Show using Banach's fixed point theorem that the equation $$f(x)=x^2+ \int_{0}^1 \frac{1}{10+|x-t|^2}(f(t)+1)^2 \ dt, \ x \in [0,1],$$ has a solution in $C([0,1])$.
Define an operator $T: C([0,1])) \to C([0,1])$ as $$Tf(x) = x^2+ \int_{0}^1 \frac{1}{10+|x-t|^2}(f(t)+1)^2 \ dt.$$
To show that $T$ is continuous we can consider $\|Tf(x) - Tg(x)\|$ for $f,g \in C([0,1])$.
One has that $$\left|x^2+ \int_{0}^1 \frac{1}{10+|x-t|^2}(f(t)+1)^2 \ dt - \left( x^2+ \int_{0}^1 \frac{1}{10+|y-t|^2}(g(t)+1)^2 \ dt \right) \right| \le \int_{0}^1 \left| \frac{1}{10+|x-t|^2} \right|\left(|f(t)+1|^2 - |g(t)+1|^2 \right) \ dt$$
and considering the sup norm $$\int_{0}^1 \left| \frac{1}{10+|x-t|^2} \right|\left(|f(t)+1|^2 - |g(t)+1|^2 \right) \ dt \le \int_{0}^1 \left| \frac{1}{10+|x-t|^2} \right|\left(\|f^2-2f+1\|_\infty -\|g^2-2g+1\|_\infty \right) \ dt$$
$$\|Tf(x) - Tg(x)\|_\infty \le \|f^2-2f+1\|_\infty -\|g^2-2g+1\|_\infty \int_{0}^1 \left| \frac{1}{10+|x-t|^2} \right|$$
but I cant get this to the form $$\|Tf(x)-Tg(x)\| \le C\|f-g\|_\infty.$$ The expression is already very nasty. Is there some tricks that can be applied here?
On the space $\mathcal C([0,1])$, $T$ is not a contraction mapping as can be seen with the sequence $\{f_n = n\}$. However, on the closed ball $B$ centered on the always vanishing map of radius $2$, we have for any $x \in [0,1]$
$$\begin{aligned} \lvert Tf(x)-Tg(x) \rvert &\le \frac{1}{10}\int_0^1 \lvert (f(t)+1)^2 - (g(t)+1)^2 \rvert dt\\ &= \frac{1}{10}\int_0^1 \lvert f(t)-g(t) \rvert \lvert f(t)+g(t) +2\rvert dt \\&\le\left(\frac{1}{10}\int_0^1 6 dt\right) \lVert f-g \rVert_\infty = \frac{3}{5}\lVert f - g \rVert_\infty \end{aligned}$$
Hence $\lVert Tf - Tg \rVert_\infty \le \frac{3}{5}\lVert f - g \rVert_\infty$ on $B$. This allows us to apply Banach fixed point theorem and to get the existence of a solution of the given equation.