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.


Consider $u_{a,b}=a\mathbf 1_{(0,b)}$ with $0\lt b\lt 1\leqslant a$, and $\ell(a)=\log(1+a^2)$.

Then $\|u_{a,b}-u_{1,b}\|_1=(a-1)b$ and $F(u_{a,b})(x)=\ell(a)\min\{x,b\}$ hence $$ \|F(u_{a,b})-F(u_{1,b})\|_1=(\ell(a)-\ell(1))\int_0^1\min\{x,b\}\mathrm dx=(\ell(a)-\ell(1))b(1-\tfrac12b). $$ Since $\ell'(1)=1$, $\|F(u_{a,b})-F(u_{1,b})\|_1\sim (a-1)b=\|u_{a,b}-u_{1,b}\|_1$ when $a\to1^+$ and $b\to0$. In particular, no inequality $\|F(u)-F(v)\|_1\leqslant c\|u-v\|_1$ may hold for every integrable functions $u$ and $v$, if $c\lt1$.

On the other hand, $\ell$ is $1$-Lipschitz hence, for every $x$ in $(0,1)$, $$ |F(u)(x)-F(v)(x)|\leqslant\int_0^1|\ell(u)-\ell(v)|\leqslant\int_0^1|u-v|=\|u-v\|_1. $$ This shows that $F$ is $1$-Lipschitz but not $c$-Lipschitz for any $c\lt1$.

  • $\begingroup$ @Did... I accepted your answer because I had your same thought and thus I believed it was correct.. incidentally today I've remade the calculations... It turned out that with your suggestions $\lim_{a,b\to 0}\|Fu\|_1=\lim_{a,b\to 0}\log(1+a^2)b(1-\frac b2)=a^2b$, while $\|u\|_1=ab$ therefore the limit of the quotient of the norm goes to $0$ in your statement.. This confuses me.. Where am I wrong? thank you very much... PS: Maybe the approach $\|Fu\lambda\|_1>\lambda\|u_\lambda\|_1$ is just wrong? $\endgroup$ – uforoboa May 29 '12 at 23:17
  • $\begingroup$ @Did.. Sorry to do that but I'm unchecking your answer temporarily but i will recheck as soon as somebody convinces me of its validity.. sorry again.. $\endgroup$ – uforoboa May 30 '12 at 0:15
  • $\begingroup$ Please do not feel sorry, my previous hint was misleading hence I am the one who should apologize. See revised version. $\endgroup$ – Did May 30 '12 at 6:54

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