<s>[INST] How did Ulam and Neumann find this solution? In the book "Chaos, Fractals and Noise - Stochastic Aspects of
Dynamics" from Lasota and Mackey the operator $P: L^1[0,1] \to L^1[0,1]$
$$ (Pf)(x) = \frac{1}{4\sqrt{1-x}} \left[ f\left(\frac{1}{2}\left(1-\sqrt{1-x}\right)\right) +  f\left(\frac{1}{2}\left(1+\sqrt{1-x}\right)\right)\right]$$
was considered, where $L^1$ denotes the space of Lebesgue integrable functions.
It is stated in the book that Ulam and Neumann showed that
$$ f^*(x) =\frac{1}{\pi\sqrt{x(1-x)}} $$
solves
$$ Pf^* = f^*,$$ in the article "On combination of stochastic and deterministic processes, Bull. Amer. Math. Soc. vol. 53 (1947) p. 1120.". I went to the library of my university and I actually found the book  "vol 53 of Bull. Amer. Math. Soc.", but this book only contains the abstract of the article, not the article itself.
So my question is: Where do I find the article - or how did they find $f^*$ ?
Edit: I am interested how Ulam and Neumann actually constructed $f^*$. 
 [/INST] I don't have the reference for the derivation, but I can proof, that $f^*$ is indeed an eigenfunction of $P$ with eigenvalue $1$, i.e. $Pf^*=f^*$.
Proof
\begin{align}
(Pf^*)(x)&= \frac{1}{4\sqrt{1-x}} \Bigl( f\Bigl(\frac12 (1-\sqrt{1-x})\Bigr)+
 f\Bigl(\frac12 (1+\sqrt{1-x})\Bigr)\Bigr) 
\\
&=\frac{1}{4\sqrt{1-x}} \Bigl(
\frac1\pi  \frac{1}{\sqrt{\frac12 (1-\sqrt{1-x})(1-\frac12 (1-\sqrt{1-x})}}\Bigr)
+
\frac1\pi  \frac{1}{\sqrt{\frac12 (1+\sqrt{1-x})(1-\frac12 (1+\sqrt{1-x})}}\Bigr)
  \Bigr) 
\\
&=\frac{1}{4\pi\sqrt{1-x}}\Bigl(
 \frac{1}{\sqrt{\frac12 -\frac12 \sqrt{1-x})(\frac12 +\frac12 \sqrt{1-x})}}\Bigr)
+\frac{1}{\sqrt{\frac12 +\frac12 \sqrt{1-x})(\frac12 -\frac12 \sqrt{1-x})}}\Bigr)
\\ &=
\frac{1}{4\pi\sqrt{1-x}}\Bigl(
 \frac{1}{\sqrt{(\frac12)^2 -(\frac12 \sqrt{1-x})^2}}\Bigr)
+ \frac{1}{\sqrt{(\frac12)^2 -(\frac12 \sqrt{1-x})^2}}\Bigr)
\\ &=
\frac{2}{4\pi\sqrt{1-x}}\Bigl(
 \frac{1}{\sqrt{(\frac12)^2 -(\frac12 \sqrt{1-x})^2}}\Bigr)
\\ &=
\frac{2}{4\pi\sqrt{1-x}}\Bigl(
 \frac{1}{\sqrt{\frac14 -\frac14({1-x})}}\Bigr)
\\ &=
\frac{2}{\pi\sqrt{1-x}}\Bigl(
 \frac{1}{\sqrt{1-({1-x})}}\Bigr)
\\ &=
\frac{2}{2\pi\sqrt{1-x}}\Bigl(
 \frac{1}{\sqrt{x}}\Bigr)=f^*
\end{align}
Okay there are a lot of brackets missing/wrong and maybe this is of no use. The message was only, that it is trivial to proof, that $Pf^*=f^*$. Simply plug in the definition of $f^*$ and use binomial expansion for $(a-b)(a+b)$.
</s>