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?

  • $\begingroup$ Basically that is correct. $\endgroup$ Aug 14, 2020 at 22:39
  • $\begingroup$ 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! $\endgroup$
    – Ruy
    Nov 27, 2020 at 16:56

1 Answer 1


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 $\tilde F'.$


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