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?