An integral functional on $C[0,1]$ is contractive Let $$P: C[0,1]\to C[0,1]; P(f)(x)= 1+kxf(x)\int_0^1 \frac{f(s)ds}{x+s},$$
where $k$ is a constant, with $|k|<\frac{1}{4\ln 2}$. Let $f_1(x)\equiv 1$ on $[0,1]$. I want to prove that $P$ is contractive using the generalized Mean Value Theorem.

Generalized Mean Value Theorem. Let $X,Y$ be Banach spaces and $F: X\to Y$ be a mapping. Then $\|F(b)-F(a)\|_Y\le K\|b-a\|_X$, with $a,b\in X$. Where $K=\sup\|DF\|_Y$, with $DF$ being the derivative of $F$.

So, in this case, I think we should use the Fréchet derivative of $P$, which is $$DP(f) = kx h(x) (f(x)+h(x))\int_0^1 \frac{f(s)ds}{x+s}$$
I think the idea is to show that $\|DP\|_\infty< 1$. But how can one do it?
$$|DP(f)|\le |k|\|h(x)\|_\infty \|f(x)+h(x)\|_\infty\left|\int_0^1 \frac{f(s)ds}{x+s}\right|$$
We don't seem to know anything about the upper bounds of either $f$ or $h$. What is it that I'm misunderstanding here?
 A: First, this operator is not contractive on all of $C[0,1]$. It does not even admit any bound of the form $\|Pf\|_\infty\le M\|f\|_\infty$, because multiplying $f$ by, say, $1000$ multiplies the left hand side by about $1000^2$ due to the quadratic dependency on $f$. 
But for your purpose, convergence of iterations starting with $f_1\equiv 1$, it would be enough to show that $P$ is contractive on sufficiently large ball. 
To that end, using the Frechét derivative is a good idea, but you didn't compute it correctly. To find $DP(f)$, replace $f$ by $f+h$ everywhere, and then extract the linear term in $h$. So, we go from 
$$1+kx(f(x)+h(x)) \int_0^1 \frac{(f(s)+h(s))\,ds}{x+s}$$
to 
$$DP(f)h = kx h(x)  \int_0^1 \frac{ f(s)\, ds}{x+s} + kx f(x) \int_0^1 \frac{ h(s)\,ds}{x+s}$$
(which is a linear operator in $h$, as it must be). If we bring $x$ inside the integrals and notice that $x/(x+s) \le 1$, it follows that 
$$\|DP(f)h\|_\infty \le  |k|  \|h\|_\infty  \int_0^1 |f(s)|\, ds + |k|  \|f\|_\infty \int_0^1 |h(s)|\,ds \le 2|k|\|h\|_\infty  \|f\|_\infty$$
Thus, the operator norm of $DP(f)$ is at most $1$ when $\|f\|_\infty \le 1/(2|k|)$.
