Does this equation $$f(u)=1+\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{1}{(u-v)^2+1}f(v)dv$$ has a bounded continuous solution? Is this solution unique? Here $f$ is defined over $\mathbb{R}$ and bounded.
Here are my attempts:
I tried the contraction mapping theorem, but I could not find the constant $p<1$ to bound the operator.
I used the Fourier transform trying to get the result. But the solution is really ridiculous(it claims that $f$ is a constant), since the fourier transform of $1$ is the Dirac delta.
So I wonder if this equation has a bounded solution or if the solution is unique. Any hints or solutions are welcomed, Thanks!
The equation has no continuous bounded solution.
Define
$$\begin{align}Tf&=\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{1}{(x-y)^2+1}f(y)dy\quad f \in C_b(\mathbb{R})\\Kf&=1+\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{1}{(x-y)^2+1}f(y)dy\quad f \in C_b(\mathbb{R})\end{align}$$
then $Tf$ is a continuous function on $\mathbb{R}$ since the kernel and $f$ are continuous , moreover, $\forall f \in C_b(\mathbb{R}), x\in \mathbb{R}$ $$\begin{align} |(Tf)(x)|&\leq\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{1}{(x-y)^2+1}|f(y)|dy\\ &\leq\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{1}{(x-y)^2+1}dy ||f||_\infty\\ &=\frac{1}{\pi} \tan^{-1}(y-x)\bigg|_{-\infty}^{\infty} ||f||_\infty\\&=||f||_\infty \end{align}$$
Hence $||Tf||_\infty\leq ||f||_\infty$. Hence $Tf \in C_b(\mathbb{R})$ and also note $T$ is a linear operator and $||T||\leq 1$.
Let $f_1(x)=1,\forall x \in\mathbb{R}$,then $||f_1||_\infty=1$ and $Tf_1 = f_1$.
Then $(Kf)(x) = f_1(x) + (Tf)(x)$. If $f\in C_b(\mathbb{R})$ is a solution of the integral equation, then $$f =f_1 +Tf \implies f =f_1 +T(f_1 +Tf)=f_1 +Tf_1 +T^2f$$
Inductively, we find $$f =f_1 +Tf_1 +\cdots+T^{n−1}f_1 +T^nf =nf_1 +T^nf,\forall n\in\mathbb{N}$$
Also, inductively, we can prove that $$||T^nf||_\infty \leq ||T||||T^{n-1}f||_\infty \leq ||T||^n||f||_\infty\leq ||f||_\infty, \forall n\in\mathbb{N}$$ So $n = ||nf_1||_\infty=||f-T^nf||_\infty \leq ||f||_\infty+||T^nf||_\infty \leq 2||f||_\infty,\forall n\in\mathbb{N}$, which leads to a contradiction that $f\in C_b(\mathbb{R})$. Therefore this integral equation has no solution on $ C_b(\mathbb{R})$.
If you are considering the integral over $[-a,a], \ 0<a<\infty$, by similar method you can show $K$ is a contraction mapping $C[-a,a]\to C[-a,a]$, hence by contraction mapping principle, you have a unique continuous bounded solution.