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.
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 $x,x’ \in B_R$).