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Here is my problem. Let $C$ be the space of continuous and nondexcreasing functions defined on $[0,1]$ and endowed with the sup norm. Take any $z\in C$ and consider the following: $$ U(x;z)=\int_{0}^{x}\left[1-\int_{s}^{1}F(z(\xi))f(\xi)d\xi\right]^{n-1}ds $$ where $F:[0,1]\rightarrow [0,1]$ 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 $n>2$. Fix the value of $x$. I would like to show that $U$ is Lipschitz continuous with respect to $z$, i.e., I'd like to show that there is a constant $K>0$ such that:

$$ ||U(x;z)-U(x;y)||\leq K||z-y|| \qquad z,y\in C $$

This is what I've done so far. For a fixed value of $x$ the above expression can be treated as a functional. I know that the Frechet derivative of this expression looks like the following:

$$ \frac{dU(z)}{dz}=-(n-1)\int_{0}^{x}\left[1-\int_{s}^{1}F(z(\xi))f(\xi)d\xi\right]^{n-2}\left(\int_{s}^{1}f(z(\xi))f(\xi)h(\xi)d\xi\right)ds $$ where $h$ is the 'perturbation' function. Since $f$ is continuous and $f(s)>0$ if $s\in[0,1]$, it must be the case that:

$$ \left|\frac{dU(z)}{dz}\right|\leq (n-1)M_0\int_{0}^{x}\left|\left[1-\int_{s}^{1}F(z(\xi))f(\xi)d\xi\right]^{n-2}\left(\int_{s}^{1}h(\xi)d\xi\right)\right|ds\\ \leq (n-1)M_0\int_{0}^{x}\left(\int_{s}^{1}|h(\xi)|d\xi\right)ds $$ where $M_0=\sup_{s\in[0,1]} |f(s)f(s)|$, and the second line follows because $f(z(s))f(s)\leq \sup_{s\in[0,1]}|f(z(s))f(s)|$ for any $z\in C$, and $\left[1-\int_{s}^{1}F(z(\xi))f(\xi)\right] \leq 1$ for any $z\in C$ and $s\in[0,1]$.

I know that the perturbation $h$ is bounded because it belongs to $C$ but any such bound depends on the choice of the perturbation and hence, the bound of the whole expression will depend on $h$. Does this means that I am not free to choose the overall bound that would allow me to show the existence of a Lipschitz constant $K$?

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

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The derivative is a linear functional of $h$, so of course it depends on $h$. What you want is a bound on the norm of this linear functional. Recall that to give a bound for linear functional means to prove an inequality of the form $|Th|\le M\|h\|$, with $M$ independent of $h$. As long as $M$ is independent of $z$, you have uniformly bounded derivative, hence the Lipschitz condition. –  user31373 Jul 10 '12 at 22:20
    
Thanks! Please correct me if I am wrong, but isn't enough that the Frecht derivative above exists to claim that the linear operator $Th$ is bounded? From the little I know about Frecht derivatives, I can recall that a functional is Frecht diff if there exists a bounded linear operator $T$ s.th. $\lim_{h\rightarrow 0}\frac{||U(z+h)-U(z)-Th||}{||h||}=0$. So, if I found the Frecht derivative, wouldn't this be sufficient to claim that $|Th|\leq M||h||$ holds true? Sorry if what I's saying is nonsense but I am truly confused at this stage. Thanks for any help you can provide me with this. –  Cristian Jul 10 '12 at 22:44
    
The Frechet derivative is a bounded functional by definition. In your post I did not see a proof that what you found is the Frechet derivative. But let's assume that it is. You still need a bound with $M$ independent of $z$; this does not come from any definitions, and you actually need some estimates. –  user31373 Jul 10 '12 at 22:51
    
Thanks again Leonid. So you're basically saying that my work so far is no god to show what I want to show. Do you have any suggestion/hint how I should proceed? I've been stuck here for quite some time and I just don't see any way out, although intuitively I would expect that my claim (about lipschitz) is true... –  Cristian Jul 10 '12 at 23:04
    
I did not say that. You did estimate $|Th|$ in terms of $h$ (I use $T$ as a shorter notation for the derivative). All that's left to do is bring it to the form $|Th|\le M\|h\|$. Recall that $\|h\|$ is the supremum of $h$, and it's easy to estimate integrals from above in terms of supremum. –  user31373 Jul 10 '12 at 23:15

1 Answer 1

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You could continue your estimates as follows: $$\left|\frac{dU(z)}{dz}\right|\leq (n-1)M_0\int_{0}^{x}\left(\int_{s}^{1}|h(\xi)|d\xi\right)ds \le (n-1)M_0\int_{0}^{x} (1-s)\|h\| ds \le (n-1)M_0 \|h\|$$ Hence, $U$ is Lipschitz with the Lipschitz constant at most $(n-1)M_0$.

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thanks! I did exactly that (before seeing your post). I am confident now of what I did! Thanks a lot! –  Cristian Jul 11 '12 at 0:17

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