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:

  1. I tried the contraction mapping theorem, but I could not find the constant $p<1$ to bound the operator.

  2. 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!

  • $\begingroup$ Just a thought: since constants are clearly not solution, and $\int_{-\infty}^\infty \frac{x}{1+x^2} dx$ is not convergent, I would be surprised if there is a polynomial or a power series solution. $\endgroup$
    – Taladris
    Mar 22 '14 at 14:20
  • $\begingroup$ @Taladris I agree. Then I should try to prove that this equation has no solution indeed. But this seems to need some effort. Would the Fourier transform proof suggests that there's no solution?(I'm not sure if this is a necessary condition for the equation to be true) $\endgroup$
    – Golbez
    Mar 22 '14 at 14:34
  • $\begingroup$ Have you tried a Taylor series for $f(1/(u^2 + 1))$ ? $\endgroup$
    – mick
    Jul 29 '14 at 20:42

The equation has no continuous bounded solution.


$$\begin{align}Tf&=\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{1}{(x-y)^2+1}f(y)dy\quad f \in C_b(\mathbb{R})\\Kf&=1+\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{1}{(x-y)^2+1}f(y)dy\quad f \in C_b(\mathbb{R})\end{align}$$

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.

  • $\begingroup$ That's a great answer! Thanks a lot! $\endgroup$
    – Golbez
    Nov 17 '14 at 3:25

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