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!
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 $$x(t) = \log(1+t) + \frac 1 5 \int^1_0 e^{-t} \cos(ts)^2 x(s)^2 ds, \,\,\,\,\ t \in [0,1].$$