Find the Taylor series for $f$ at $t= 0$ where $f(t) = a + \int_0^tsin(t-s)f(s)ds$.

Question

Let $a, b\in \mathbb{R}$ and $b\notin 0$. Suppose that there is exist $f\in C([-b, b])$ such that, for all $t\in [-b,b]$, $$f(t) = a + \int_0^tsin(t-s)f(s)ds$$

1- Show that if $f$ exist, then $f\in C^{\infty}([-b, b])$.

2- Find the Taylor series for $f$ at $t= 0$.

I know that from the given equation, by using Laplace transform, the function $f(t)$ is easily obtained. However, I am trying to solve this question without using Laplace transform as it is given to me as a question of real analysis.

Answer 1

1. Write $$ f(t) = a + \sin(t) \int_{0}^{t} \cos(s) f(s) ds - \cos(t) \int_{0}^{t} \sin(s) f(s) ds $$ If $f$ is continuous on $[-b, b]$, then $\int_{0}^{t} \cos(s) f(s) ds$ is $C^{1}([-b,b])$ (and the same for the other integral). This yields that $f$ is actually $C^1$ (as it is equal to the sum of $C^1$ functions). Then you keep iterating this argument to get the same regularity as $\sin$ and $\cos$, i.e. $C^{\infty}$.
2. Using $$ f(t) = a + \sin(t) \int_{0}^{t} \cos(s) f(s) ds - \cos(t) \int_{0}^{t} \sin(s) f(s) ds $$ we have by continuity that $f(0) = a$ (as the integrals give 0). Differentiating the relation yields $$ f'(t) = \cos(t) \int_{0}^{t} \cos(s) f(s) ds + \sin(t) \int_{0}^{t} \sin(s) f(s) ds $$ and $f'(0) = 0$. Next, $$ f''(t) = -\sin(t) \int_{0}^{t} \cos(s) f(s) ds + f(t) + \cos(t) \int_{0}^{t} \sin(s) f(s) ds $$ and $f''(0) = a$.