# Combining Cayley transform and Fourier series

If one has a function $$F(x)$$ defined on the real line ($$x \in \mathbb{R}$$) then one can study it by means of its Fourier transform. Because $$\mathbb{R}$$ is not compact one has a Fourier integral rather than a Fourier series (assuming $$F$$ is sufficiently nice that it can be expressed as such). However $$\mathbb{R}$$ can be mapped by the Cayley transform to the unit circle $$C: x \mapsto \frac{i-x}{i+x}$$ and so composing $$F$$ with the Cayley transform one can define $$F$$ on the unit circle by $$F \circ C^{-1}$$. One then can compute Fourier coefficients $$a_n = \frac{1}{2\pi} \int_{\mathbb{T}} F \circ C^{-1}(e^{i\theta}) e^{-in\theta} \, d\theta = \frac{1}{\pi} \int_{\mathbb{R}} F(x) \left( \frac{i+x}{i-x} \right)^n \frac{1}{1+x^2} \, dx$$ $$F \circ C^{-1}$$ is of course not defined at $$-1$$ but $$\{ -1 \}$$ is a set of measure zero and so we could let $$F \circ C^{-1}$$ take some arbitrary value at this point.

If $$F$$ is continuously differentiable, and if $$F(\infty) = F(-\infty)$$ and $$F^\prime(\infty) = F^\prime(-\infty)$$ then it would seem the Fourier series of $$F \circ C^{-1}$$ converges pointwise uniformly on the circle and so the series $$\sum_{n \in \mathbb{Z}} a_n \left( \frac{i-x}{i+x} \right)^n$$ should converge pointwise uniformly to $$F(x)$$ on $$\mathbb{R}$$. Is this analysis correct and has this approach ever been studied before?

• Basically that is correct. Aug 14, 2020 at 22:39
• Looks good to me but I'd say this is not so useful as not many functions on $\mathbb R$ that show up in applications have those properties. But, still, this is a very smart observation!
– Ruy
Nov 27, 2020 at 16:56

Since $$F$$ is defined on $$\mathbb R$$ and its Cayley transform $$C$$ is defined on the circle $$\mathbb T,$$ I think it might be good to define $$\tilde F = F\circ C^{-1}.$$
Note that with your conditions on $$F,$$ $$\tilde F$$ extends to a function continuous on $$\mathbb T$$ from the $$F(\infty)=F(-\infty)$$ condition. Secondly, since $$F'(\infty)=F'(-\infty),$$ then $$F'(\infty)=0.$$ If not, then $$|F(x)|$$ must grow to $$\infty$$ as $$x\to \pm \infty.$$ That violates $$F(\infty)=F (-\infty).$$ Thus, intuitively, $$F$$ is close to being constant on $$\{|x|>R\}$$ if $$R$$ is large.
On the convergence question: I tried but did not answer this fully. One thing we can say: The Fourier series of $$\tilde F$$ converges at each point of $$\mathbb T\setminus \{-1\}.$$ That's because i) $$\tilde F= F\circ C^{-1}$$ is bounded on $$\mathbb T,$$ hence has a well defined Fourier series, and ii) $$\tilde F$$ is differentiable at every point of $$\mathbb T\setminus \{-1\}.$$ Here I'm using the fact that if $$g$$ is an $$L^1$$ function on the circle, and $$g$$ is differentiable at some $$e^{it},$$ then the Fourier series of $$g$$ converges to $$g(e^{it}).$$
We can say more: The Fourier series of $$\tilde F$$ converges uniformly to $$\tilde F$$ on the arc $$\{e^{it}: t\in [-\pi+\delta,\pi -\delta]\},$$ for any $$\delta, 0<\delta<\pi.$$ That's because $$\tilde F$$ is $$C^1$$ on such arcs. (Actually the result will hold if $$\tilde F$$ is merely Lipschitz on such arcs.)
Now, if we knew $$\tilde F$$ had a bounded derivative on $$\mathbb T\setminus \{-1\},$$ then with the other conditions we have on $$F,$$ we could view $$\tilde F$$ as a function in $$C^1(\mathbb T).$$ This would give the desired uniform convergence.
I looked at two examples: $$F(x)=x/(1+x^2)$$ and $$F(x)=1/(1+x^2).$$ If I've done the calculations correctly, the first example gives $$\tilde F'$$ blowing up at $$-1.$$ The second one gives a bounded $$\tilde F'.$$