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Let $C[0,1]$ be the space of continuous and nondecreasing functions with the sup norm. Moreover, let $f[0,1]\rightarrow \mathbb{R}$ be continuous and positive, i.e., $f(s)>0,\,s\in[0,1]$. Take any two element $z,h$ in $C$. I would like to put an upper bound on the following expression: $ \int_{0}^{1}f(z(x))h(x)dx $ that is independent of $z$ and $h$, i.e., I would to claim that there exists some $M>$ such that for every $z,h\in C$, $ \int_{0}^{1}f(z(x))h(x)dx\leq M $

So far, I know that since $f$ is continuous, positive and defined on a compact set, $|f(s)|\leq \max_{s\in[0,1]} f(s)=M_{0}>0$, which leaves me with: $ \int_{0}^{1}f(z(x))h(x)dx\leq M_{0}\left|\int_{0}^{1}h(s)ds\right| $ But here is my question. I know that the last integral term is bounded, but this bound depends on the particular choice of $h$. How can I get rid (if possible) of this last restriction? That is, can I put a bound on the integral term that is independent of $h$? If not, what would I need it to make it work?

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    It solves this problem, but not the one I pointed out in my answer, as I actually said in my answer. =) But it's okay that you wrote your definitions ; at least we can say things about them. If you don't, we can't tell you're bugged there or something like that.2012-07-10

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You cannot put a bound on your integral that is independent of the choice of $z$ and $h$ (even if you get rid of the problem I mentioned in my comment) because of the following counter examples : Take $f(x) = 1$ everywhere, and take $h(x) = M$ everywhere. Then $ \int_0^1 f(z(x)) h(x) dx = \int_0^1 M dx = M. $ If you would've bounded your expression with no dependence on $z$ and $h$, your bound would need to be bigger than $M$ for any $M > 0$, which doesn't make sense.

If you can allow a dependence on $h$ in the underlying problem (because I am assuming there is one), then perhaps there is something to work out. But if there is no underlying problem, then I believe there is no general bound to be expected at all.

Hope that helps,

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    @Cristian : It does, but I am still missing some details to be of any help. Perhaps what you could do is edit your question by adding **EDIT** : under your current question and then add the details about the Frechet derivative. I will probably be able to help since I know about those.2012-07-11