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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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When you take any two element $z,h \in C[0,1]$, you need to assume that $z([0,1]) \subseteq [0,1]$, or otherwise your integral is not well-defined since you would be trying to compute $f(t)$ with $t = z(x) \notin [0,1]$. –  Patrick Da Silva Jul 10 '12 at 20:55
    
yea Patrict. You're completely right. I should've been more careful with my definitions. Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be such that $f(s)>0$ if $s\in[0,1]$ and $f(s)=0$ if $s\notin [0,1]$. I guess this solve that problem, doesn't it? –  Cristian Jul 10 '12 at 21:02
    
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. –  Patrick Da Silva Jul 10 '12 at 21:07

1 Answer 1

up vote 1 down vote accepted

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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The underlying problem is as follows. I computed the Frecht derivative of an expression, which gives me something like the integral expression I wrote above. The $h$ function in the integral is the perturbation function you used when you compute this derivative. What I want is to put a bound on the expression that is independent of my choice of $z$ because I want to use this as an input to how that a function is Lipschitz continuous with respect to the function $z$...does any of this make any sense to you? –  Cristian Jul 10 '12 at 21:07
    
@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. –  Patrick Da Silva Jul 11 '12 at 0:07

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