I was asked to study the following map: let $I=[0,1]$ and let $f(s)=\log(1+s^2)$ for any $s\in\mathbb R$. For every $u\in L^1(I,\mathbb R)$ we set $(F(u))(x)=\int_0^xf(u(t))\mathrm d t.$
First of all I had shown that $F$ maps $L^1(I,\mathbb R)$ into itself. Then the second point was to show that $F$ is Lipschitz continuous from $L^1(I,\mathbb R)$ into itself with Lipschitz constant less than or equal than $1$, and I've done that.
Finally the problem asked to show that the Lipschitz constant couldn't be less than $1$. My first approach was to use the Caccioppoli Banach lemma to derive a sort of a contradiction. However if you study the integral equation $\begin{cases}u(x)=\int_0^x\log(1+u(t)^2)\mathrm d t\\ u(0)=0,\end{cases}$ then by unicity of the solution one sees that $u\equiv 0$ is the only fixed point... if I am not mistaken.
Then i tried to find a sequence of function $u_\lambda\in L^1(I,\mathbb R)$, with $\lambda<1$ such that, $\|Fu_\lambda\|_{L^1}>\lambda\|u_\lambda\|_{L^1}$, but i had no success.
Can anybody help me please? thank you...
EDIT: As I have written in the comment after DId answer, I don't think that his method matches my idea of proceeding in the exercise.. as it is written, it sounds troublesome to me.. Can anyone help me in finishing the problem?
Ps: sorry for accepting the answer, but as I wrote in the comment, I rushed in checking the detail because that kind of functions were also my first candidate to solve the exercise. However, let me again thank Did for answering.