# Is Fourier transform defined on $L^p(\mathbb{R})$ only for $p \in [1, 2]$?

Is Fourier transform defined on $L^p(\mathbb{R})$ only for $p \in [1, 2]$?

From Lieb and Loss's Analysis, they extend the definition of Fourier transform from $L^1(\mathbb{R})$ to $L^p(\mathbb{R}), p \in (1, \infty)$, using $$\| FT(f) \|_q \leq C_{p,q} \|f\|_p$$ which they said only holds when $p \in (1,2]$. Does that mean that FT cannot be defined on $L^p(\mathbb{R})$ with $p \in (2, \infty)$, possibly via other means?

Thanks and regards!

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I recall having heard that the Fourier transform cannot be defined in $L^p$ space with $p>2$ if we want to confine ourselves to function spaces. That is, for any $p>2$ there exists a function $f\in L^p$ such that the distributional Fourier transform (the same Pavel mentions below) $\hat{f}$ is not a function, meaning that it does not belong to $L^1_{\text{loc}}$. But I cannot prove this and I cannot point you to any reference, so please don't take me too seriously. – Giuseppe Negro Dec 30 '12 at 19:46

Fourier transform is a linear operator. When defining an operator, we should think not only of the domain space, but also of the target. It is true that Fourier transform is not a bounded operator from $L^p(\mathbb R)$ to $L^q(\mathbb R)$ ($1/p+1/q=1$) when $p>2$. The issue is that large $p$ allows the function to have a pretty heavy "tail", which on the Fourier side results in bad local behavior.
But it is possible to define the Fourier transform of any function in $L^p(\mathbb R)$, by considering $f$ as a tempered distribution. Then the transform becomes an automorphism of the topological vector space of tempered distributions. In fact, the Wikipedia article on Fourier transform covers this, and links to an article on distributions.
Related MSE question: Fourier transform in $L^p$
The key fact is that $L^p \subset L^1_\mathrm{loc}$. – Nate Eldredge Dec 30 '12 at 20:12
@NateEldredge Yes, but $L^1_{\rm loc}$ by itself would not be enough to get a tempered distribution. – user53153 Dec 30 '12 at 20:18
Thanks! +1. I need to spend some time to understand distributions and FT on them. But let me ask this question. Is Fourier series also defined firstly as $L^1([a,b]) \to l^{\infty}$, and then extended to $L^p([a,b]) \to l^q, p \in (1, 2], 1/p + 1/q =1$? I didn't find the answer in the Wikipedia article for FS, and some books I have. So references are also appreciated. – Tim Dec 31 '12 at 2:07
The comparison with Fourier series in an imperfect one and here is why : $\mathcal{L}_p\big([a,b]\big) \subseteq \mathcal{L}_1\big([a,b]\big)$ for all $1 < p \leq \infty$ so it suffices to define the Fourier series for $\mathcal{L}_1\big([a,b]\big)$ and the extension to $\mathcal{L}_p\big([a,b]\big)$ is immediate. However $\mathcal{L}_p(\mathbb{R}) \not\subseteq \mathcal{L}_1(\mathbb{R})$. – M.G May 27 at 17:46