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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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    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.2012-07-10

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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!2012-07-11