Prove there is a unique continuous function satisfying this integral equation

This is a question from an old real analysis qual:

Prove that there is a unique continuous function $f:[0,1] \to \mathbb{R}$ such that $$f(x) = \cos x + \int_0^x f(y)e^{-y}dy$$ for $x \in [0,1]$

I haven't seen any problems like this before and I'm not really sure where to start.

Define a bounded operator on $C[0,1]$ (the Banach space of continuous functions on $[0,1]$ with supremum norm $\|\cdot\|_\infty$) by $Tf(x) = \cos(x) + \int_0^x f(y)e^{-y}dy$.

Observe that $\| Tf - Tg\|_\infty \leq \|f-g\|_\infty \int_0^1e^{-y}dy = (1-\frac{1}{e})\|f-g\|_\infty$, so that $T$ is a contraction. By the Contraction Mapping Theorem, there is a unique fixed point of $T$.

• nice use of the contraction mapping theorem (+1) – robjohn Jun 4 '13 at 20:54

We derivate and we find the differential equation $$f'(x)=-\sin x+ f(x)e^{-x}\quad \text{with}\quad f(0)=1$$ which has a unique solution by Picard–Lindelöf theorem.

• Picard-Lindelof only gives local uniqueness. There's a bit more to do here. – Chris Janjigian Jun 4 '13 at 20:48
• @ChrisJanjigian There's a unique maximal solution. – user63181 Jun 4 '13 at 21:12
• Right, but you need to show that the maximal interval of uniqueness includes the entire interval $[0,1]$. – Chris Janjigian Jun 4 '13 at 21:23

Differentiating and rearranging, we get $$f(x)e^{-x}-f'(x)=\sin(x)$$ With an integrating factor of $g(x)=e^{e^{-x}}$, where $g'(x)=-g(x)e^{-x}$ we get $$(f(x)g(x))'=-\sin(x)g(x)$$ Then, we simply integrate and divide by $g(x)$ to get $$f(x)=\frac{e}{g(x)}-\frac1{g(x)}\int_0^x\sin(t)g(t)\,\mathrm{d}t$$ The constant of integration was chosen so that $f(0)=1$.