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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.

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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 $T$.

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

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

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