Let
$T:f\mapsto (x\mapsto \frac{2}{5}\int_0^1 (x^2+t^5)f(t) dt + \sin(x))$ for any $x\in[0,1]$, $f\in C([0,1])$.
I want to show that that there is a uniqu $\tilde{f}$ that solves that equation $f(x)=\frac{2}{5}\int_0^1 (x^2+t^5)f(t) dt + \sin(x)$, i.e. by using the BNF to show there is a $C\in[0,1]$ such that $||Tf-Tg||
By definition, I get
$||Tf-Tg||\leq\frac{4}{5}||f-g||+\sin(1) \leq \sin(1)(||f-g||+1)$
What am I doing wrong?