Trigonometric integral inequality Let us consider the integral equation
\begin{equation}
    f(x)=\lambda \int_{0}^{\pi} \cos (x-y) f(y) \hspace{1mm}dy+g(x), \quad x \in[0, \pi],
\end{equation}
where $f$ is an unknown function on $[0, \pi]$, $g(x)$ is a given continuous function on $[0, \pi]$ and $\lambda$ is a given real constant. Prove that the equation has a unique solution $f \in C[0, \pi]$ for each $\lambda \neq \frac{2}{\pi}$.
 A: The function $y\mapsto |\cos(x-y)|$ is $\pi$-periodic. Therefore, its integral over an interval of length $\pi$ does not depend on the endpoints. In particular :
\begin{align}
\int_0^\pi|\cos(x-y)|\text dy &= \int_{x-\pi/2}^{x+\pi/2} |\cos(y-x)|\text dy\\
&= \int_{-\pi/2}^{\pi/2} \cos(y)\text dy \\
&= 2
\end{align}
A: The question is incorrect! Let $x=0$ and calculate the integral you find it equals $2$!
A: Since
$$
\cos (x-y) =\cos x \cos y + \sin x \sin y,
$$
we have that the equation in question is equivalent to
$$
    f(x)=\lambda \cos x \int_0^\pi f(y)\cos y \hspace{1mm}dy + \lambda \sin x \int_0^\pi f(y)\sin y \hspace{1mm}dy + g(x).
    $$
We see that $f$ is the function of the form
\begin{equation}
f (x) =a\cos x + b \sin x +g (x)
\label{fab02_009}
\end{equation}
where
$$
    a=\lambda  \int_0^\pi f(y)\cos y \hspace{1mm}dy \text{ und } b=\lambda  \int_0^\pi f(y)\sin y \hspace{1mm}dy
    $$
Adding this to the other equation gives the equation for $a$ and $b:$
$$
\begin{aligned}
a \cos x+b \sin x=& \lambda \cos x \int_{0}^{\pi} f(y) \cos y \hspace{1mm}dy+\lambda \sin x \int_{0}^{\pi} f(y) \sin y \hspace{1mm}\hspace{1mm}dy \\
=& \lambda \cos x \int_{0}^{\pi}(a \cos y+b \sin y+g(y)) \cos y \hspace{1mm}dy \\
&+\lambda \sin x \int_{0}^{\pi}(a \cos y+b \sin y+g(y)) \sin y \hspace{1mm}dy .
\end{aligned}
$$
Let us note:
$$
\begin{aligned}
\int_{0}^{\pi} \cos(y)^{2} \hspace{1mm}dy&=\int_{0}^{\pi} \frac{1+\cos 2 y}{2} \hspace{1mm}dy=\frac{\pi}{2} \\
\int_{0}^{\pi} \sin(y)^{2}  \hspace{1mm}dy&=\int_{0}^{\pi} \frac{1-\cos 2 y}{2} \hspace{1mm}dy=\frac{\pi}{2}
\end{aligned}
$$
and
$$
\int_{0}^{\pi} \cos y \sin y \hspace{1mm}dy=\frac{1}{2} \int_{0}^{\pi} \sin 2 y \hspace{1mm}dy=0 .
$$
It follows that:
$$
\begin{aligned}
\int_{0}^{\pi}(a \cos y+b \sin y+g(y)) \cos y \hspace{1mm}dy &=a \int_{0}^{\pi} \cos(y)^{2} \hspace{1mm}dy+b \int_{0}^{\pi} \sin y \cos y \hspace{1mm}dy+\int_{0}^{\pi} g(y) \cos y \hspace{1mm}dy \\
&=a \frac{\pi}{2}+A,
\end{aligned}
$$
where:
$$
A=\int_{0}^{\pi} g(y) \cos y \hspace{1mm}dy
$$
and
$$
\begin{aligned}
\int_{0}^{\pi}(a \cos y+b \sin y+g(y)) \sin y dy &=a \int_{0}^{\pi} \cos y \sin y \hspace{1mm}dy+b \int_{0}^{\pi} \sin(y)^{2} \hspace{1mm}dy+\int_{0}^{\pi} g(y) \sin y \hspace{1mm}dy \\
&=b \frac{\pi}{2}+B
\end{aligned}
$$
where:
$$
B=\int_{0}^{\pi} g(y) \sin y \hspace{1mm}dy.
$$
So we get the equation for $a$ and $b$:
$$
a \cos x+b \sin x=\lambda\left (a \frac{\pi}{2}+A\right) \cos x+\lambda\left (b \frac{\pi}{2}+B\right) \sin x
$$
i. e.
$$
\begin{aligned}
a&=\lambda\left (a \frac{\pi}{2}+A\right) \\
b&=\lambda\left (b \frac{\pi}{2}+B\right).
\end{aligned}
$$
For $\lambda \neq \frac{2}{\pi}$ we get the unique solution:
$$
\begin{aligned}
a&= \frac{\lambda A}{1-\lambda\frac{\pi}{2}} \\
b&=\frac{\lambda B}{1-\lambda\frac{\pi}{2}}.
\end{aligned}
$$
