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I am going through some lecture notes on Fourier transforms (here) and it is stated without proof (example 2.16 on page 29) that the general solution to the equation
$$x f(x) = a$$ is given by $$f(x) = a/x + b\, \delta(x)$$ and the general solution to
$$x^2 f(x) = a$$ is given by $$f(x) = a/x^2 + b \delta(x)/x + c \delta(x) + d\delta'(x).$$ I don't really understand how $\delta$ appears here. Could anyone please give a hint/proof?

I know that $\delta$ is a distribution so I tried integrating out both sides of these equations w.r.t $x$, but I failed to show that they give similar results. Also, I think an expression like $\delta(x)/x$ is ambiguous except if integrated against another function that can possibly cancel out the denominator.

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    $\begingroup$ Even more general solutions exist, but wouldn't be "suitably well-behaved" (see the first sentence of Sec. 2.10). So you'll need to understand how that term is defined therein. Example 2.16 cites two sources proving the first result; no doubt they'd be explicit. $\endgroup$ – J.G. Jul 16 at 10:23
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    $\begingroup$ Could you give an example of these 'even more general solutions'? $\endgroup$ – SaMaSo Jul 16 at 10:38
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First, if $f_1$ and $f_2$ are solutions to $Tf=g,$ where $T$ is some linear operator and $g$ is given, then $f_1-f_2$ is a solution to $Tf=0.$ Therefore we will study $x f(x)=0.$ It can easily be shown in distribution theory that $x\delta(x)=0,$ $x^2\delta(x)=0,$ and $x^2\delta'(x)=0,$ but since you're studying Fourier transforms I will give an explanation using Fourier transforms:

Take the equation $x f(x) = 0$ and apply the Fourier transform to both sides. You get $i\hat{f}'(\xi) = 0.$ This is a differential equation with solutions $\hat{f}(\xi) = C,$ where $C$ is a constant. Taking the inverse Fourier transform gives us $f(x) = C\delta(x).$

Likewise, $x^2 f(x) = 0$ transforms to $-\hat{f}''(\xi)=0$ with solutions $\hat{f}(\xi) = A\xi+B,$ i.e. $f(x) = -iA\delta'(x)+B\delta(x).$

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    $\begingroup$ Thank you this is helpful. Just one question: what happens to the $b \delta(x)/x$ term in the second solution? $\endgroup$ – SaMaSo Jul 16 at 11:34
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    $\begingroup$ $\delta(x)/x$ is not a welldefined distribution, but can best be interpreted as $-\delta'(x)$ since $x\delta' = (x\delta)' - x'\delta = 0' - 1\delta = -\delta.$ $\endgroup$ – md2perpe Jul 16 at 12:19
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So, informally, the Dirac $\delta$ is zero everywhere except at $0$ and has integral $1$. So, informally, $\delta$ is infinite at $0$, therefore $\delta$ is not admitted by traditional analysis. In regular analysis, given $x.f(x) = a$, we divide both sides by $x$ to obtain $f(x) = a/x$ but we can add any number (say $b$) times $\delta(x)$ on to $a/x$ as when $x$ is not zero any number times $\delta(x)$ is just $0$ and when $x$ is $0$ then $x.f(x)$ is still zero and so adding $b. \delta (x)$ on to $a/x$ does not change the truth of the fact that $x.f(x)=a$.

Now perhaps the other solution will make sense but it might help to know that if $\delta^{'}(x)$ is the derivative of the Dirac function then $\delta^{'}(x)=-\delta(x)/x$ so $\delta^{'}$ is 'even more infinite' than $\delta(x)$ :).

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    $\begingroup$ That's a good point. I meant it informally but you are right; t would be accurate to say $x.\delta^{'}(x) =\delta{x}$. $\endgroup$ – Simon Terrington Jul 16 at 11:51
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    $\begingroup$ Yes, actually this notation was bothering me, so now I understand why some people use it. $\endgroup$ – LL 3.14 Jul 16 at 11:59
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    $\begingroup$ OK. So the Dirac $\delta$ makes sense when we consider $x.f(x)$, but is it also correct to attribute $\delta$ to $f(x)$ only? Because once $f$ is calculated here, it's product with any other arbitrary function could be considered $\endgroup$ – SaMaSo Jul 16 at 12:54
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To complete the good answer given by md2perpe, you just need to get one particular solution of the equations. In dimension $1$ however, $1/x$ and $1/x^2$ are not locally integrable functions, and so one should define them as principal values (and one sometimes writes $\mathrm{P}(\tfrac{1}{x}) = \mathrm{pv.}(\tfrac{1}{x})$ and $\mathrm{fp.}(\tfrac{1}{x^2})$ for principal value and finite part). For any smooth and compactly supported function $\varphi$, they are defined by $$ \langle\mathrm{P}(\tfrac{1}{x}),\varphi\rangle = \int_{\mathbb{R}} \frac{\varphi(x)-\varphi(0)}{x}\,\mathrm{d} x $$ which can also be written $\langle\mathrm{P}(\tfrac{1}{x}),\varphi\rangle = \lim_{\varepsilon\to 0}\int_{|x|>\varepsilon} \frac{\varphi(x)}{x}\,\mathrm{d} x$. One can easily verify that $$ x\, \mathrm{P}(\tfrac{1}{x}) = 1 $$

So the general solution for the first equation is $$ f(x) = a \, \mathrm{P}(\tfrac{1}{x}) + b \, \delta_0 $$

In the same way, one can define $$ \langle\mathrm{pf.}(\tfrac{1}{x^2}),\varphi\rangle = \int_{\mathbb{R}} \frac{\varphi(x)-\varphi(0)- x \varphi'(0)}{x^2}\,\mathrm{d} x $$ and then the general solution of the second equation is $$ f(x) = a \, \mathrm{pf.}(\tfrac{1}{x^2}) + b \, \delta_0 + c \, \delta_0' $$

Edit: $\delta_0(x)/x$ has no clear meaning in distribution theory. However, as indicated by Simon Terrington, one could define $\delta_0(x)/x = -\delta_0'(x)$ since it is one of the solution of the equation $$ x\,g(x) = -\delta_0(x). $$ The general solution being $g(x) = -\delta_0' + c\, \delta_0$. It is better to use just $\delta_0'$.

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    $\begingroup$ Could you please elaborate more on the definition of $pf. (1/x^2)$? I don't understand why there is $\phi'(0)$ in the numerator of the integrand. Is this constructed the same way as Cauchy principal value? $\endgroup$ – SaMaSo Jul 16 at 12:51
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    $\begingroup$ @dedekindCuttage. Not in the same way, but in a similar way. If you think of the Maclaurin expansion of $\varphi$ then you can see that for the Cauchy principal value the constant term is removed, and for the finite part both the constant and the linear terms are removed so make the division work. $\endgroup$ – md2perpe Jul 16 at 13:38
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$\def\a{\alpha}$Here is another approach.

We begin by taking the Fourier transform of each side of the original equation: \begin{align*} x f(x) &= a \\ \int_{-\infty}^\infty x f(x) e^{ikx}dx &= \int_{-\infty}^\infty a e^{ikx}dx \\ -i\frac{\partial}{\partial k} \int_{-\infty}^\infty f(x) e^{ikx}dx &= 2\pi a\delta(k) \\ \hat f'(k) &= 2\pi i a \delta(k) \end{align*} This differential equation can be solved by standard methods, with the result \begin{align*} \hat f(k) &= 2\pi i a\Theta(k) + c, \end{align*} where $\Theta$ is the Heaviside step function.

All that remains is to perform the inverse transform: \begin{align*} f(x) &= \frac{1}{2\pi}\int_{-\infty}^\infty (2\pi i a \Theta(k)+c)e^{-ikx}dk \\ &= ia\int_{-\infty}^\infty \Theta(k)e^{-ikx}dk + c\frac{1}{2\pi}\int_{-\infty}^\infty e^{-ikx}dk \\ &= i a\left(-\frac{i}{x}+\pi\delta(x) \right)+c\delta(x) \\ &= \frac{a}{x} + b\delta(x), \end{align*} where $b=ia\pi+c$.

Aside: Fourier transform of the Heaviside step function

\begin{align*} \int_{-\infty}^\infty \Theta(k)e^{-ikx}dk &= \lim_{\a\rightarrow 0^+} \int_{-\infty}^\infty \Theta(k)e^{-\a k}e^{-ikx}dk \\ &= \lim_{\a\rightarrow 0^+} \int_0^\infty e^{-k(\a+i x)}dk \\ &= \lim_{\a\rightarrow 0^+} \left.\frac{-e^{-k(\a+i x)}}{\a+i x}\right|_0^\infty \\ &= \lim_{\a\rightarrow 0^+} \frac{1}{\a+i x}\\ &= \lim_{\a\rightarrow 0^+} \left(\frac{\a}{\a^2+x^2}-i\frac{x}{\a^2+x^2}\right) \\ &= \pi \delta(x) - \frac{i}{x}. \end{align*} In the last step we use that $\a/(\pi(\a^2+x^2))$ is a standard nascent delta function.

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