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My question concerns the convolution theorem from harmonic analysis. In particular, I would like to know whether one can extend the result to a particular case.

We will define the Fourier transform of a Schwartz function $\varphi \in \mathcal{S}(\mathbb{R})$ by \begin{equation} \mathcal{F}(\varphi)(\xi):=\int_\mathbb{R} \varphi(x) \mathrm{e}^{-2 \pi \mathrm{i}x\xi} \, \mathrm{d}x. \end{equation} Moreover we define the convolution of two Schwartz functions $\varphi, \psi \in \mathcal{S}(\mathbb{R})$ by \begin{equation} \varphi \ast \psi (x) := \int_\mathbb{R} \varphi(y)\psi(x-y) \, \mathrm{d}y. \end{equation} With these definitions one can prove the convolution theorem:

Let $f,g \in \mathcal{S}(\mathbb{R})$. Then, \begin{equation} \mathcal{F}(f \cdot g) = \mathcal{F}(f) \ast \mathcal{F}(g). \end{equation}

One can prove the same for $f,g \in L^1(\mathbb{R})$ with $\mathcal{F}(f),\mathcal{F}(g) \in L^1(\mathbb{R})$. Using the notion of tempered distributions we can additionally extend this theorem quite easily:

We first set $\varphi^\#(x):=\varphi(-x)$, for $\varphi \in \mathcal{S}(\mathbb{R})$. Now, let $T \in \mathcal{S}^\prime(\mathbb{R})$ and $\varphi \in \mathcal{S}(\mathbb{R})$, we can then define $T \ast \varphi \in \mathcal{S}^\prime(\mathbb{R})$ and $\varphi \cdot T \in \mathcal{S}^\prime(\mathbb{R})$ by \begin{equation} \langle T \ast \varphi, \psi \rangle := \langle T, \varphi^\# \ast \psi \rangle, \qquad \langle \varphi \cdot T , \psi \rangle := \langle T, \varphi \cdot \psi \rangle, \end{equation} where $\psi \in \mathcal{S}(\mathbb{R})$ and $\langle \cdot , \cdot \rangle$ denotes the duality pairing in the Schwartz space. We also define the Fourier transform of a tempered distribution $T \in \mathcal{S}^\prime(\mathbb{R})$ by \begin{equation} \langle \mathcal{F}(T),\psi \rangle := \langle T, \mathcal{F}(\psi) \rangle, \end{equation} which gives us another tempered distribution $\mathcal{F}(T) \in \mathcal{S}(\mathbb{R})$. The convolution theorem now becomes:

Let $T \in \mathcal{S}^\prime(\mathbb{R})$ and $\varphi \in \mathcal{S}(\mathbb{R})$. Then, \begin{equation} \mathcal{F}(\varphi \cdot T) = \mathcal{F}(T) \ast \mathcal{F}(\varphi). \end{equation}

The Main Question:

Let us denote by $\mathcal{C}_0(\mathbb{R})$ the continuous functions on $\mathbb{R}$ vanishing at infinity, let $\sigma > 0$ and $p \in [1,\infty)$. Let us take $f,g \in \mathcal{C}_0(\mathbb{R}) \subset L^\infty(\mathbb{R}) \subset \mathcal{S}^\prime(\mathbb{R})$ and $\mathfrak{f}, \mathfrak{g} \in L^p([-\sigma,\sigma]) \subset L^1([-\sigma,\sigma])$, such that \begin{equation} \mathcal{F}(f) = \mathfrak{f}, \qquad \mathcal{F}(g) = \mathfrak{g}. \end{equation}

Does \begin{equation} \mathcal{F}(f \cdot g) = \mathfrak{f} \ast \mathfrak{g} \tag{1} \end{equation} hold?

I tried to use density of the Schwartz functions in the $L^p$ spaces, but did not manage to get very far. In summary my question is:

  1. Does $(1)$ hold?
  2. How would one proceed to prove $(1)$, in the case it holds?
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1 Answer 1

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Your identity is equivalent (since the Fourier transform is bijective on the space of tempered distributions) to $$ \mathcal{F}^{-1}(\mathfrak{f} \ast \mathfrak{g}) = f \cdot g. $$

But since $\mathfrak{f}, \mathfrak{g} \in L^1 ([-\sigma, \sigma])\subset L^1 (\Bbb{R})$, you have (by the usual convolution theorem for $L^1$, but applied to $\mathcal{F}^{-1}$ instead of $\mathcal{F}$ (the proof is the same)): $$ \mathcal{F}^{-1}(\mathfrak{f} \ast \mathfrak{g}) = \mathcal{F}^{-1}(\mathfrak{f}) \cdot \mathcal{F}^{-1}(\mathfrak{g}) = f \cdot g, $$ as desired.

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  • $\begingroup$ Really nice, thank you!! $\endgroup$ Commented Oct 24, 2016 at 8:27

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