# Lipschitz constant on a functional

Let $C$ be the space of continuous and nondecreasing functions defined on $[0,1]$ and endowed with the sup norm. Let $T:C\rightarrow C$ be a continuous mapping, and consider the following expression: $$U(Tz(x);z)=\int_{0}^{Tz(x)}\left[\int_{0}^{1}F(z(\xi))f(\xi)d\xi+\int_{s\in \Gamma(t,z)}\left\{F(t)-F(z(s))\right\}f(s)ds\right]^{n-1}dt$$ where $\Gamma(t,z)=\{s:[0,1]|s\geq z(t)\}$, and $F:[0,1]\rightarrow [0,1]$ is continuously differentiable and increasing; $F'=f$, $f(s)>0, s\in[0,1]$ and $f(s)=0,\,s\notin [0,1]$, $x\in[0,1]$, and $n>2$.

Suppose that I've managed to show that there exists a constant $K$ such that $$||U(Tz(x);z)-U(Ty(x);y)||\leq K||z-y|| \qquad z,y\in C$$ holds true.

What I want is to make a claim about $T$. Particularly, I'd like to show whether is true or not that the following hold:

$$||Tz(x)-Ty(x)||\leq C||z-y|| \quad z,y\in C$$ where $C$ is a constant independent of $z$ and $y$.

Does anyone has any idea how I can prove/disprove this? I have no clue whatsoever on how to proceed, and I do need some help to get through it. Can someone help me or point me the right direction on how to tackle this problem?

Any help/suggestion/insight/reference is greatly appreciated it!

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In the earlier question we saw that $|U(Ty(x);z)-U(Ty(x);y)|\le M\|z-y\|$. By the triangle inequality, $|U(Tz(x);z)-U(Ty(x);z)|\le (M+K)\|z-y\|$. On the other hand, $$|U(Tz(x);z)-U(Ty(x);z)|\ge \left|\int_{Ty(x)}^{Tz(x)} \left[1-\int_{s}^{1}F(z(\xi))f(\xi)d\xi\right]^{n-1}ds \right| \\ \ge |Tz(x)-Ty(x)|\min_{[Ty(x),Tz(x)]} \left[1-\int_{s}^{1}F(z(\xi))f(\xi)d\xi\right]^{n-1}$$ So if you can estimate the minimum from below by a positive number, you win. In other words, you need an upper bound on $\int_{s}^{1}F(z(\xi))f(\xi)d\xi$ which is less than $1$.
Leonid: sorry for the earlier confusion in the expression I gave you (the one in the question now is the correct one). The early question you mentioned allowed me to show the existence of the constant $K$ (using some equalities between the expressions)...but I am stuck here...and it's been almost two months thinking about it without any success... – Cristian Jul 11 '12 at 23:21
@Cristian Well, some of what I said still makes sense: if the integrand is bounded from below, then the difference of $U$ is bounded from below in terms of $|Tz(x)-Ty(x)|$, which gives a handle on a latter quality. But this is the point where I'm tuning out. You asked 7 questions on MSE, all with similar-looking, convoluted and unmotivated integral estimates. I've had enough. – user31373 Jul 11 '12 at 23:30