Prove that there exists a unique continuous function $ f:[0,1]\to\mathbb R$ such that $f(x)=\sin x+\int_0^1 f(y)\exp(-x-y-1)dy$

Question

Prove that there exists a unique continuous function $ f:[0,1]\to\mathbb R$ such that $f(x)=\sin x+\int_0^1 f(y)\exp(-x-y-1)dy$.

This is a homework question from my functional analysis class, and here are some of my thoughts.

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Let $T:C[0,1]\to C[0,1]$ be defined by $T(f)(x)=\sin x+\int_0^x f(y)\exp(-x-y-1)dy$

Let $d_{C[0,1]}(f_1,f_2)=\max|f_1(x)-f_2(x)|$, where $x\in[0,1]$, $\forall f_1, f_2\in C[0,1].$

Now I am supposed to look for some relation between $d_{C[0,1]}\bigl(T(f_1),T(f_2)\bigl)$ and $d_{C[0,1]}(f_1,f_2)$ and somehow to prove that $T$ is a contraction.

However, when I do $d_{C[0,1]}\bigl(T(f_1),T(f_2)\bigl)=|T(f_1)(x)-T(f_2)(x)|$, I don't know exactly what the integral part should be like.

I guessed like $d_{C[0,1]}\bigl(T(f_1),T(f_2)\bigl)=|T(f_1)(x)-T(f_2)(x)|\le d_{C[0,1]}(f_1,f_2) M\int_0^1 \exp(-y-1)dy$, where $M=\max_{[0,1]}|e^x|=e$. Yet I don't expect this to be correct.

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Could anyone teach me how to properly handle the integral part, like "why should it be like this"?

Answer 1

$|Tf_1(x)-Tf_2(x)| \leq \int_0^{1} e^{-1-y}dy \|f_1-f_2\|$ and $\int_0^{1} e^{-1-y}dy=\frac 1 e (1-\frac 1 e) \in (0,1)$. Hence $\|Tf_1-Tf_2\|\leq c\|f_1-f_2\|$ for all $f_1$ and $f_2$. By contraction mapping Theorem $T$ has unique fixed point and this fixed point is the solution you want.