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For Fourier transformation defined from $L^p(\mathbb{R}^n)$ to $L^q(\mathbb{R}^n)$, $p\in [1, 2]$, and $1/p+1/q=1$,

I heard that when $p=2$, FT is bijective.

Is $p=2$ iff

  • FT is bijective,
  • FT is injective, or/and
  • FT is surjective?

Thanks and regards!

Tim
  • 47,382
  • All $L^p$ functions are tempered distributions, and the Fourier transform is an automorphism of the space of tempered distributions, so it's always injective. – Daniel Fischer Jul 18 '13 at 20:57
  • Is the fourier transform well defined in the way you write it? Because On $L^1 (\mathbb{R})$ the Fourier transform is in $BC(\mathbb{R})$. – Dominic Michaelis Jul 18 '13 at 21:01
  • @DominicMichaelis: What is $BUC(\mathbb R)$? The range of FT on $L^1(\mathbb R)$ is $L^\infty(\mathbb R)$, isn't it? – Tim Jul 18 '13 at 21:05
  • @Tim Sorry it is Only $BC(\mathbb{R})$ those are the Bounded Continuous functions which are a subset of $L^\infty $ for sure, but for $p>1$ how do you know that the integral converges at all (over which the FT is defined? – Dominic Michaelis Jul 18 '13 at 21:07
  • @DominicMichaelis: For $p \in [1,2]$, $FT(L^p) \subseteq L^q$ with $1/p+1/q=2$, doesn't it? – Tim Jul 18 '13 at 21:11

2 Answers2

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The Fourier transform is always injective. In greatest generality, it's bijective as a map on the space of tempered distributions. Indeed, the transform of distribution $T$ is the distribution $\widehat T$ such that $\widehat{T}\varphi = T\widehat \varphi$ for all test functions $\varphi$ in the Schwartz space. The Fourier transform is a linear isomorphism of the Schwartz space onto itself. Therefore, $T\mapsto \widehat T$ is a bijection on the space of tempered distributions.

The Fourier transform is not surjective as a map $L^p\mapsto L^q$ for $1\le p<2$. When $p=1$, this follows from the fact that the Fourier transform of an integrable function is continuous. (Alternatively, from the fact that $L^1$ is separable by $L^\infty$ isn't.) For $1<p<2$ one can argue as follows: if $g\in L^q$ is a Fourier transform of $f\in L^p$, then $\widehat g\in L^p$ (because $\mathcal{F}^{-1}$ is just $\mathcal F$ composed with a flip). Thus, if $\mathcal{F}:L^p\to L^q$ were surjective, it would follows that $\mathcal F$ maps $L^q$ to $L^p$ when $q>2$. The latter is known to be false (see here).

40 votes
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2

Adding to the answer before that the Fourier transform $T : L^2 \rightarrow L^2$ is indeed surjective, the answer is: the Fourier transform is bijective iff $p=2$ (considering it on all $L^p, p \in [1,2]$).

The most important igredient to show this is Plancherel's Theorem (unitarity of $T$) which is first used to extend $T$ continuously from $L^1$ to $L^2$ and later to show invertibility of $T$ on all of $L^2$.

Mage
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