In this post, we study Fourier transform properties and give some applications. In fact, this transformation helps in converting partial differential equations to ODE. A simple method to solve the heat equation is the Fourier transform technique.
The Riemann-Lebesgue Lemma
We denote by $L^1(\mathbb{R})$ the Lebesgue space of measurable functions $f:\mathbb{R}\to \mathbb{R} $ such that\begin{align*} \|f\|_1^2:=\int_{\mathbb{R}} |f(x)|dx<\infty.\end{align*}
Also, we recall that $\|\cdot\|_1$ is a norm on $ L^1(\mathbb{R})$ and that $( L^1(\mathbb{R}), \|\cdot\|_1 ) $ is a Banach space, no reflexive. On the other hand, the space of continuously differentiable functions with compact support $\mathcal{C}_c^1(\mathbb{R})$ is dense in the space $ L^1(\mathbb{R}) $.
As for any $t,x\in\mathbb{R}$, we have $|e^{-itx}f(x)|=|f(x)|$, then for $f\in L^1(\mathbb{R})$, the function $(x\mapsto e^{-itx}f(x) )\in L^1(\mathbb{R}) $. We then have the following definition:
Definition: For $f\in L^1(\mathbb{R}) $ we define the Fourier transform of $f$ by \begin{align*}\hat{f}(t)=\int_{\mathbb{R}} e^{-itx}f(x) dx,\qquad t\in \mathbb{R}.\end{align*}
Clearly $f\mapsto\hat{f}$ is a linear application on $ L^1(\mathbb{R})$. We have the following Riemann-Lebesgue result.
Theorem: The Fourier transform is an application from $ L^1(\mathbb{R})$ to $\mathcal{C}_0(\mathbb{R})$, where \begin{align*} \mathcal{C}_0(\mathbb{R}) :=\{f\in \mathcal{C}(\mathbb{R}) : \lim_{|t|\to +\infty}f(t)=0\}.\end{align*}
Proof: Here we use the density of $\mathcal{C}_c^1(\mathbb{R})$ is dense in the space $ L^1(\mathbb{R}) $. Hence, in the first step, we will prove the result for $f\in \mathcal{C}_c^1(\mathbb{R}) $. In this case, and by using integration by part, we get for any $t\neq 0,$ \begin{align*}\hat{f}(t)=\frac{1}{it}\int_{\mathbb{R}}e^{-itx}f'(x)dx.\end{align*}This implies that\begin{align*} | \hat{f}(t) |\le \frac{1}{|t|}|f’|_1.\end{align*}
If $f\in L^1(\mathbb{R}) ,$ by density there exists a sequence $(f_n)_n\subset \mathcal{C}_c^1(\mathbb{R}) $ that converges to $f$ in $ L^1(\mathbb{R}) $. Thus, for any $\varepsilon>0,$ there exists $m\in\mathbb{N}$ such that $\|f_m-f\|_1\le \frac{\varepsilon}{2}$. Moreover, remark that\begin{align*} \hat{f}(t) = \hat{f_m}(t) + \hat{f}(t) – \hat{f_m}(t),\end{align*} so that \begin{align*} | \hat{f}(t) |&\le \frac{1}{|t|}\|f_m’\|_1+\|f-f_m\|_1\cr & \le \frac{1}{|t|}\|f_m’\|_1 + \frac{\varepsilon}{2}. \end{align*} This ends the proof.
Fourier Transform in the Schwartz space
We define the Schwartz space by\begin{align*}\mathcal{S}(\mathbb{R})=\{f\in \mathcal{C}^\infty(\mathbb{R}):\forall p,q\in\mathbb{N}&,\;\exists C_{p,q}>0:\cr & |x^pf^{(q)}(x)|\le C_{pq},\; \forall x\in\mathbb{R}\}.\end{align*} The first remark is that $ \mathcal{S}(\mathbb{R}) \subset L^1(\mathbb{R})$ continuously and densely. Hence we can prove all properties of the Fourier transform in $\mathcal{S}(\mathbb{R})$. These properties are translated to $L^1(\mathbb{R})$ by density.
As example we have $(x\mapsto e^{-x^2})\in \mathcal{S}(\mathbb{R}) $. On the other hand, we can easily prove that if $f\in \mathcal{S}(\mathbb{R}) ,$ then also $f’$ and $x\mapsto xf(x)$ are in $ \mathcal{S}(\mathbb{R}) $.
Theorem: If $f\in \mathcal{S}(\mathbb{R}) $ then the Fourier transform $\hat{f}\in \mathcal{S}(\mathbb{R}) $.
