Chapter H Problem Sheet 8 (Lectures 16-17)
H.1 Dilation formula
For \(c>0\) and \(u\in L^1(\mR )\) show that
\[ \mathcal {F}\left (x\mapsto u(cx)\right )= \omega \mapsto \frac {1}{c}\left (\mathcal {F}(u)\right )\left (\frac {\omega }{c}\right ). \]
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Solution. We have (using the substitution \(y=cx\))
\[ \mathcal {F}\left (x\mapsto u(cx)\right ) =\frac {1}{\sqrt {2\pi }}\int _{-\infty }^\infty \e ^{-i\omega x}u(cx)\,dx =\frac {1}{\sqrt {2\pi }}\int _{-\infty }^\infty \e ^{-i\omega y/c}u(y)\,\frac {1}{c}\,dy =\frac {1}{c}\left (\mathcal {F}(u)\right )\left (\frac {\omega }{c}\right ). \]
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H.2 Fourier transform of general Gaussian
Use Example 14.10 and Problem H.1 to obtain the formula in Remark 14.11.
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Solution. From Example 14.10 we have that \(\mathcal {F}\left (x\mapsto \e ^{-x^2/2}\right )=\omega \mapsto \e ^{-\omega ^2/2}\). Applying Problem H.1 with \(c=\sqrt {a}\) gives
\[ \mathcal {F}\left (x\mapsto \e ^{-(\sqrt {a} x)^2/2}\right )=\omega \mapsto \frac {1}{\sqrt {a}}\mapsto \e ^{-(\omega /\sqrt {a})^2/2}, \]
which is
\[ \mathcal {F}\left (x\mapsto \e ^{-ax^2/2}\right )=\omega \mapsto \frac {1}{\sqrt {a}}\mapsto \e ^{-\omega /(2a)}, \]
as claimed in Remark 14.11. □
H.3 Fourier transform of sine
Let \(a\in \mR \). Show that the Fourier (-Schwartz) transform of \(\sin (ax)\) equals \(\sqrt {2\pi }~\dfrac {\delta _a-\delta _{-a}}{2i}\).
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Solution. We have
\[ \sin (ax)=\frac {\e ^{iax}-\e ^{-iax}}{2i}, \]
and using Example 17.17 (applied with \(a\) and \(-a\)) then gives
\[ \mathcal {F}(x\mapsto \sin (ax))=\sqrt {2\pi }\frac {\delta _a-\delta _{-a}}{2i}. \]
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H.4 Distributional dilation formula
For \(b>0\) and a function \(f\) define
\[ (D_bf)(x):=f(bx), \]
and for a distribution \(u\) define the distribution \(\tilde {D}_au\) by
\[ (\tilde {D}_au)(\varphi ):=\frac {1}{a}u\left (D_{1/a}\varphi \right ). \]
(a) Show that if \(u\) is the regular distribution corresponding to the locally integrable function \(f\), then \(\tilde {D}_au\) is the regular distribution corresponding the function \(D_af\).
(b) Show that we have
\[ \mathcal {F}\tilde {D}_au=\frac {1}{a}\tilde {D}_{1/a}\mathcal {F}u. \]
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Solution. (a) We have for the regular distribution corresponding to the function \(D_af\)
\[ T_{D_af}(\varphi ) =\int _\mR (D_af)(x)\varphi (x)\,dx =\int _\mR f(ax)\varphi (x)\,dx. \]
Using the change of variables \(y:=ax\) gives
\[ \int _\mR f(ax)\varphi (x)\,dx =\int _\mR f(y)\varphi (a^{-1}y)\,a^{-1}dy =\frac {1}{a}T_f\left (D_{1/a}\varphi \right ), \]
where \(T_f\) is the regular distribution corresponding to the function \(f\). Therefore
\[ T_{D_af}(\varphi )=\frac {1}{a}T_f\left (D_{1/a}\varphi \right ), \]
and since this holds for all \(\varphi \in \mathcal {D}\) we have
\[ T_{D_af}=\frac {1}{a}T_fD_{1/a}. \]
From the definition of \(\tilde {D}_a\) we then have \(T_{D_af}=\tilde {D}_aT_f\) as desired.
(b) We have (using the definition of Fourier transform and dilation for distributions):
\[ \left (\mathcal {F}\tilde {D}_au\right )(\varphi ) =\tilde {D}_au\left (\hat {\varphi }\right ) =\frac {1}{a}u\left (D_{1/a}\hat {\varphi }\right ). \]
Using the dilation formula for functions we have
\[ \frac {1}{a}D_{1/a}\hat {\varphi }=\mathcal {F}D_a\varphi . \]
Therefore
\[ \left (\mathcal {F}\tilde {D}_au\right )(\varphi ) =u\left (\mathcal {F}D_a\varphi \right ). \]
Using the definition of Fourier transform and dilation (with \(a\) replaced by \(1/a\)) for distributions we have
\[ u\left (\mathcal {F}D_a\varphi \right ) =\left (\mathcal {F}u\right )\left (D_a\varphi \right ) =\frac {1}{a}\left (\tilde {D}_{1/a}\mathcal {F}u\right )(\varphi ), \]
so that
\[ \left (\mathcal {F}\tilde {D}_au\right )(\varphi ) =\frac {1}{a}\left (\tilde {D}_{1/a}\mathcal {F}u\right )(\varphi ), \]
and since this holds for all \(\varphi \in \mathcal {S}\) we have
\[ \mathcal {F}\tilde {D}_au=\frac {1}{a}\tilde {D}_{1/a}\mathcal {F}. \]
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