Integral equations for research in quantum mechanics In my research, I have reached a point where I need to find a nonnegative function which satisfies the following properties:

*

*It's an even function:

$$f(-\theta) = f(\theta)$$


*It's normalised as:

$$\int^\pi_0f(\theta)d\theta = \frac{\pi}{2}$$


*For any $\phi$ in the interval $[0,\pi]$,the following property is satisfied:

$$\int^\pi_0[f(\theta)+f(\theta - \phi)-2f(\theta)f(\theta - \phi)]d\theta = \pi\sin^2\frac{\phi}{2}$$
I would be very much in your debt if you help me find such a function.
 A: This falls into the problem of computing convolutional square roots, modulo the flaws already outlined by Cameron Williams.
A more sensible problem would be; compute a function $f$ on the circle (that is, a $2\pi$ periodic function) that satisfies $f(\theta)=f(-\theta)$ and such that
$$\tag{1}
f\ast f (\phi) = \pi \cos^2\frac \phi 2.$$
Here
$$
f\ast f(\phi):= \int_{-\pi}^\pi f(\phi-\theta)f(\theta)\,d\theta.$$
The normalisation condition is already encoded in (1); indeed, evaluating (1) at $\phi=0$,
$$
f\ast f(0)= 2\int_0^\pi f^2(\theta)\, d\theta = \pi.$$
Convolution square roots have been discussed before on Math.StackExchange. (Here there is an answer of Julián Aguirre (who, unfortunately, recently died). In the comments to that question, I have linked a related problem, answered by Terry Tao. All of these links, however, describe the problem on $\mathbb R$. I doubt they can be useful to the OP).
A: There are two functions that satisfy the updated conditions.
Let us start by writing $f(\theta)$ as a Fourier series involving only cosines (which are even functions):
\begin{equation}
f(\theta)=\sum_{n=0}^{+\infty}f_n\cos(n\theta)
\end{equation}
the first condition is immediately satisfied. Let us impose the second condition:
\begin{equation}
\int_{0}^{\pi}d\theta \,\,f(\theta)=\sum_{n=0}^{+\infty}f_n\int_{0}^{\pi}\cos(n\theta)=\sum_{n=0}^{+\infty}f_n\frac{\sin{(\pi n)}}{n} \stackrel{!}{=}\frac{\pi}{2}
\end{equation}
this expression vanishes except when $n=0$, in that case
\begin{equation}
\frac{\sin{(\pi n)}}{n}=\pi
\end{equation}
or, in other words, $f_0=\frac{1}{2}$. Hence,
\begin{equation}
f(\theta)=\frac{1}{2}+\sum_{n=1}^{+\infty}f_n\cos(n\theta)
\end{equation}
Now we have to look at condition 3. We have

\begin{equation}
\int_{0}^{\pi}d\theta\,\,\cos\left(n(\theta-\phi)\right)=\frac{\sin(n\phi )}{n}[1-(-1)^n],
\end{equation}
\begin{equation}
\begin{split}
\int_0^{\pi}d\theta\,\,\left(\frac{1}{2}+\sum_{n= 1}^{+\infty}f_n\cos(n\theta)\right)\left(\frac{1}{2}+\sum_{m=1}^{+\infty}f_m\cos(m(\theta-\phi))\right) \\ \\
=\frac{\pi}{4}+\frac{1}{2}\sum_{n=1}^{+\infty}\frac{\sin(n\phi )}{n}\left[1-(-1)^n\right]
+\sum_{n,m\neq0}f_nf_m\int_0^{\pi}d\theta\,\,\cos(n\theta)\cos(m(\theta-\phi))
\end{split}
\end{equation}

The last integral is
\begin{equation}
\int_0^{\pi}d\theta\,\,\cos(n\theta)\cos(m(\theta-\phi))=\frac{m}{m^2-n^2}[\sin(m\phi )-(-1)^n\sin(m(\phi-\pi))]
\end{equation}

Let us study the limit $n\to m$
\begin{equation}
\lim_{m\to n}\frac{m}{m^2-n^2}[\sin(m\phi )-(-1)^n\sin(m(\phi-\pi))]=\frac{\pi}{2}\cos\left(n\phi\right)
\end{equation}

Hence, imposing condition 3 means that
\begin{equation}
\frac{\pi}{2}+\frac{\pi}{2}+\sum_{n=1}^{\infty}f_n \frac{\sin(\phi n)}{n}[1-(-1)^n]-2\bigg\{\frac{\pi}{4}+\frac{1}{2}\sum_{n=1}^{+\infty}f_n\frac{\sin(n\phi)}{n}[1-(-1)^n]\\
+\frac{\pi}{2}\sum_{n=1}^{+\infty}f^2_n\cos(n\phi)+\sum_{n\neq m\neq 0}f_nf_m \frac{m}{m^2-n^2}\sin(m\phi )[1-(-1)^{n+m}]\bigg\}\stackrel{!}{=}
\pi\sin^2\left(\frac{\phi}{2}\right)
\end{equation}

Now the right hand side can be written as
\begin{equation}
\pi\sin^2\left(\frac{\phi}{2}\right)=\frac{\pi}{2}(1-\cos\phi)
\end{equation}
Therefore we have obtained the equation
\begin{equation}
\begin{split}
\pi\sum_{n=1}^{+\infty}f^2_n\cos(n\phi)+2\sum_{n\neq m\neq 0}f_nf_m \frac{m}{m^2-n^2}\sin(m\phi )[1-(-1)^{n+m}]=\frac{\pi}{2}\cos(\phi)
\end{split}
\end{equation}

We can satisfy this equation if $f^2_1\pi=\frac{\pi}{2}$ and if $f_{k}=0$ whenever $k>1$. This means $f_1=\pm \frac{1}{\sqrt{2}}$.
We conclude that the functions that you are looking for are
\begin{equation}
f(\theta)=\frac{1}{2}\pm\frac{1}{\sqrt{2}}\cos\theta
\end{equation}
If you want $f$ to be non negative, then just choose the plus sign.
