Show that there is a unique function on the interval that solves:

Question

$$f(x) = e^{-2x} + \int_0^\infty e^{-2x-2y} \sin(x-y)f(y) \, dy$$

I can't get a good bound on $e^{-2x-2y} \sin(x-y)$ so that I can apply Banach.

Answer 1

Define the operator $T$ on an admissable class of functions (I'll use $V = \{ f: [0, \infty) \to \mathbb R : f \text{ is continuous and bounded} \}$ which is a Banach space with the supremum norm, $\|\cdot\|_\infty$) by $$(Tf)(x) = e^{-2x} + \int^\infty_0 e^{-2x-2y} \sin(x-y) f(y) dy, \,\,\,\,\, x \ge 0.$$ Then for $f,g \in V$, $x \in [0\infty)$, $$\lvert (Tf)(x) - (Tg)(x) \rvert \le \int^\infty_0 e^{-2x} e^{-2y} \lvert \sin(x-y)\rvert \lvert f(y) - g(y)\rvert dy.$$ But $e^{-2x} \le 1, \lvert \sin(x-y) \rvert \le 1$ and $\lvert f(y) - g(y) \rvert \le \| f - g\|_\infty$. Thus $$\lvert (Tf)(x) - (Tg)(x) \rvert \le \| f- g\|_\infty \int^\infty_{0} e^{-2x} dx = \frac{1}{2} \|f - g\|_\infty.$$ Since this holds for all $x$, we can pass to the supremum to see $$\|Tf - Tg \| \le \frac{1}{2} \| f - g\|_\infty.$$ This shows that $T$ is a contraction. It is straight-forward to show that $Tf \in V$ whenever $f \in V$; that is $T$ maps $V$ to $V$ (indeed, proving this uses basically the same steps used above to prove that $T$ is a contraction). So the Banach Fixed Point theorem gives existence of a fixed point.