitrary $L^2$ functions $g \in L^2$ that $\|g\|_{L^2} \asymp \| g \|_{L^q}$, which easily implies $q = 2$, by considering e.g. $g = \sum_{n=1}^N f_n$ with $f_n$ as above.


EDIT: There seems to be an error in my calculations.

The fact that $\mathcal F$ is no isomorphism between $L^p$ spaces for $p \neq 2$ can be proven using the same argument that is used to determine the sobolev conjugate.

Suppose that for $p,p'$ with $\frac{1}{p} + \frac{1}{p'} = 1$ the Fourier transform is an isomorphism between $L^p$ and $L^{p'}$. Then for all $f \in \mathcal S$, where $\mathcal S$ denotes the Schwartz space, we have $$ \|f\|_p = \|\mathcal{F}^{-1} \mathcal{F} (f)\|_{p} \leq \|\mathcal{F}^{-1}\|_{L^{p'} \to L^p} \|\mathcal{F}(f)\|_{p'} = C \, \|\mathcal F (f)\|_{p'}. $$

Now, consider for $\lambda > 0$ the dilation $\delta^\lambda f$ defined by $$ \delta^\lambda f(x):= f(\lambda x). $$ We then have $$ \lambda^{-\frac{n}{p}} \|f\|_p = \|\delta^\lambda f\|_p \leq C \; \|\mathcal{F}(\delta^\lambda f)\|_{p'} = C \, \lambda^{-n} \|\delta^{\frac{1}{\lambda}} \, \mathcal{F}(f) \|_{p'} = C \, \|\mathcal{F}(f)\|_{p'}. $$ Letting $\lambda \to 0$ shows that there is no such constant $C > 0$.

Here is the error I made...

The second line of inequalities should read $$ \lambda^{-\frac{n}{p}} \|f\|_p = \|\delta^\lambda f\|_p \leq C \; \|\mathcal{F}(\delta^\lambda f)\|_{p'} = C \, \lambda^{-n} \|\delta^{\frac{1}{\lambda}} \, \mathcal{F}(f) \|_{p'} = C \, \lambda^{-n + \frac{n}{p'}}\|\mathcal{F}(f)\|_{p'}, $$ which after reordering of the $\lambda$-terms just gives the first inequality. Just in the case of the sobolev conjugate, this proof only shows that if $\mathcal F$ is an isomorphism then $p,p'$ necessarily have to be conjugate.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .