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I am trying to compute the Fourier transform of the Heaviside step function in the space of tempered distributions. $$\langle \mathscr{F}(H), \varphi \rangle \stackrel{\mathrm{\mathscr{S}'(\mathbb{R})}}{=} \langle H, \mathscr{F}(\varphi) \rangle $$ Since $H(x)$ is a moderate function (at most polynomial divergence approaching infinity) I can write (Can I? I know that in $\mathscr{D}'$ I can since the distribution is regular but I'm not entirely sure about $\mathscr S '$)

$$= \int_{-\infty}^\infty H(x) \left( \int_{-\infty}^\infty \varphi(y) e^{-2 \pi i x y} dy \right) dx \stackrel{(1)}{=} \lim_{\varepsilon \to 0^+} \int_{0}^\infty e^{-\varepsilon x^2} \left( \int_{-\infty}^\infty \varphi(y) e^{-2 \pi i x y} dy \right) dx \\ \stackrel{(2)}{=} \lim_{\varepsilon \to 0^+} \int_{-\infty}^\infty \varphi(y) \left( \int_{0}^\infty e^{-\varepsilon x^2} e^{-2 \pi i x y} dx \right) dy $$

and that after some integration (completing the square in the exponent, changing the variable, integrating the inner integral and changing the variable again) becomes

$$= \lim_{\varepsilon \to 0^+} \frac{1}{2\sqrt{\pi}} \int_{-\infty}^\infty \varphi \left(\frac{\sqrt{\varepsilon}}{\pi}u\right) e^{-u^2}du \stackrel{(3)}{=} \frac{1}{2} \varphi(0) = \frac{1}{2} \langle \delta, \varphi \rangle $$

In (1) I am using the monotone convergence theorem (edit: as was pointed out in the comments dominated convergence here as well), in (3) the dominated convergence theorem and in (2) Fubini's theorem.

Usually one manages to lose the delta part along the way but I managed to lose the $ \sim \frac{1}{x}$ part.

Where does this approach go wrong?

VaNa
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  • Why do you want to use this approach? There is arguably a better way to solve this problem by using the identity $H'=\delta_0$ and then via some symmetry argument getting rid of extra undefiined constants in the final expression. – TZakrevskiy Jan 16 '17 at 15:25
  • The step (1) can not be done by monotone convergence theorem without some additional argument, because this theorem applies to non-negative-valued functions, and here you have a complex-valued function. You could split it into real and imaginary parts, but they could be negative... Again, splitting them into positive and negative parts leads to unnecessary difficulties. – TZakrevskiy Jan 16 '17 at 21:21
  • However, you can safely apply the dominated convergence there (separately for real and imaginary parts). – TZakrevskiy Jan 16 '17 at 21:29
  • Actually that $H' = \delta$ approach was the first one that came to my mind but I couldn't quite get through the division problem as $\xi \mathscr F (H)(\xi) =\frac{1}{2 \pi i}$ obviously does not imply in $\mathscr{S}'$ that $\mathscr F (H)(\xi)=\frac{1}{2 \pi i \xi}$. The problem being at 0 and that generates the $\delta$ but I don't see why there can't be $\delta^{(n)}$ (there's a theorem that when a distribution has $\mathrm{supp} = {0}$ it can be written as a sum of derivatives of deltas) or how one really derives the correct result. – VaNa Jan 17 '17 at 16:47
  • This is an easy exercise; let a distribution $T$ satisfy the equation $xT=0$. By that theorem $T = \sum_{k=0}^N a_k \delta^{(k)}0$. Now take test functions $\phi{p}$ such that $\phi_p^{(p)}(0) = 1 $ and $\phi_p^{(q)}(0)=0$ for $q\ne p$. Apply the distribution $xT$ to $\phi_p$ and see what you can deduce from that. – TZakrevskiy Jan 17 '17 at 16:52
  • With the method in the previous comment I got inspired by this question

    http://math.stackexchange.com/questions/1353607/fourier-transform-of-the-heaviside-function

    but there's the problem with the derivatives of deltas and how the deltas got there in the first place.

     – VaNa Jan 17 '17 at 16:54
  • @TZakrevskiy Ok, I think I have it. One can check that (like in the linked question) $\mathrm{v.p.}\frac 1x$ is a solution to $xT = 1$ and from your excercise it follows that $a_k=0$ for $k>0$ so from $1 = 1 + 0$ and linearity of the equation yields $T = \mathrm{v.p.}\frac 1x + c \delta$ where $c$ can be easily fixed. Thank you! – VaNa Jan 17 '17 at 19:08

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The devil is in the details. When you write the integral $$\int_0^{\infty} \exp(-(x-iy)^2)$$ you can not simply say that $$\int_0^{\infty} \exp(-(x-iy)^2) = \frac 12 \int_{-\infty}^{\infty} \exp(-(x-iy)^2)=\frac{\sqrt \pi}{2},$$because it is false.

In fact, you obtain the imaginary error-function $\operatorname {erfi}$: $$\int_0^{\infty} \exp(-(x-iy)^2) = \frac 12 \int_{-\infty}^{\infty} \exp(-(x-iy)^2)=\frac 12 \sqrt \pi (1+i\operatorname {erfi}(y)),$$ and this additional complex-valued term would - after some manipulations - yield you the principal value of $\frac 1x$.

In my opininion, a safer way to deduce the Fourier transform would be to use the $sign$ function: $$H(x) = \frac 12 (1+ sign(x)),$$ and for the sign function you write that $( sign (x))' = \delta _0$, hence $i \xi F[ sign (x)](\xi) = F[\delta_0] = 1$. Then you can apply the well-known idea about the division by $\xi$ to get $$F[ sign (x)](\xi) = -i\,pv(1/\xi) + C\delta_0.$$ The sign function is odd, hence its Fourier transform must also be odd, so $C=0$.

Finally, $$F[H] = F\left[\frac 12 (1+ sign(x))\right] = \frac 12 (\delta_0 - i\,pv(1/\xi)).$$ The exact constants depend, of course, on your normalisation of the Fourier transform.

TZakrevskiy
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    Yes, it was eventually pointed out by my lecturer that the error was there. I just overlooked that the coordinate transform of $e^{-(x-iy)^2} \mapsto e^{-u^2}$ would also shift the contour of the integral into the complex numbers. I've tried to integrate it and ended up with the $erfi$ and all that mess. HOWEVER I don't seem to be aquainted with the well-known idea about the division by $\xi$. Can you please explain that or point me to a source? My thought about the division is it can be thought of as multiplication by a $\mathcal{C}^\infty$ function $1/\xi$ on $\mathbb{R}\setminus {0}$ – VaNa Jan 17 '17 at 17:10