Unique solution of an integral equation in $L^1[0,1]$

Question

Let $h\in L^1[0,1]$. Prove that there is a unique solution (almost everywhere) of the following integral equation: $$f(x)=h(x)+\frac{1}{2}\int_0^x\log(1+f(y)^2)dy$$ The idea is to use the fixed-point theorem for Banach spaces. So let's consider the map: $$T:L^1[0,1]\to L^1[0,1],f(x)\mapsto h(x)+\frac{1}{2}\int_0^x\log(1+f(y)^2)dy$$ I want to prove that this is a contraction map. So i want to find a constant $C<1$ such that $||T(f)-T(g)||_1\leq C||f-g||_1$. $$||T(f)-T(g)||_1=\frac{1}{2}\int_0^1|\int_0^x\log(1+f(y)^2)-\log(1+g(y)^2)dy|dx$$ Taking the absolute value under the integral and then changing the order of integration: $$||T(f)-T(g)||_1\leq \frac{1}{2}\int_0^1\int_0^x|\log(1+f(y)^2)-\log(1+g(y)^2)|dydx$$ So $$||T(f)-T(g)||\leq \frac{1}{2}\int_0^1(1-y)|\log(1+f(y)^2)-\log(1+g(y)^2)|dy$$ But from here i'm not able to find the exstimation.

Answer 1

First idea: Use the mean value theorem on $g (x)= \ln (1 + x^2) $ to get $$g (x)-g (y) = \frac {2\xi}{1+\xi^2} (x-y) $$ and estimate the fraction.