I have the definition of Fourier transform $$\hat f(\lambda) = \int_{\infty}^\infty f(t) \exp(- i \lambda t) dt$$ and have proved the following lemmas:

  • $\hat E(x) = \sqrt{2 \pi} E(x)$ where $E(x) = \exp(- \tfrac{1}{2} x^2)$
  • Let $f_y(x) = f(x-y)$ then $\hat f_y(\lambda) = \hat f(\lambda) \exp(- i \lambda y)$
  • Let $\varphi(\lambda) = R E (R \lambda)$ then $\hat \varphi (\lambda) = \hat E(\frac{\lambda}{R})$
  • $\int \hat f(x) g(x) dx = \int f(x) \hat g(x)$
  • $\widehat{f\star g} = \hat f \cdot \hat g$

and I want to show that $$\hat {\hat f}(x) = 2 \pi f(-x).$$

I think the idea of the proof is to use the third lemma with a shift of $f$ anda scale of $E$, so that when you take the limit of the scale it tends towards $1$ and you are just left with $f$ but I just can't make any thing work out correctly.

  • I think we also have $f \star \varphi \to f$ as $R \to \infty$.
  • $\begingroup$ Are you allowed to use fourier inversion? $\endgroup$ – user27126 Nov 20 '12 at 18:27
  • 4
    $\begingroup$ @Sanchez, I think that's what I'm trying to prove $\endgroup$ – sperners lemma Nov 20 '12 at 18:35
  • $\begingroup$ In your definition of the Fourier transforms one of the $t$ factors should be $i$. $\endgroup$ – Lukas Geyer Nov 20 '12 at 18:50
  • $\begingroup$ @LukasGeyer, thank you I have corrected it now $\endgroup$ – sperners lemma Nov 20 '12 at 18:56
  • $\begingroup$ What does the Fourier transform here act on? Integrable functions? Continuous functions? Compact support? $\endgroup$ – Lukas Geyer Nov 20 '12 at 19:24

The strongest point $\bullet$ is your fourth, because it contains two arbitrary functions. Therefore we are free to choose a suitable special $g$ and still can prove something about an arbitrary $f$. In the situation at hand choose $$g(x):=E(x/a)\ ,\qquad a>0\ .$$ Then $$\hat g(\lambda)=\int_{-\infty}^\infty E(x/a)\ e^{-i\lambda x}\ dx=a \int_{-\infty}^\infty E(t)\ e^{-i\lambda a t}\ dt=a \hat E(\lambda a)=a\sqrt{2\pi}E(\lambda a)\ .$$ It follows from your fourth $\bullet$ that $$\eqalign{\int_{-\infty}^\infty \hat f(x) E(x/a)\ dx&=\int_{-\infty}^\infty \hat f(x) g(x)\ dx=\int_{-\infty}^\infty f(x) \hat g(x)\ dx \cr&=\sqrt{2\pi}\int_{-\infty}^\infty f(x) a E(a x)\ dx \cr &=\sqrt{2\pi}\int_{-\infty}^\infty f\bigl({t\over a}\bigr) E(t)\ dt\ ,\cr}$$ and this is valid for every $a>0$. Letting $a\to\infty$ under reasonable assumptions about $f$ the left side of the last formula converges to $\int_{-\infty}^\infty \hat f(x)\ dx$ and the right side to $2\pi f(0)$.

This shows that the conjectured formula is true for $x=0$. Replacing $f$ by $f_b(x):=f(x-b)$ one easily shows that the stated formula is in fact true for all $x$.


Taking the limit $R \rightarrow \infty$ is an efficient way to show the claim under suitable assumptions. However, you might want to apply the result to the box function or to the impulse response \begin{equation} f(t) = \begin{cases} Ae^{-\alpha t}\cos(\omega_0 t) &, \ t > 0 \\ \frac{1}{2} A &, \ t = 0 \\ 0 &, \ t < 0 \end{cases} \end{equation} of an oscillator circuit. Then the transform have to be extended, that is done usually with tools from functional analysis. However then the expression of Fourier transform is not necessarily available anymore on the boundary of the domain of definition. For example the transform of the box function $\chi_{[-\pi,\pi]}(t)$ is $2\pi \textrm{sinc}(\omega)$ that doesn't have a convergent Lebesgue integral expression of Fourier transform anywhere. Otherwise $\textrm{sinc}(\omega) e^{-i\omega t}$ would be absolutely integrable, that is not true. The transform of the impulse response $f$ above is \begin{equation} \mathscr{F}f(\omega) = \frac{1}{2}\bigg(\frac{1}{i(\omega-\omega_0) + \alpha} + \frac{1}{i(\omega+\omega_0) + \alpha}\bigg) \ . \end{equation} If $\omega_0 = 0$, the imaginary part of $\mathscr{F}f(\omega)$ is $\frac{-\omega}{\omega^2 + \alpha^2}$, that doesn't again have a convergent Lebesgue integral expression of Fourier transform anywhere. Also the expression of the transform with improper Riemann-integral of the first kind from $-\infty$ to $\infty$ diverges at $0$. That's why you might want to use the improper Riemann-integral of the first kind from $0$ to $\infty$ or Cauchy principal value in the definition of the transform. Assume that \begin{eqnarray} \mathscr{F}f(\lambda) = \int_0^\infty (f(t) e^{-i\lambda t} + f(-t) e^{i\lambda t}) dt \ . \end{eqnarray} Then \begin{eqnarray} \mathscr{F}f(\lambda) = PV \int_{-\infty}^\infty f(t) e^{-i\lambda t} dt \ . \end{eqnarray} Assume that $f \in L^1$ is piecewise differentiable with finite number of pieces on every finite interval and satisfies \begin{eqnarray} f(x) = \frac{1}{2} (f(x-)+f(x+)) \ , \end{eqnarray} where \begin{eqnarray} f(x+) & = & \lim_{t \rightarrow x^+} f(t) \\ f(x-) & = & \lim_{t \rightarrow x^-} f(t) \ . \end{eqnarray} Then $f(x+)$ is right differentiable and $f(x-)$ left differentiable. Define \begin{eqnarray} h_1(t) & = & \frac{f(-x-t+)-f(-x+)}{-t}, \ t < 0 \ , \\ h_2(t) & = & \frac{f(-x-t-)-f(-x-)}{-t}, \ t > 0 \ . \end{eqnarray} Then \begin{eqnarray} -th_1(t-) & = & f(-x-t+)-f(-x+), \ x \leq 0 \ , \\ -th_2(t+) & = & f(-x-t-)-f(-x-), \ x \geq 0 \ . \end{eqnarray} We calculate \begin{eqnarray} \mathscr{F} \mathscr{F} f(x) & = & PV \int_{-\infty}^\infty PV \int_{-\infty}^\infty f(t) e^{-i\lambda t} dt e^{-i x \lambda} d\lambda = \lim_{M \rightarrow \infty} \int_{-M}^M \int_{-\infty}^\infty f(t) e^{-i\lambda t} dt e^{-i x \lambda} d\lambda \\ & = & \lim_{M \rightarrow \infty} \int_{-\infty}^\infty \int_{-M}^M f(t) e^{i\lambda(-x-t)} d\lambda dt \\ & = & \lim_{M \rightarrow \infty} \int_{-\infty}^\infty f(t) \int_{-M}^M e^{i\lambda(-x-t)} d\lambda dt = \lim_{M \rightarrow \infty} \int_{-\infty}^\infty f(-x-t) \int_{-M}^M e^{i\lambda t} d\lambda dt \\ & = & \lim_{M \rightarrow \infty} \int_{-\infty}^0 f(-x-t) \int_{-M}^M e^{i\lambda t} d\lambda dt + \lim_{M \rightarrow \infty} \int_0^\infty f(-x-t) \int_{-M}^M e^{i\lambda t} d\lambda dt \\ & = & \lim_{M \rightarrow \infty} \int_{-\infty}^0 (f(-x-t+) - f(-x+)) \int_{-M}^M e^{i\lambda t} d\lambda dt + \frac{2\pi}{2} f(-x+) \\ & & + \lim_{M \rightarrow \infty} \int_0^\infty (f(-x-t-) - f(-x-)) \int_{-M}^M e^{i\lambda t} d\lambda dt + \frac{2\pi}{2} f(-x-) \\ & = & \lim_{M \rightarrow \infty} \int_{-\infty}^{-M_1} (f(-x-t+) - f(-x+)) \int_{-M}^M e^{i\lambda t} d\lambda dt \\ & & + \lim_{M \rightarrow \infty} \int_{M_1}^\infty (f(-x-t-) - f(-x-)) \int_{-M}^M e^{i\lambda t} d\lambda dt \\ & & + \lim_{M \rightarrow \infty} \int_{-M_1}^0 -th_1(t-) \int_{-M}^M e^{i\lambda t} d\lambda dt + \lim_{M \rightarrow \infty} \int_0^{M_1} -th_2(t+) \int_{-M}^M e^{i\lambda t} d\lambda dt \\ & & + 2\pi \frac{1}{2} (f(-x+)+f(-x-)) \\ & = & 2\lim_{M \rightarrow \infty} \int_{-\infty}^{-M_1} (f(-x-t+) - f(-x+)) \frac{\sin(Mt)}{t} dt \\ & & + 2\lim_{M \rightarrow \infty} \int_{M_1}^\infty (f(-x-t-) - f(-x-)) \frac{\sin(Mt)}{t} dt \\ & & + 2\lim_{M \rightarrow \infty} \int_{-M_1}^0 -h_1(t-) \sin(Mt) dt + 2\lim_{M \rightarrow \infty} \int_0^{M_1} -h_2(t+) \sin(Mt) dt \\ & & + 2\pi f(-x) \\ & = & 2\lim_{M \rightarrow \infty} \int_{-\infty}^{-M_1} \frac{f(-x-t+)}{t} \sin(Mt) dt - 2f(-x+) \lim_{M \rightarrow \infty} \int_{-\infty}^{-M_1} \frac{\sin(Mt)}{Mt} Mdt \\ & & + 2\lim_{M \rightarrow \infty} \int_{M_1}^\infty \frac{f(-x-t-)}{t} \sin(Mt) dt - 2f(-x-) \lim_{M \rightarrow \infty} \int_{M_1}^\infty \frac{\sin(Mt)}{Mt} Mdt \\ & & + 2\pi f(-x) \\ & = & - 2f(-x+) \lim_{M \rightarrow \infty} \int_{-\infty}^{-M_1 M} \frac{\sin(t)}{t} dt - 2f(-x-) \lim_{M \rightarrow \infty} \int_{M_1 M}^{\infty} \frac{\sin(t)}{t} dt + 2\pi f(-x) \\ & = & 2\pi f(-x) \ . \end{eqnarray} The order of integration is changed by Fubini's theorem. We may also add and substract $f(-x+)$ and $f(-x-)$ under the integral. The integrals from $-\infty$ to $0$ and $0$ to $\infty$ of $\int_{-M}^M e^{i\omega t} d\omega$ are both equal to $\pi$. We can apply the Riemann-Lebesgue -lemma because of the finite intervals $[-M_1,0]$ and $[0,M_1]$ and bounded integrands. The Riemann-Lebesgue lemma is applied also on intervals $(-\infty,-M_1]$ and $[M_1,\infty)$ where the integrands are in $L^1$. The last dimit is a consequence of the existence of integral of $\frac{\sin(x)}{x}$.

Note that if $f$ can be written as Fourier integral \begin{equation} f = \mathscr{F}g \end{equation} of an $L^1$ function $g$, that it piecewise differentiable and satisfies also $g(\lambda) = \frac{1}{2}(g(\lambda+)+g(\lambda-))$ and has finite number of pieces on every finite interval, then \begin{eqnarray} \mathscr{F}\mathscr{F}f(t) & = & \mathscr{F} \mathscr{F} \mathscr{F} g(t) = PV \int_{-\infty}^\infty \mathscr{F} \mathscr{F} g(\lambda) e^{-i\lambda t} d\lambda = PV \int_{-\infty}^\infty 2\pi g(-\lambda) e^{-i(-\lambda)(-t)} d\lambda \\ & = & 2\pi PV \int_{-\infty}^\infty g(\lambda) e^{-i\lambda(-t)} d\lambda = 2\pi \mathscr{F}g(-t) = 2\pi f(-t) \ . \end{eqnarray}

This was another approach to transform of transform. Now we are able to transform the transform of impulse response of an oscillator circuit. We can also transform the transform of a finite sum of translated box functions, that are obtained for example as the output of D/A conversion. Also transform of transform of $2\pi \textrm{sinc}(t)$ converges and equals to $2\pi 2\pi \textrm{sinc}(-t) = 4\pi^2 \textrm{sinc}(t)$.


My attempt so far..

I define a dilation operator $D_\varepsilon g(\lambda) = \varepsilon g(\varepsilon \lambda)$ or $g(\varepsilon \lambda)$ I'm not sure which would be best.

As $\varepsilon \to 0$ the dilation $D_\varepsilon E$ should act like dirac delta.

$$\begin{eqnarray} \hat {\hat f}(x) &=& \int \hat f (\lambda) e^{i x \lambda} d\lambda = \lim_{\varepsilon \to 0} \int \hat f(\lambda) &e^{- i x \lambda} D_\varepsilon E (\lambda) d \lambda \\ &&& \text{(call this $g_x(\lambda)$)} \\ &=& \int f(y) \hat g(y)dy = \sqrt{2\pi} \lim_{\varepsilon \to 0} \int f(y) E\left(\tfrac{y-x}{\varepsilon}\right)dy \end{eqnarray}$$

I want to conclude it's $ = f(x)$ but I can't because either way I define dilation I get problem with epsilons.

my computation of $\hat g_x(\lambda)$:

$$= \int D_\varepsilon E (t) e^{ixt} e^{-iyt} dt = \frac{\varepsilon}{\varepsilon} \int E(s) \exp\left({-i\frac{y-x}{\varepsilon}s}\right)ds = \hat E\left(\frac{y-x}{\varepsilon}\right) = \sqrt{2 \pi}E\left(\frac{y-x}{\varepsilon}\right)$$


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