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Given a continuous non-periodic function, its Fourier transform is defined as:

$$f(x) = \int_{-\infty}^\infty c(k) e^{ikx} dk, \ \ \ \ \ \ \ \ \ \ \ \ \ c(k) = \frac{1}{2\pi} \int_{-\infty}^\infty f(x) e^{-ikx} dx$$

The problem is proving this is true by evaluating the integral when $c(k)$ is plugged into the equation for $f(x)$.

This ends up with a long integral:

$$f(x) = \int_{-\infty}^\infty \left( \frac{1}{2\pi} \int_{-\infty}^\infty f(x') e^{-ikx'} dx' \right) e^{ikx} dk$$

I'm not sure really how to proceed from here. I moved the $e^{ikx}$ into the inner integral, which I figured was fine since it's constant relative to $x'$.

$$f(x) = \int_{-\infty}^\infty \left( \frac{1}{2\pi} \int_{-\infty}^\infty f(x') e^{ik(x-x')} dx' \right) dk$$

I tried to kill at least one of the integrals by seeing if something evaluated to a Dirac Delta but I can't seem to get that result. I also tried integrating by parts, but that led me nowhere.

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  • $\begingroup$ First: the Fourier transform of a continuous non-periodic function is NOT defined in that way. Second: it is difficult to say something sensible about a ill posed question... $\endgroup$ – Bob Aug 4 '18 at 11:47
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    $\begingroup$ You are trying to prove the inversion formula (which is valid if $f$ and $c$ are both integrable. You cannot do this by just plugging in the expression for $c$ into the first formula. This is a non-trivial result and you can get a proof from any book on Fourier Analysis $\endgroup$ – Kavi Rama Murthy Aug 4 '18 at 11:49
  • $\begingroup$ @Bob the $c(k)$ is the FT of $f(x)$ in this question. And it's indeed one of the many definitions of the FT. $\endgroup$ – Botond Aug 4 '18 at 11:50
  • $\begingroup$ @Botond sure, for example, f(x)=x is a continuous non-periodic function. How $c(k)$ is defined for such a function? $\endgroup$ – Bob Aug 4 '18 at 11:51
  • $\begingroup$ @Botond that is distributional Fourier transform and it is NOT defined through that integral $\endgroup$ – Bob Aug 4 '18 at 11:55
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Let $f\in L^1(\mathbb{R}^d)$ such that $\widehat{f}\in L^1(\mathbb{R}^d)$ and:

$$ \forall x\in\mathbb{R}^d, \forall \epsilon\in\mathbb{R}, \phi(x)=\pi^{-\frac{d}{2}}\exp(-|x|^2), \phi_\epsilon(x)=\frac{1}{\epsilon^d}\phi\left(\frac{1}{\epsilon}x\right)\\ $$ Lemma:$$ \mathcal{F}(\exp(-a|\cdot|^2))(\xi)=\left(\frac{\pi}{a}\right)^{\frac{d}{2}}\exp\left(-\frac{|\xi|^2}{4a}\right) $$ Then: $$ \begin{equation} (2\pi)^{-d}\int\limits_{\mathbb{R}^d}\exp(i\langle\xi, x\rangle)\widehat{f}(\xi)\mathrm{d}\xi=\lim\limits_{\epsilon\to0}\int\limits_{\mathbb{R}^d}\exp(i\langle\xi,x\rangle)\widehat{\phi_\epsilon}(x)\widehat{f}(\xi)\mathrm{d}\xi \end{equation} $$

By Fubini theorem: $$ (2\pi)^{-d}\int\limits_{\mathbb{R}^d}\exp(i\langle\xi, x\rangle)\widehat{\phi_\epsilon}(\xi)\widehat{f}(\xi)\mathrm{d}\xi=(2\pi)^{-d}\int\limits_{\mathbb{R}^d\times\mathbb{R}^d}\exp(i\langle\xi, x\rangle)\widehat{\phi_\epsilon}(\xi)f(y)\mathrm{d}\xi\mathrm{d}y $$ So: $$ (2\pi)^{-d}\int\limits_{\mathbb{R}^d}\exp(i\langle\xi, y-x)\widehat{\phi_\epsilon}(\xi)\mathrm{d}\xi=(2\pi)^{-d}\left(\frac{4\pi}{\epsilon^2}\right)^{\frac{d}{2}}\exp\left(-\frac{|x-y|^2}{\epsilon^2}\right)=\phi_\epsilon(x-y) $$ And: $$ \forall \epsilon>0, (\phi_\epsilon\star f)(x)=(2\pi)^{-d}\int\limits_{\mathbb{R}^d}\exp(i\langle x, \xi\rangle)\widehat{\phi_\epsilon}(\xi)\widehat{f}(\xi)\mathrm{d}\xi $$ Let $(\epsilon_n)_{n\in\mathbb{N}}$ be a sequence of reals, $\epsilon>0$, that converge to 0. Then $\lim\limits_{n\infty}\|\phi_{\epsilon_n}\star f-f\|_{L^1(\mathbb{R}^d)}=0$, there is a subsequence such that for allmost all $x\in\mathbb{R}^d$, $$ f(x)=(2\pi)^{-d}\int\limits_{\mathbb{R}^d}\exp(i\langle\xi, x\rangle)\widehat{f}(\xi)\mathrm{d}\xi $$

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Start by assuming $f$ is absolutely integrable. Then $$ \frac{1}{2\pi}\int_{-R}^{R}\int_{-\infty}^{\infty}f(x')e^{-ikx'}dx' e^{ikx}dk \\ = \frac{1}{2\pi}\int_{-\infty}^{\infty}f(x')\int_{-R}^{R}e^{ik(x-x')}dk dx' \\ = \frac{1}{\pi}\int_{-\infty}^{\infty}f(x')\frac{\sin(R(x-x'))}{x-x'}dx'. $$ The kernel $K(x,x')=\frac{1}{\pi}\frac{\sin(R(x-x'))}{x-x'}$ has a principal value integral in $x'$ equal to $1$. So, if $f$ is differentiable at $x$, then the above may be written as $$ \lim_{R\rightarrow\infty}\frac{1}{\pi}\int_{-R}^{R} \frac{f(x')-f(x)}{x-x'}\sin(R(x-x'))dx'+f(x), $$ and the limit of the integral is $0$ by the Riemann-Lebesgue lemma. So, you can a sense in which $\frac{1}{2\pi}\int_{-R}^{R} e^{ik(x-x')}dk \rightarrow \delta_{x}(x')$ as $R\rightarrow\infty$.

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From your work: $$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x')e^{ik(x-x')} \mathrm{d}x' \mathrm{d}k$$ $$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x')e^{ik(x-x')} \mathrm{d}k\mathrm{d}x'$$ $$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x')\int_{-\infty}^{\infty} e^{ik(x-x')} \mathrm{d}k\mathrm{d}x'$$ $$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x')\int_{-\infty}^{\infty} e^{-ik(x'-x)} \mathrm{d}k\mathrm{d}x'$$ $$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x')2\pi \delta(x'-x)\mathrm{d}x'$$ $$f(x)=\int_{-\infty}^{\infty}f(x')\delta(x'-x)\mathrm{d}x'$$

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