Fourier transform of a product in $L^1$

Question

There is a little thing that I do not understand, about the Fourier transform of a product of functions in $L^1$ (and only in this space), with the relation ${\mathcal F}(f g)(\lambda) = \mathcal{F}(f)\star \mathcal{F}(g) (\lambda)$ (and not the easier relation ${\mathcal F}(f \star g)(\lambda) = \mathcal{F}(f)(\lambda)\mathcal{F}(g) (\lambda)$ ).

Let us note $\mathcal{F}(f)$ the Fourier transform of a function $f$, if this transformation exists.

I can demonstrate that for any $f\in L^1(\mathbb{R})$, its Fourier transform exists.

I will now consider 2 cases:

1. For any $f,g\in L^1$ with $\mathcal{F}(g)\in L^1$, therefore I can show $fg\in L^1$. Thus, $\mathcal{F}(fg)$ exists. Plus, the inverse Fourier transform of $\mathcal{F}(g)$ exists. And in this case we can write: \begin{eqnarray*} {\mathcal F}(f g)(\lambda) &= & \int_{-\infty}^{+\infty} e^{-2 i \pi \lambda t} f(t) \int_{-\infty}^{+\infty} e^{2 i \pi u t} \mathcal{F}(g)(u) du dt \tag{1}\\ &= & \int_{-\infty}^{+\infty} \mathcal{F}(g)(u) \int_{-\infty}^{+\infty} e^{-2 i \pi (\lambda - u) t} f(t) dt du \\ & = & \int_{-\infty}^{+\infty} \mathcal{F}(g)(u) \mathcal{F}(f)(\lambda - u) du \\ &= & {\mathcal F}(f) \star {\mathcal F}(g) (\lambda) . \end{eqnarray*} From this relation we can deduce that $\mathcal{F}(f)\star \mathcal{F}(g)$ exists, so everyting is okay.

2. In the second case, I want to know if with $f,g\in L^1$ and $fg \in L^1(\mathbb{R})$, it is possible to prove the same relation: ${\mathcal F}(f g)(\lambda) = \mathcal{F}(f)\star \mathcal{F}(g) (\lambda)$? Because the first step in (1) uses the inverse Fourier transform, which is not necessary existing in my second case.

I assume that the answer is yes, since I saw some people using this relation with some functions that are in $L^1$ such that there Fourier transforms are not in $L^1$ (if there is not mistakes). But I would like to prove it.

Answer 1

If I understand correctly question (1), you want check if this holds:

Assume $f,g\in L_1(\mathbb{R}^n)$ and $\widehat{f}\in L_1(\mathbb{R}^n)$. Then $f\cdot g\in L_1(\mathbb{R}^n)$, and \begin{align} \widehat{f\cdot g}=\widehat{f}*\widehat{g}\tag{1}\label{one} \end{align}

You are only concern with $n=1$, but the result holds for any dimension. Here is a proof that it does:

The Fourier inversion theorem for $L_1$ implies that $f$ is almost surely equal to function (which we still denote by $f$) in $\mathcal{C}_\infty(\mathbb{R}^n)\cap L_1(\mathbb{R}^n)$ and $$ f(\boldsymbol{x})=\int_{\mathbb{R}^n}\widehat{f}(\boldsymbol{y})e^{2\pi i \boldsymbol{x}\cdot\boldsymbol{y}}\,d\boldsymbol{y} $$ See for example, for example Rudin, W., Real and Complex Analysis, McGraw Hill, 1984, Chapter 9 (in one dimension), or Jones, F. Legesgue Integration on Euclidean space, Jones and Bartlett, 2002, section 13.B (in higher dimension).

Then $f\cdot g\in L_1(\mathbb{R}^n)$ and so,

\begin{align} \widehat{f\cdot g}(\boldsymbol{t})=\int_{\mathbb{R}^n}f(\boldsymbol{x}) g(\boldsymbol{x}) e^{-2\pi\boldsymbol{x}\cdot\boldsymbol{t}}\,d\boldsymbol{x}=\int_{\mathbb{R}^n} g(\boldsymbol{x}) e^{-2\pi\boldsymbol{x}\cdot\boldsymbol{t}}\Big(\int_{\mathbb{R}^n}\widehat{f}(\boldsymbol{y})e^{2\pi i \boldsymbol{x}\cdot\boldsymbol{y}}\,d\boldsymbol{y}\Big) \,d\boldsymbol{x} \end{align} Since $(\boldsymbol{x},\boldsymbol{y})\mapsto g(\boldsymbol{x})\widehat{f}(\boldsymbol{y})$ is in $L_1(\mathbb{R}^n\times\mathbb{R}^n)$, we can apply Fubini's theorem to obtain \begin{align} \widehat{f\cdot g}(\boldsymbol{t})= \int_{\mathbb{R}^n}\widehat{f}(\boldsymbol{y}) \Big(\int_{\mathbb{R}^n} g(\boldsymbol{x})e^{-2\pi i \boldsymbol{x}\cdot(\boldsymbol{t}-\boldsymbol{y})}\,d\boldsymbol{x}\Big) \,d\boldsymbol{y}=\int_{\mathbb{R}^n}\widehat{f}(\boldsymbol{y}) \widehat{g}(\boldsymbol{t}-\boldsymbol{y})\,d\boldsymbol{y}=\widehat{f}*\widehat{g}(\boldsymbol{t}). \end{align}

Remark: The equality in \eqref{one} is pointwise equality among continuous functions.

Question 2 is correct and follows from the Fourier inverse theorem.

Answer 2

If $f,g \in L^1(\mathbb{R})$, then $f\star g \in L^1$ and $\frac{1}{\sqrt{2\pi}}\widehat{f\star g}=\widehat{f}\widehat{g}$. The proof that $f\star g\in L^1$ follows nicely from integration properties of positive functions: \begin{align} \int\left|\int f(x-t)g(t)dt\right| dx&\le \int\int |f(x-t)||g(t)|dtdx \\ &= \int\int|f(x-t)||g(t)|dxdt \\ &=\int\int|f(x-t)|dx |g(t)|dt \\ &= \int\int|f(y)|dy |g(t)|dt \\ &= \|f\|_{L^1}\|g\|_{L^1}. \end{align} Once you know that $f\star g\in L^1$, then you can Fourier transform, and perform a similar switch of integration to obtain the stated convolution property.