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Let $a$ be a positive real number and let $f$ be defined as follows:

$$ f(x)=\frac{1}{|x|^2-a}, \ \ x \in \mathbb{R}^3 $$

As I understand the normal inverse Fourier transform of $f$ is not defined since $f$ is not absolute integrable. Why is the inverse Fourier transform of $f$ defined, when we look at $f$ as a Distribution?

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    $f$ doesn't define a distribution, you need to slap a principal value or something similar on it to get a distribution. And once one has done that, it's easy to see that it's a tempered distribution. Since the Fourier transform is an automorphism of $\mathscr{S}(\mathbb{R}^n)$, its transpose (which we also call Fourier transform) is an automorphism of $\mathscr{S}'(\mathbb{R}^n)$, the space of tempered distributions. So every tempered distribution is the Fourier transform of a tempered distribution. – 2017-02-04
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    Ok, thank you. Actually this question comes from this problem: I want to solve a PDE and have transformed it to the equation $(x^2-a)\hat{u}(x)=\hat{\delta_0}=1$. Now I devided and got the equation from the above question (is it actually alowed to devide, since the "1" is the fourier transform of the delta distribution?). If I have then a function with singularities, does that always mean that I have to work with a principle value? – 2017-02-04
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    You can divide if you get something sensible. Formally, the question is whether for a distribution $T$ and a nice enough function $f$ there exists a distribution $S$ such that $T = f\cdot S$, where the distribution $f\cdot S$ is defined by $(f\cdot S)[\varphi] = S[f\cdot \varphi]$. Roughly, that means $\varphi \mapsto T[\varphi/f]$ must be well-defined and define a distribution. Here, we have $T$ given by $T[\varphi] = \int \varphi$. – 2017-02-04
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    If $f$ is a function such that $1/f(x)$ is locally integrable, things work fine, and no principal value is required. But with $f(x) = \lvert x\rvert^2 - a$, $f$ vanishes on the sphere of radius $\sqrt{a}$ and $1/f(x)$ isn't locally integrable. But the bad parts of $1/f$ cancel out, so the principal value $$\varphi \mapsto \lim_{\varepsilon \to 0} \int_{\bigl\lvert\lvert x\rvert - \sqrt{a}\bigr\rvert > \varepsilon} \frac{\varphi(x)}{\lvert x\rvert^2-a}\,dx$$ exists and gives us a nice tempered distribution. – 2017-02-04
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    Ah ok, so if we are in 1D and $a=1$ then the distribution would be $$L \in \mathcal{S}(\mathbb{R})' \to \mathbb{C}, L\phi = \lim_{\epsilon \to 0}{\int_{||x|-1|>\epsilon}{\frac{\phi(x)}{x^2-1}dx}}$$ right? Is there any way to get a nicer representation of this distribution? – 2017-02-04
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    In several articles about the above PDE the people do the following: To the constant $a$ they add $i\epsilon$ to get an $L^1$ function with singularities outside the real axis. And then they just say that $u(x)=\lim_{\epsilon \to 0}\mathcal{F}^{-1}(\frac{1}{x^2-(a+i\epsilon)})$ where $\mathcal{F}^{-1}$ is the normal inverse F.T. I have the feeling that this has something to do with the principle value, but I can't see it. – 2017-02-04
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    It's similar. For the principal value, we avoid the bad behaviour of $1/f$ at the sphere $\lvert x\rvert = \sqrt{a}$ by setting the values to $0$ on a small spherical shell, and then take the limit letting the thickness of the shell tend to $0$. In this method, we avoid the bad behaviour by adding a small purely imaginary value to the denominator, thus we never divide by $0$, and then letting the added constant tend to $0$. Unsurprisingly, the limit is the same distribution for both approaches. – 2017-02-04
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    First of all, thanks a lot for your help. I continued working on the problem today and it seams like it's enough to restrict oneself on the 1D-case with a=1. I tried to write down your solution rigorous but a last step is missing: I made the following approach: – 2017-02-05
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    Let $Q \in \mathcal{S}'(\mathbb{R})$ with $Q\phi = \lim_{\epsilon \to 0}{\int_{||x|-1|>\epsilon}{\frac{\phi(x)}{x^2-1}dx}}$ and let $L_f$ be the distribution associated to a normal integrable function, i.e. $L_f\phi=\int_{\mathbb{R}}{f(x)\phi(x)dx}$. Set $g_\epsilon(x)=\frac{1}{x^2-(1+i\epsilon)}$. We want to show that $\lim_{\epsilon \to 0}{L_{g_\epsilon}} = Q$, i.e. $\lim_{\epsilon \to 0}{L_{g_\epsilon}\phi} = Q\phi \ \ \forall \phi \in \mathcal{S}(\mathbb{R})$. Is everything right up to here? – 2017-02-05
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    Now I have to show that $\lim_{\epsilon \to 0}{\int_{\mathbb{R}}{g_\epsilon(x)\phi(x)dx}} = \lim_{\epsilon \to 0}{\int_{\mathbb{R}}{\frac{\phi(x)}{x^2-(1+i\epsilon)}dx}} \stackrel{?}{=} \lim_{\epsilon \to 0}{\int_{||x|-1|>\epsilon}{\frac{\phi(x)}{x^2+1}}}$. I don't know how to show the last equation rigorously. – 2017-02-05
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    Right so far on your part, but I was wrong. The limits aren't the same. Note that away from $\pm 1$, $\frac{1}{x^2-1}$ behaves well. So choose a small $\delta > 0$ and split the integral into three parts, over $[-1-\delta,-1+\delta]$, over $[1-\delta,1+\delta]$, and the rest. The third part is dealt with by dominated convergence. On $[1-\delta,1+\delta]$ write $\phi(x) = \phi(1) +\bigl(\phi(x)-\phi(1)\bigr)$. Since $\lvert\phi(x)-\phi(1)\rvert \leqslant \lvert x-1\rvert\cdot\lVert\phi'\rVert_{\infty}$, again by dominated convergence we have – 2017-02-05
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    $$\lim_{\epsilon\to 0}\int_{1-\delta}^{1+\delta}\frac{\phi(x)-\phi(1)}{x^2-(1+i\epsilon)}\,dx = \int_{1-\delta}^{1+\delta}\frac{1}{x+1}\cdot\frac{\phi(x)-\phi(1)}{x-1}\,dx = \lim_{\epsilon\to 0} \int_{\epsilon < \lvert x-1\rvert < \delta} \frac{\phi(x)-\phi(1)}{x^2-1}\,dx.$$ For the constant term $\phi(1)$, we make a partial fraction decomposition, $\frac{1}{x^2-1} = \frac{1}{2}\bigl(\frac{1}{x-1} - \frac{1}{x+1}\bigr)$ and $\frac{1}{x^2-(1+i\epsilon)} = \frac{1}{2\rho(\epsilon)}\bigl(\frac{1}{x-\rho(\epsilon)} - \frac{1}{x+\rho(\epsilon)}\bigr)$ – 2017-02-05
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    where $\rho(\epsilon)$ is the square root of $1+i\epsilon$ with positive real part. The terms with $+$ in the denominator are regular ($\rho(\epsilon) \to 1$) and tend to the same limit $\int_{1-\delta}^{1+\delta}\frac{dx}{x+1}$. But for the remaining term we get different limits. By symmetry, $$\int_{\epsilon < \lvert x-1\rvert < \delta} \frac{dx}{x-1} = 0$$ for all $0 < \epsilon < \delta$. – 2017-02-05
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    However, with e.g. the principal branch of the logarithm $$\begin{aligned}\int_{1-\delta}^{1+\delta} \frac{dx}{x-\rho(\epsilon)} &= \log (\delta + (1-\rho(\epsilon)) - \log (-\delta + (1-\rho(\epsilon))\\ &= \log (\delta - i\epsilon/2+O(\epsilon^2)) - \log (-\delta - i\epsilon/2 + O(\epsilon^2)) \to \pi i.\end{aligned}$$ Things are similar on $[-1-\delta,-1+\delta]$, and so the two limits differ in the end by $\dfrac{\pi i}{2}\bigl(\phi(1) + \phi(-1)\bigr)$ [if I haven't made a sign error]. – 2017-02-05
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    Ah ok, I see. I thought that the above problem solving $(x^2-a)\hat{u}(x)=\hat{\delta_0}=1$ is not that hard, but it seams like it's not easy ... For example in this script ( http://www.cv.titech.ac.jp/~hiro-lab/lecture/Vibration_Wave_Analysis/text-4.pdf , equation 106-108) and in other books they just devide by $x^2-a$ and say that $u(x)=\lim_{\epsilon \to 0}\mathcal{F}^{-1}(\frac{1}{x^2-(a+i\epsilon)})$. Everything else is just an application of the residue theorem or (when we are in higher dimensions) coordinate transformations. – 2017-02-05
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    But I can't really see why we can add an $\epsilon$ and say that the solution is the above limit. Do you have any other suggestions or some kind of intuitive idea? – 2017-02-05
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    Well, I guess the idea is that if $u_{\epsilon}$ is the solution to $\Delta u_{\epsilon} + (a+i\epsilon) u_{\epsilon} = -\delta_0$, then $u_{\epsilon}$ tends to a solution of $\Delta u + au = -\delta_0$ as $\epsilon \to 0$. I don't know if that only looks like it may be true or there's some real reason to expect that. I've happily forgotten what little I once knew about regularity of PDEs. – 2017-02-05

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