I am not certain about a step in the following argument where I use the frechet derivative.


Let $F$ be a continuously differentiable function with founded derivative (let $f$ be its derivative) satisfying $F(s)=0$ for all $s\leq 0$ and $F(s)=1$ for all $s\geq 1$. Consider the following operator: $$ (K\rho)(t)=1-\int_{t}^{1}F(\rho(\tau))dF(\tau) $$ with $t\in[0,1]$ and $\rho$ a continuous and nondecreasing function defined on $[0,1]$. The frechet derivative of K at $\rho$ is: $$ (DF_{\rho})(h)=-\int_{t}^{1}f(\rho(\tau))h(\tau)f(\tau)d\tau $$

Let $\theta\in[0,1]$ be an scalar and $\rho$ and $\phi$ be two nondecreasign and continuous functions on $[0,1]$. Define $\rho^\theta=\theta \rho(t)+(1-\theta)\phi(t)$. Then, $$ (K\rho^\theta)(t)=1-\int_{t}^{1}F(\rho^\theta(\tau))dF(\tau) $$

I want to compute $\frac{dK\rho^\theta}{d\theta}$. What I have is the following: $$ \frac{dK\rho^\theta}{d\theta}=-(DF_{\rho^\theta})(h)\frac{d\rho^\theta}{d\theta}\\ =-(DF_{\rho^\theta})(h)[\rho-\phi] $$


Since $(DF_\rho)(h)$ is a bounded linear operator, it satisfies $||(DF_\rho)(h)||\leq C||h||$ for all $h$ continuous and nondecreasing (here $||\cdot||$ is the sup norm). Then, this must also hold for the constant function $h\equiv 1$ and hence, $||(DF_\rho)(1)||\leq C$. Is the following claim valid? $$ \left|\frac{dK\rho^\theta}{d\theta}\right|\leq \left|(DF_{\rho^\theta})(1)\right|\left|\rho - \phi\right|\\ \leq \left|\rho - \phi\right|\left||(DF_{\rho^\theta})(1)\right||\\ \leq \left|\rho - \phi\right|C $$

Thanks for your help/suggestions!


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