# Intuition behind the proof of the Inverse Fourier Transform?

I am interested in the proof of the Inverse Fourier Transform for absolutely integrable real valued functions.

The proof I have read asks you to consider an auxiliary function $g_{a}(x)$ defined as

$$g_{a}(x) = \int_{-\infty}^{\infty} e^{-a \lvert t\rvert}e^{2\pi i t x}\operatorname{d}\!t = \frac{2a}{4\pi^{2}x^{2}+a^{2}}.$$

It then proceeds to prove that $\lim_{a\rightarrow 0 }f*g_{a}(x) = f(x)$ and therefore shows that $\hat{\hat{f}}(x) = f(-x)$ by the dominated convergence theorem.

My question is as follows: Where did $g$ come from?

So far, I've noticed that $g(x)$ is the form of a polynomial of degree two higher than the numberator, so perhaps there is a relationship between this expression and hyperbolic cotangent or the digamma function?

I've also noticed that $g_{a}(x)$ is a probability distribution function which looks a lot like the Cauchy Distribution. As a consequence to this, it feels like computing the Fourier transform of $g$ is going to be somewhat related to the Cauchy Distribution's characteristic function?

That being said, I have a feeling that I am just looking for similarities which are not useful at all. Can I gain any useful insights from those two observations? If not, how can I reasonably arrive at this integral equation for $g_{a}(t)$ "naturally"?

As for my meaning of "naturally", consider a standard epsilon-delta proof. When you let $\epsilon>0$ be given and consider $\delta = f(\epsilon)$, most of the intuition behind the proof itself was completely hidden by considering such a $\delta$ if you never show how you arrived at that $\delta$. In a similar vein, how do I reasonably arrive at $g_{a}(x)$?

• The "intuition" essentially boils down to $\int e^{ik(x-y)} dk = \delta(x-y) (2\pi)$. There's nothing special about your $g_a(x)$. Any function converging to a delta function would do the job. Moreover the delta function is a probability distribution. This explains why usually one chooses $g_a(x)$ to be a probability density.
– lcv
Mar 9 '15 at 7:45
• Usually, one tries to use a function for which one can calculate the (inverse) Fourier transform easily. One then shows $f \ast g_a \to f$, where the $g_a$ are $L^1$-normalized dilates of $g$. For reasonable choices of $g$, we get $\widehat{g_a} \to 1$ pointwise. Mar 9 '15 at 21:25

If you consider the operator $A=\frac{1}{i}\frac{d}{dx}$ on the domain consisting of periodic and differentiable functions on $[-\pi,\pi]$, then you can consider, as Cauchy did, $(\lambda I-A)^{-1}f$. The led to a expression with poles at $\lambda=0,\pm 1,\pm 2,\cdots$ and residue at $N$ equal to $\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)e^{-int}dt e^{inx}$. Cauchy noted that the sum of all such residues is the Fourier series sfor $f$. Cauchy then looked at trading all of the residues in the finite plane for a single residue at $\infty$, which would appear to be
$$\lim_{\lambda\rightarrow\infty}\lambda(\lambda I-A)^{-1}f=\lim_{\lambda\rightarrow\infty}\frac{\lambda}{\lambda I-A}f = f.$$ This thought was never fully developed in Cauchy's lifetime, but it did pave the way for general and workable proofs about a century later for all kinds of classical expansions in orthogonal functions, not just the exponentials.
The same analysis works for the operator $A=\frac{1}{i}\frac{d}{dx}$ on $(-\infty,\infty)$ as well, and this leads to equating the above with an integral surrounding the real axis: $$\frac{1}{2\pi i}\int_{-\infty}^{\infty}\{ ((t+i\epsilon)I-A)^{-1}f-((t-i\epsilon)I-A)^{-1}f \}dt$$ This integral gives your integral with $a=\epsilon$ convolved with $f$, and it boils down to the Poisson integral representation of a harmonic function. The details are little tedious, but natural in this context.