Remark 17.4. We have been deliberately vague about the sense of continuity in Definition 17.3 since it is complicated, but will never be an issue.

Remark 17.7. In Definition 16.9 we defined multiplication of a distribution \(u\) with a \(C^\infty \) function \(\psi \). If \(u\) is tempered, then \(\psi u\) need not be tempered. However, if \(\psi \in C^\infty _{\rm pol}(\mR )\) and \(\varphi \in \mathcal {S}(\mR )\) we have \(\psi \varphi \in \mathcal {S}(\mR )\) by Lemma 17.6 and from this it follows that if \(\psi \in C^\infty _{\rm pol}(\mR )\) and \(u\in \mathcal {S}'(\mR )\), then \(\psi u\in \mathcal {S}'(\mR )\).

Remark 17.10. For a regular distribution corresponding to \(f\in L^1(\mR )\), the Fourier–Schwartz transform is consistent with the Fourier transform since

\[ T_{\hat {f}}(\varphi ) =\int _{-\infty }^\infty \hat {f}(\omega )\varphi (\omega )\,d\omega =\int _{-\infty }^\infty f(x)\hat {\varphi }(x)\,dx, =\widehat {T_f}(\varphi ), \]

since

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

and

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

and these two iterated integrals are equal by Fubini’s Theorem.

Example 17.14. We will show that \(\mathcal {F}(\delta _a)=\omega \mapsto \frac {1}{\sqrt {2\pi }}\e ^{-ia\omega }\).

We have for \(\varphi \in \mathcal {S}\) (the first equality is the definition of the Fourier transform of a tempered distribution, the second equality is the definition of the Dirac delta, the third equality uses the definition of the Fourier transform of an \(L^1\) function and the fourth uses the definition of regular distribution)

\[ \hat {\delta }_a(\varphi )=\delta _a(\hat {\varphi })=\hat {\varphi }(a)=\frac {1}{\sqrt {2\pi }}\int _{-\infty }^\infty \varphi (x)\e ^{-iax}\,dx=T_{\frac {1}{\sqrt {2\pi }}\e ^{-iax}}(\varphi ). \]

Since this equality holds for all \(\varphi \in \mathcal {S}\) we have \(\hat {\delta }=\omega \mapsto \frac {1}{\sqrt {2\pi }}\e ^{-ia\omega }\).

Example 17.15. We show that for \(f\in L^1(\mR )\) we have \(\mathcal {F}^{-1}f=\mathcal {F}\left (\omega \mapsto f(-\omega )\right )\).

We have (using the change of variable \(\omega =-\eta \))

\[ (\mathcal {F}^{-1}f)(x)=\frac {1}{\sqrt {2\pi }}\int _{-\infty }^\infty \e ^{i\eta x}f(\eta )\,d\eta =\frac {1}{\sqrt {2\pi }}\int _{-\infty }^\infty \e ^{-i\omega x}f(-\omega )\,d\omega =\mathcal {F}\left (\omega \mapsto f(-\omega )\right ). \]

Example 17.16. For \(u\) a tempered distribution, we define the inverse Fourier transform of \(u\) by \((\mathcal {F}^{-1}u)(\varphi )=u\left (\mathcal {F}^{-1}\varphi \right )\). We show that \((\mathcal {F}^{-1}u)(\varphi )=u\left (\mathcal {F}\left (\omega \mapsto \varphi (-\omega )\right )\right )\).

We have (applying Example 17.15 to \(\varphi \))

\[ (\mathcal {F}^{-1}u)(\varphi ) =u\left (\mathcal {F}^{-1}\varphi \right ) =u\left (\mathcal {F}\left (\omega \mapsto \varphi (-\omega )\right )\right ). \]

Example 17.17. We show that for \(a\in \mR \) we have that \(\mathcal {F}\left (\frac {1}{\sqrt {2\pi }}\e ^{iax}\right )\) equals \(\delta _a\).

By Example 17.14 we have

\[ \mathcal {F}(\delta _a)=\frac {1}{\sqrt {2\pi }}\e ^{-ia\omega }. \]

Applying the inverse Fourier transform to both sides and using Example 17.16 gives

\[ \delta _a =\mathcal {F}^{-1}\left (\frac {1}{\sqrt {2\pi }}\e ^{-ia\omega }\right ) =\mathcal {F}\left (\frac {1}{\sqrt {2\pi }}\e ^{ia\omega }\right ), \]

and changing the dummy variable \(\omega \) to \(x\) gives the result.

Example 17.18. We show that for \(a\in \mR \) we have that \(\mathcal {F}\left (\cos (ax)\right )\) equals \(\sqrt {2\pi }~\dfrac {\delta _a+\delta _{-a}}{2}\).

We have

\[ \cos (ax)=\frac {\e ^{iax}+\e ^{-iax}}{2}, \]

and using Example 17.17 (applied with \(a\) and \(-a\)) then gives

\[ \mathcal {F}(x\mapsto \cos (ax))=\sqrt {2\pi }\frac {\delta _a+\delta _{-a}}{2}. \]