Proving T is a contraction Trying to sure up some holes in my proof for the following:
Let $I=[0,1]$ and define $T:C(I)\rightarrow C(I)$ by
$$(Tf)(x)=x+\int_{0}^{x}(x-t)f(t)dt \quad \bigl(x\in I, f\in C(I)\bigr).$$
Show that $T$ is a contraction on $C(I)$ (with uniform metric). 
$$\begin{align}
\bigl\lvert(Tf)(x)-(Tg)(x)\bigr\rvert &= \biggl\lvert x+\int_{0}^{x}(x-t)f(t)dt-x+\int_{0}^{x}(x-t)g(t)dt\biggr\rvert \\
&= \biggl\lvert\int_{0}^{x}(x-t)f(t)dt-\int_{0}^{x}(x-t)g(t)dt\biggr\rvert \\
&\le d_\infty(f,g)\int_{0}^{x}(x-t)dt$
\end{align}$$
[can someone explain succinctly why this particular step is true].
Clearly, $\int_{0}^{x}(x-t)dt$ evaluates to a number between 0 and 1/2 (since $x\in I$) leading to $\lvert(Tf)(x)-(Tg)(x)\rvert\le \frac{1}{2}d_\infty(f,g)\Longrightarrow T$ is a contraction on $C(I)$.
Does this look to be adequate?
By definition, we know that T has a unique fixed point. My book claims it is $\sinh(x)$. How do I show this?
 A: The step in question is why does it follow that 
\begin{align*}
\left | \int_0^x (x-t) f(t) dt - \int_0^x (x - t) g(t) dt \right | \leq d_\infty(f,g)
\int_0^x (x-t) dt.
\end{align*}
This follows easily from the defintion of $d_\infty(f,g) = \sup_{s \in [0,1]}|f(s) - g(s)|$ and the the estimate
\begin{align*}
\left | \int_0^x (x-t) f(t) dt - \int_0^x (x - t) g(t) dt \right | &=
\left | \int_0^x (x-t) [f(t) - g(t)] dt\right | \\
&\leq \int_0^x (x-t) |f(t) - g(t)| dt \\
&\leq \int_0^x (x-t) \sup_{s \in [0,1]}|f(s) - g(s)| dt \\
&= d_\infty(f,g) \int_0^x (x-t)dt.
\end{align*}
To see that the fixed point $f(x) = \sinh x$, observe that a fixed point satisfies (by definition)
\begin{align*}
f(x) = x + \int_0^x (x-t) f(x) dx.
\end{align*}
Differentiating the above expression twice and using the fundamental theorem of calculus yield
\begin{align*}
f''(x) = f(x).
\end{align*}
Moreover, $f(0) = 0$, and $f'(0) = 1$.  The general solution to the above ODE is 
$f(x) = a e^x + b e^{-x}$ for some constants $a,b$.  The initial conditions
$f(0) = 0$, $f'(0) = 1$ imply that $a = 1/2$, $b = -1/2$ so that 
\begin{align*}
f(x) = \frac{1}{2} e^x - \frac{1}{2} e^{-x} = \sinh x.
\end{align*} 
A: We have the inequality
$$ \left\lvert \int_a^b f \right\rvert \leqslant \int_a^b \lvert f \rvert, $$
basically as a generalisation of the triangle inequality. Applying this to your integral gives
$$ \left\lvert \int_0^x (x-t) (f(t)-g(t)) \, dt \right\rvert \leqslant \int_0^x (x-t) \lvert f(t)-g(t) \rvert \, dt. $$
If $u$ is a nonnegative function on $(a,b)$ and $\lvert v(x) \rvert \leqslant M$ for almost every $x$ in $(a,b)$, then $ -Mv(x) \leqslant u(x)v(x) \leqslant Mv(x) $ for almost every $x$ in $(a,b)$ and hence, integrating this inequalities,
$$ -M\int_a^b u(x) \, dx \leqslant \int_a^b u(x)v(x) \, dx \leqslant M\int_a^b u(x) \, dx. $$
You can use this on your integral, with $u(t)=x-t$ and $v(t)=\lvert f(t)-g(t) \rvert$, to get to your $d_{\infty}(f,g) \int_0^x (x-t) \, dt$ (here $M$ being the distance $d_{\infty}(f,g)$, of course).
The rest looks fine. As far as finding the fixed point goes, perhaps you'll spot it more easily in differential form: note that if
$$ f(x) = x + \int_0^x (x-t)f(t) \, dt, $$
then $f(0)=0$, $f'(0) = 1+0=1$, and differentiating,
$$ f'(x) = 1 + (x-x)f(x) + \int_0^x f(t) \, dt = 1 + \int_0^x f(t) \, dt \\
f''(x) = f(x) $$
Now, by the uniqueness theorem for ODEs, the only solution to these equations is $\sinh{x}$, as required.
A: Succinctly, we have for $x\in I=[0,1]$
$$\begin{align}
\left|\int_0^x (x-t)\left(f(t)-g(t)\right)\,dt\right|&\le\int_0^x (x-t)\left|f(t)-g(t)\right|\,dt \tag 1\\\\
&\le \sup_{t\in [0,x]}\left|f(t)-g(t)\right|\int_0^x (x-t)\,dt \tag 2\\\\
&\le\frac12\,\sup_{t\in [0,x]}\left|f(t)-g(t)\right| \tag 3
\end{align}$$
since $f$ and $g$ (and hence $f-g$) are in $C(I)$.
In arriving at $(1)$, we used the Triangle inequality.
In going from $(1)$ to $(2)$, we used the MVT for integrals.
In arriving at $(3)$, we used the fact the $\int_0^x(x-t)\,dt=\frac{x^2}{2}\le\frac12$.

To show that $\sinh x$ is a fixed point, we see that 
$$\begin{align}
x+\int_0^x (x-t)\sinh t\,dt&=x+x\left.\cosh t\right|_0^x-\left.(t\cosh t-\sinh t)\right|_0^x\\\\
&=\sinh x
\end{align}$$
and we are done!
A: You have 
$$\left|\int_0^x (x - t) (f(t) - g(t))\ dt \right|.$$
I assume your notation $d_\infty(f, g)$ stands for the $L^\infty$ norm $\|f - g\|_\infty$. In this case I can shed some light on the next step. Assume for convenience that $f$ and $g$ are continuous (the next step holds regardless, I just think it's more intuitive if they are). Then we can crash through with the absolute values to get 
$$\left|\int_0^x (x - t) (f(t) - g(t))\ dt \right| \le \int_0^x \left| (x - t) \right| \left| f(t) - g(t) \right|\ dt\ .$$
But the second integrand we can bound by $|f(t) - g(t)| \le \max_{t\in (0,x)} |f(t) - g(t)| = \| f - g \|_\infty$. Thus 
$$ \int_0^x \left| (x - t) \right| \left| f(t) - g(t) \right|\ dt \le \|f - g\|_\infty \int_0^x (x - t)\ dt\ .$$
I dropped the absolute values on the integrand because it is clearly non-negative.
Everything else looks good.
To show that $\sinh(x)$ is a fixed point, simply set $f(x) := \sinh(x)$ and evaluate $\left(Tf\right)(x)$. Presumably, you'll get $\sinh(x)$ as your answer.
