Chapter 14 The Fourier transform

Define the space

\[ L^1(\mR ):=\left \{u:\mR \to \mC : \int _{-\infty }^\infty |u(x)|\,dx<\infty \right \}. \]

  • Definition 14.1. The Fourier transform of a function \(u\in L^1(\mR )\) is the function \(\hat {u}:\mR \to \mC \) defined by

    \[ \hat {u}(\omega ):=\frac {1}{\sqrt {2\pi }}\int _{-\infty }^\infty \e ^{-i\omega x}u(x)\,dx. \]

  • Remark 14.2. The above version of the Fourier transform is the unitary angular frequency one. Other normalization conventions are in use; all such Fourier transforms have essentially the same properties, but various multiplicative constants are different.

  • Remark 14.3. Depending on what is most convenient, we will denote the Fourier transform of \(u\) either by \(\hat {u}\) (as above), or by \(\mathcal {F}(u)\).

  • Theorem 14.4 (Riemann–Lebesgue). The Fourier transform \(\hat {u}\) of \(u\in L^1(\mR )\) is a continuous function \(\mR \to \mC \) with \(\lim _{|\omega |\to \infty }\hat {u}(\omega )=0\).

  • Remark 14.5. Not every continuous function \(f\) with \(\lim _{|\omega |\to \infty }f(\omega )=0\) is the Fourier transform of an \(L^1(\mR )\) function. A counterexample is

    \[ f(\omega )= \begin {cases} \frac {\omega }{\ln (2)}&|\omega |\leq 1,\\ \frac {1}{\ln (1+\omega )}&\omega >1,\\ \frac {-1}{\ln (1-\omega )}&\omega <-1. \end {cases} \]

  • Theorem 14.6 (Fourier Inversion Formula). If \(u\in L^1(\mR )\) and \(\hat {u}\in L^1(\mR )\), then we have

    \[ u(x)=\frac {1}{\sqrt {2\pi }}\int _{-\infty }^\infty \e ^{i\omega x}\hat {u}(\omega )\,d\omega . \]

  • Definition 14.7. For \(f\in L^1(\mR )\) the function \(u:\mR \to \mC \) defined by

    \[ u(x)=\frac {1}{\sqrt {2\pi }}\int _{-\infty }^\infty \e ^{i\omega x}f(\omega )\,d\omega , \]

    is called the inverse Fourier transform of \(f\).

  • Theorem 14.8 (Plancherel). The restriction of the Fourier transform to the dense space \(L^1(\mR )\cap L^2(\mR )\) extends by continuity to a bijection \(L^2(\mR )\to L^2(\mR )\). This extension is isometric in the sense that \(\|u\|_{L^2(\mR )}=\|\hat {u}\|_{L^2(\mR )}\). Its inverse is given by the continuous extension of the restriction to \(L^1(\mR )\cap L^2(\mR )\) of the inverse Fourier transform.

  • Remark 14.9. If all the terms make sense, we have

    \[ \mathcal {F}(u')=i\omega \mathcal {F}(u), \qquad \mathcal {F}(u)'=\mathcal {F}(-ix u), \]

    It follows that (again: if all terms make sense)

    \[ \mathcal {F}(u'')=-\omega ^2\mathcal {F}(u). \]

  • Example 14.10. We show that the Fourier transform of \(u(x)=\e ^{-x^2/2}\) is \(\hat {u}(\omega )=\e ^{-\omega ^2/2}\).

    We have \(u'(x)=-xu(x)\) by the chain rule. Fourier transforming both sides gives (using that \((-i)^2=-1\))

    \[ \mathcal {F}(u')=-i\mathcal {F}(-ixu), \]

    and using the relation between differentiation and Fourier transformation (Remark 14.9) then gives

    \[ i\omega \,\hat {u}=-i\frac {d}{d\omega }\hat {u}. \]

    We further have

    \[ \hat {u}(0)=\frac {1}{\sqrt {2\pi }}\int _{-\infty }^\infty \e ^{-x^2/2}\,dx=1, \]

    where we used the known integral \(\int _{-\infty }^\infty \e ^{-x^2/2}\,dx=\sqrt {2\pi }\). Therefore \(\hat {u}\) satisfies the initial value problem

    \[ (\hat {u})'=-\omega \hat {u},\qquad \hat {u}(0)=1. \]

    Let \(f(\omega ):=\e ^{-\omega ^2/2}\). By the chain rule we have \(f'(\omega )=-\omega f(\omega )\) and we clearly have \(f(0)=1\). We conclude that \(f\) is a solution of

    \[ f'(\omega )=-\omega f(\omega ),\qquad f(0)=1. \]

    Since \(\hat {u}\) and \(f\) satisfy the same initial value problem, by uniqueness (from MA30062 Analysis of Nonlinear Ordinary Differential Equations) we have that \(\hat {u}=f\), i.e. \(\hat {u}(\omega )=\e ^{-\omega ^2/2}\).

  • Remark 14.11. Slightly more generally we have that for \(a>0\) the Fourier transform of \(\e ^{-ax^2/2}\) equals \(\frac {1}{\sqrt {a}}\e ^{-\omega ^2/(2a)}\).

  • Definition 14.12. The convolution of \(f\in L^1(\mR )\) and \(g\in L^2(\mR )\) is the function

    \[ (f\ast g)(t)=\int _{-\infty }^\infty f(\theta )g(t-\theta )\,d\theta . \]

  • Remark 14.13. When \(f\) and \(g\) are both zero on \((-\infty ,0)\), then

    \[ (f\ast g)(t)=\int _0^t f(\theta )g(t-\theta )\,d\theta , \]

    a formula which you may recognize from MA20220 Ordinary Differential Equations and Control.

  • Theorem 14.14 (Convolution Theorem). We have for \(f\in L^1(\mR )\) and \(g\in L^2(\mR )\) that \(f\ast g \in L^2(\mR )\) and

    \[ \widehat {f\ast g}=\sqrt {2\pi }~\widehat {f}~\widehat {g}. \]