Trying to show a certain map is a contraction I'm having trouble understanding each step of the inequality shown below:
$$\Large\begin{eqnarray}
\left\|F(x)-F(x')\right\|
&=&  \left\|\:\int_0^1 DF_{(x-x')t+x'}      \cdot      (x-x') \:dt\: \right\|
\\[1.5ex]
&\leq& \int_0^1 \left\|DF_{(x-x')t+x'}\right\|\: \left\|x-x'\right\| \: dt
\\[1.5ex]
&\leq& \left(\sup_{z \in B_R} \left\|DF_z\right\|\right) \left\|x-x'\right\|
\end{eqnarray}$$
[Transcribed from this image.]
To give some context, this is an inequality that will be used in the proof of the inverse function theorem.  (The full document can be viewed here.)
Here, $F$ is a continuously differentiable function on $\mathbb{R}^n$ to $\mathbb{R}^n$, and $B_R$ denotes the closed ball of radius $R >0$ centered at the origin.  The norms involved are the Euclidean norm for the vectors and the operator norm for the linear operator $DF_z$.
My attempt: I have an idea that the mean-value theorem in several variables is involved, but I can't apply it to $F$ directly since it is a function to $\mathbb{R}^n$.  I believe I can finish it off if I can prove the first equality, $\| F(x) - F(x') \| = \| \int_{0}^{1}DF_{(x-x')t+x'}(x-x') dt \|$ and this one: $\| \int_{0}^{1}DF_{(x-x')t+x'}(x-x') dt \| \leq \int_{0}^{1} \| DF_{(x-x')t+x'} (x-x') \| dt$, but the trouble I am having is that the integral is computed componentwise, so I'm unsure how to go about this.
 A: First line to second line: triangle inequality for integrals.  See https://math.stackexchange.com/a/103783/123905 for a quick discussion.
Second line to third line: This is the ML inequality (maximum of integrand times length of integration interval) together with the definition of the norm as the maximum on the unit ball.
A: Actually, this isn’t the mean value theorem, it really is simpler than that.
Consider the function $t \in [0,1] \longmapsto F((1-t)x+tx’)\in \mathbb{R}^n$. It is differentiable enough, and its derivative at $t$ is $DF_{(1-t)x+tx’}\cdot (x’-x)$. By the fundamental theorem of calculus (it works when a finite-dimensional vector space instead of $\mathbb{R}$ or $\mathbb{C}$ is the range and the derivative is continuous – basically just apply it to each coordinate) $\int_0^1{DF_{(1-t)x+tx’}\cdot (x’-x)dt}=F(x’)-F(x)$. To get the first equality, just apply the norm to each side.
For the first inequality, you need to prove that if $f: [0,1] \rightarrow \mathbb{R}^n$ is continuous, $\|\int_0^1{f(t)dt}\|\leq \int_0^1{\|f(t)\|dt}$. The simplest way to see this (in finite dimension at least) is to check that for any continuous linear form $L$ of norm at most one, $L\left(\int_0^1{f(t)dt}\right) \leq \int_0^1{\|f(t)\|dt}$. This is easier because you can convince yourself easily with Riemann sums that LHS can also be written as $\int_0^1{L(f(t))dt}$.
The last inequality stems from the fact that the line segment $[x,x’]$ is a subset of $B_R$ (as long as $R$ is chosen so that $x,x’ \in B_R$).
