Show that the integral equation has a solution on a suitable subset of $C[0,1]$ Show that the integral Equation 
$$ 
x(t) = \ln(1+t)+\frac{1}{5}\int_0^1 e^{-t}\cos(ts)^2x(s)^2\, \mathrm{d}s
$$
has a solution on a suitable subset of $C[0,1]$ using the contraction mapping theorem. 
I have seen a few proofs similar to this, but often the subsets are either given, or they seem to be chose out of no where, such as in this solution: Existence of integral equation solution. Any help would be appreciated!
 A: The reason that the subsets seem like they are chosen arbitrarily is because they are chosen slightly arbitrarily. Let's define the operator $T: C[0,1] \to C[0,1]$ by $$(Tx)(t) = \log(1+t) + \frac 1 5 \int^1_0 e^{-t} \cos(ts)^2 x(s)^2 ds, \,\,\,\,\,\,\,\,\, t \in [0,1], \,\,\,\,\, x \in C[0,1].$$ $($I leave you to show that this indeed a map from $C[0,1] \to C[0,1]).$ We need to choose a complete subspace of $C[0,1]$ and show that $T$ is a contraction on this subspace. Since any closed subset of a complete space is complete, we simply need to choose a closed subset and the most natural closed subsets of $C[0,1]$ are the balls: $$B_r = \{ x \in C[0,1] : \,\,\, \| x \|_\infty \le r\}, \,\,\,\,\, r \ge 0.$$ We show that for suitable $r \ge 0$, we have $T : B_r \to B_r$ and $T$ is a contraction. 
Take $x \in B_r$ and $t \in [0,1]$. We see \begin{align*}\lvert (Tx)(t) \rvert &= \left \lvert \log(1+t) + \frac 1 5\int^1_0 e^{-t} \cos(ts)^2 x(s)^2 ds \right \rvert \\
&\le \log(2) +\frac  1 5\int^1_0 \lvert e^{-t} \rvert \lvert \cos(ts) \rvert^2 \lvert x(s) \rvert^2 ds \\
&\le \log(2) +\frac 1 5 \int^1_0 \|x \|^2_\infty ds = \log(2) + \frac{\| x \|^2_\infty}{5} \le \log(2) + \frac{r^2}{5}. 
\end{align*} Passing to the supremum, we have $$\| Tx \|_\infty \le \log(2) + \frac{r^2}{5}.$$ We want $\| Tx \|_\infty \le r$ so we need to choose $r \ge 0$ such that $\log(2) + \frac{r^2}{5} \le r$. Looking at the graph of $p(r) = r^2 - r + \log(2)$, we notice that this is negative for $0.85 \lesssim r \lesssim 4.2$  so for any $r$ in that range we have $T: B_r \to B_r$.
Next note that for any $x,y \in B_r$ and $t \in [0,1]$ \begin{align*}
\lvert (Tx)(t) - (Ty)(t) \rvert &= \left \lvert \frac 1 5 \int^1_0 e^{-t} \cos(ts)^2 x(s)^2 ds- \frac 1 5 \int^1_0 e^{-t} \cos(ts)^2 y(s)^2 ds \right \rvert \\
&\le \frac 1 5 \int^1_0 \lvert e^{-t}\rvert \lvert \cos(ts) \rvert^2 \lvert x(s)^2 - y(s)^2 \rvert ds \\ 
&\le \frac 1 5 \int^1_0 \lvert x(s) + y(s) \rvert \lvert x(s) - y(s) \rvert ds \\
& \le \frac{1}{5} \| x+y \|_{\infty} \| x-y \|_{\infty}\\
&\le \frac{\| x \|_{\infty} + \| y\|_{\infty}}{5} \| x-y \|_{\infty} \le \frac{2r}{5} \|x-y\|_{\infty}.
\end{align*} Passing to the supremum gives $$\| Tx - Ty\|_{\infty} \le \frac{2r}{5} \|x-y\|_\infty$$ whence $T$ is a contraction on $B_r$ so long as $2r/5 <1$ or $r < 5/2$. 
Choosing any $r$ which satisfies both bounds, the Banach fixed point theorem gives existence of a unique fixed point in $B_r$. Thus for example, there is a unique fixed point of $T$ in $B_2$ and this fixed point will satisfy $$x(t) = \log(1+t) + \frac 1 5 \int^1_0 e^{-t} \cos(ts)^2 x(s)^2 ds, \,\,\,\,\ t \in [0,1].$$
