Fourier Transform on the Real Line — Lecture Notes
Purpose & Convention
These notes fix a single convention for the Fourier transform on \(\mathbb{R}\) (the \(2\pi i\) convention), state the inversion formula, prove the differentiation and multiplication rules, and apply them to a first-order ODE and to the heat equation. We also formalize the space \(L^1(\mathbb{R})\) and prove the Riemann–Lebesgue theorem.
Definition and Basic Integrability
For \(f:\mathbb{R}\to\mathbb{C}\) define the Fourier transform
\[
\widehat{f}(\xi)=\int_{\mathbb{R}} f(x)\,e^{-2\pi i x\xi}\,dx.
\]
If \(f\in L^1(\mathbb{R})\), then the integral is absolutely convergent:
\[
\Bigl|\!\int_{\mathbb{R}} f(x)\,e^{-2\pi i x\xi}\,dx\Bigr|
\le \int_{\mathbb{R}} |f(x)\,e^{-2\pi i x\xi}|\,dx
= \int_{\mathbb{R}} |f(x)|\,dx < \infty.
\]
Integrable Functions \(L^1(\mathbb{R})\)
A (Lebesgue) measurable function \(f:\mathbb{R}\to\mathbb{C}\) belongs to \(L^1(\mathbb{R})\) if
\[
\|f\|_{1}:=\int_{\mathbb{R}} |f(x)|\,dx < \infty.
\]
Elements of \(L^1\) are equivalence classes modulo equality almost everywhere.
(i) If \(f\in L^1\) then \(|\widehat f(\xi)|\le \|f\|_1\) for all \(\xi\) (immediate from the triangle inequality).
(ii) \(\widehat f\) is uniformly continuous on \(\mathbb{R}\), since
\[
|\widehat f(\xi+\eta)-\widehat f(\xi)|
=\Big|\!\int_{\mathbb{R}} f(x)\,e^{-2\pi i x\xi}\big(e^{-2\pi i x\eta}-1\big)\,dx\Big|
\le \int_{\mathbb{R}} |f(x)|\,|e^{-2\pi i x\eta}-1|\,dx \xrightarrow[\eta\to0]{} 0
\]
by dominated convergence with dominator \(2|f|\in L^1\).
if \((\tau_h f)(x):=f(x+h)\), then \(\|\tau_h f-f\|_1\to 0\) as \(h\to 0\).
Hint: Approximate \(f\) in \(L^1\) by a continuous compactly supported function.
Inversion Formula and Notation
Inversion (without proof). For functions that are smooth and decay rapidly (e.g., Schwarz functions),
\[
f(x)=\int_{\mathbb{R}} \widehat{f}(\xi)\,e^{2\pi i \xi x}\,d\xi .
\]
\[
\mathcal{F}^{-1}\widehat{f}(x)=\int_{\mathbb{R}} \widehat{f}(\xi)\,e^{2\pi i \xi x}\,d\xi .
\]
Riemann–Lebesgue Theorem (Fourier Transform of \(L^1\))
Theorem. If \(f\in L^1(\mathbb{R})\), then \(\widehat f\) is bounded and uniformly continuous, and
\[
\lim_{|\xi|\to\infty}\widehat f(\xi)=0.
\]
Equivalently, \(\widehat f\in C_0(\mathbb{R})\).
Proof. Boundedness and uniform continuity were recorded above. It remains to prove
\(\widehat f(\xi)\to 0\) as \(|\xi|\to\infty\). Fix \(\xi\ne 0\) and set \((\tau_h f)(x):=f(x+h)\).
For any \(h\in\mathbb{R}\) we have the identity
\[
\int_{\mathbb{R}} \big(f(x)-f(x+h)\big)e^{-2\pi i \xi x}\,dx
= \big(1-e^{2\pi i \xi h}\big)\,\widehat f(\xi).
\]
Indeed, substituting \(y=x+h\) in the second integral gives
\(\int f(x+h)\,e^{-2\pi i \xi x}\,dx = e^{2\pi i \xi h}\widehat f(\xi)\).
Hence
\[
\widehat f(\xi)=\frac{1}{1-e^{2\pi i \xi h}}\int_{\mathbb{R}} \big(f(x)-f(x+h)\big)e^{-2\pi i \xi x}\,dx.
\]
Taking absolute values,
\[
|\widehat f(\xi)|\le \frac{\|\tau_h f – f\|_1}{|1-e^{2\pi i \xi h}|}.
\]
Now choose \(h=\frac{1}{2\xi}\) (for negative \(\xi\), this is negative), so that
\(e^{2\pi i \xi h}=e^{\pi i}=-1\) and \(|1-(-1)|=2\). Therefore,
\[
|\widehat f(\xi)|\le \tfrac12\,\|\tau_{1/(2\xi)} f – f\|_1.
\]
As \(|\xi|\to\infty\), we have \(h=1/(2\xi)\to 0\), and by translation continuity in \(L^1\),
\(\|\tau_h f – f\|_1\to 0\). Thus \(\widehat f(\xi)\to 0\) as claimed. \(\square\)
\(\widehat f'(\xi)= -2\pi i\,\widehat{(x f)}(\xi)\), hence \(\widehat f’\in C_0(\mathbb{R})\) as well.
Operator Rules: Differentiation and Multiplication
Differentiation Rule. If \(f,f’\in L^1\) and the boundary term vanishes, then
$$ \mathcal{F}(f’)(\xi) = (2\pi i\,\xi)\,\mathcal{F}f(\xi). $$
$$
\int_{\mathbb{R}} f'(x)\,e^{-2\pi i x\xi}\,dx
= \big[f(x)e^{-2\pi i x\xi}\big]_{-\infty}^{\infty}
– \int_{\mathbb{R}} f(x)\,(-2\pi i\xi)\,e^{-2\pi i x\xi}\,dx
= 2\pi i\,\xi \int_{\mathbb{R}} f(x)\,e^{-2\pi i x\xi}\,dx.
$$
Multiplication by \(x\). If \(x f(x)\in L^1(\mathbb{R})\), then
$$
\mathcal{F}(x f(x))(\xi) = \frac{i}{2\pi}\,\frac{d}{d\xi}\,\mathcal{F}f(\xi).
$$
$$
\frac{d}{d\xi}\widehat{f}(\xi)
= \frac{d}{d\xi}\!\int_{\mathbb{R}} f(x)\,e^{-2\pi i \xi x}\,dx
= \int_{\mathbb{R}} f(x)\,(-2\pi i x)\,e^{-2\pi i \xi x}\,dx
= -2\pi i \int_{\mathbb{R}} x f(x)\,e^{-2\pi i \xi x}\,dx.
$$
Hence
$$
\mathcal{F}(x f(x))(\xi)
= -\frac{1}{2\pi i}\,\frac{d}{d\xi}\widehat{f}(\xi)
= \frac{i}{2\pi}\,\frac{d}{d\xi}\,\mathcal{F}f(\xi).
$$
\(\mathcal{F}(f’)=(2\pi i\,\xi)\,\widehat f\),
\(\mathcal{F}(x f)=\dfrac{i}{2\pi}\,\partial_\xi \widehat f\).
Example: First-Order Linear ODE
Problem. Solve \(u'(x)+a\,u(x)=f(x)\) for \(a>0\).
Taking Fourier transforms,
\[
(2\pi i\,\xi)\,\widehat{u}(\xi)+a\,\widehat{u}(\xi)=\widehat{f}(\xi)
\quad\Rightarrow\quad
\widehat{u}(\xi)=\frac{\widehat{f}(\xi)}{a+2\pi i\,\xi}.
\]
Inverting,
\[
u(x)=\int_{\mathbb{R}} \frac{\widehat{f}(\xi)}{a+2\pi i\,\xi}\,e^{2\pi i \xi x}\,d\xi .
\]
Example: Heat Equation in One Dimension
Problem. Solve
\[
\frac{\partial u}{\partial t}(x,t)-\kappa\,\frac{\partial^2 u}{\partial x^2}(x,t)=0,
\qquad \kappa>0,
\]
using the Fourier transform in \(x\) (write \(\widehat{u}(\xi,t)=\mathcal{F}_x u(\xi,t)\)).
Compute
\[
\mathcal{F}\!\left(\frac{\partial u}{\partial t}-\kappa\,\frac{\partial^2 u}{\partial x^2}\right)
=\frac{\partial \widehat{u}}{\partial t}(\xi,t)-\kappa(2\pi i\xi)^2\,\widehat{u}(\xi,t)
=\frac{\partial \widehat{u}}{\partial t}(\xi,t)+4\kappa\pi^2\xi^2\,\widehat{u}(\xi,t)=0.
\]
Hence
\[
\frac{\partial \widehat{u}}{\partial t}(\xi,t)
=-4\kappa\pi^2\xi^2\,\widehat{u}(\xi,t).
\]
Separating variables gives
\[
\ln \widehat{u}(\xi,t)=-4\kappa\pi^2\xi^2\,t+C(\xi),
\quad\text{so}\quad
\widehat{u}(\xi,t)=e^{C(\xi)}\,e^{-4\kappa\pi^2\xi^2 t}.
\]
Since \(\widehat{u}(\xi,0)=e^{C(\xi)}\), inversion yields
\[
u(x,t)=\int_{\mathbb{R}} \widehat{u}(\xi,0)\,e^{-4\kappa\pi^2\xi^2 t}\,e^{2\pi i \xi x}\,d\xi .
\]