Lipschitz continuity of vector-valued function

[psie]

TL;DR Summary

I have a question about a proof on Lipschitz continuity of a vector-valued function.

I'm reading Ordinary Differential Equations by Andersson and Böiers, although this is more related to multivariable calculus. There is a Lemma regarding Lipschitz continuity which I have a question about. Below $\pmb{f}:\mathbf{R}^{n+1}\to \mathbf{R}^n$ is a vector-valued function defined by $\pmb{f}=\pmb{f}(t,\pmb{x})$, where $\pmb x$ is a vector in $\mathbf{R}^n$ that depends on $t$.

Lemma. Assume that $\Omega\subseteq \mathbf{R}\times\mathbf{R}^n$ is a convex and bounded set, and that the function $\pmb{f}$ is continuously differentiable in a neighborhood of $\overline{\Omega}$. Then $\pmb{f}$ is Lipschitz continuous in $\Omega$.

Proof. The line segment between two points $(t,\pmb{x})$ and $(t,\pmb{y})$ in $\Omega$ is contained in $\Omega$ by the assumption of convexity. Hence \begin{align} \pmb{f}(t,\pmb{x})-\pmb{f}(t,\pmb{y})&=\int_0^1\frac{\mathrm{d}}{\mathrm{d}s}\pmb{f}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))\mathrm{d}s \tag1 \\ &=\int_0^1\sum_{i=1}^n\frac{\partial \pmb{f}}{\partial x_i}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))(x_i-y_i)\mathrm{d}s,\tag2\end{align} by the chain rule. Set $K=\max_i\sup_{\Omega}\left|\frac{\partial f}{\partial x_i}\right|$ (which exists by assumption). Then $$|\pmb{f}(t,\pmb{x})-\pmb{f}(t,\pmb{y})|\leq\int_0^1 nK|\pmb{x}-\pmb{y}|\mathrm{d}s=nK|\pmb{x}-\pmb{y}|.\tag3$$

Here's my understanding of the notation. In $(2)$, $\frac{\partial \pmb{f}}{\partial x_i}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))$ is a vector evaluated at $(t,\pmb{y}+s(\pmb{x}-\pmb{y}))$. $\frac{\partial}{\partial x_i}$ denotes the partial derivative of $\pmb{f}(t,x_1,\ldots,x_n)$ with respect $x_i$ (a better notation in $(2)$ would probably be $\frac{\partial}{\partial z_i}$, as $\pmb{x}$ is already being used to denote a vector). More precisely, if $f_1,\ldots,f_n$ are the components of $\pmb{f}$, then $$\frac{\partial \pmb{f}}{\partial x_i}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))=\begin{pmatrix}
\frac{\partial f_1}{\partial x_i}(t,\pmb{y}+s(\pmb{x}-\pmb{y})) \\ \vdots \\ \frac{\partial f_n}{\partial x_i}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))
\end{pmatrix}.\tag4$$

My question is that I do not really understand how $(3)$ is obtained. For the sake of brevity, I will write $\frac{\partial \pmb{f}}{\partial x_i}$ when I mean $\frac{\partial \pmb{f}}{\partial x_i}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))$. Suppose $n=2$. Taking the norm ($\ell^2$ norm to be exact) of $(1)$ and using $\left\lVert\int_a^b \pmb{f}\right\rVert\le\int_a^b\left\lVert\pmb{f}\right\rVert$, we get
\begin{align}
\lVert\pmb{f}(t,\pmb{x})-\pmb{f}(t,\pmb{y})\rVert_2&=\left\lVert\int_0^1\sum_{i=1}^2\frac{\partial \pmb{f}}{\partial x_i}(x_i-y_i)\mathrm{d}s\right\rVert_2 \tag5\\
&\leq \int_0^1\left\lVert\sum_{i=1}^2\frac{\partial \pmb{f}}{\partial x_i}(x_i-y_i) \right\rVert_2 \mathrm{d}s \tag6
\end{align}
Next, using the triangle inequality, the norm inside the integral gets bounded by
\begin{align}
\left\lVert\sum_{i=1}^2\frac{\partial \pmb{f}}{\partial x_i}(x_i-y_i) \right\rVert_2 &\leq\left\lVert \frac{\partial \pmb{f}}{\partial x_1}(x_1-y_1)\right\rVert_2+\left\lVert \frac{\partial \pmb{f}}{\partial x_2}(x_2-y_2)\right\rVert_2. \tag7
\end{align}
The above terms on the right-hand side equal
\begin{align}
\sqrt{\left(\frac{\partial f_1}{\partial x_1}\right)^2+\left(\frac{\partial f_2}{\partial x_1}\right)^2}|x_1-y_1|+\sqrt{\left(\frac{\partial f_1}{\partial x_2}\right)^2+\left(\frac{\partial f_2}{\partial x_2}\right)^2}|x_2-y_2|.\tag8
\end{align}
Letting $K=\max\limits_{i,j\in\{1,2\}}\sup\limits_{\Omega}\left|\frac{\partial f_i}{\partial x_j}\right|$, $(8)$ gets bounded by
\begin{align}
\sqrt{2}K|x_1-y_1|+\sqrt{2}K|x_2-y_2|=\sqrt{2}K(|x_1-y_1|+|x_2-y_2|) \tag9
\end{align}
which is clearly not equal to $nK|\pmb{x}-\pmb{y}|$. I suspect I am doing something wrong, but I'm unable to spot it.

 

[pasmith]

A basic upper bound for the value of an integral of $F: [a,b] \to \mathbb{R}$ is $$\int_a^b F(s)\,ds \leq (b- a)\sup_{s \in [a,b]} F(s).$$ By extension, given $$I = \int_0^1 \sum_{i=1}^n\|\mathbf{F}_i(t,s)\||x_i - y_i|\,ds$$ we can obtain $$\begin{split} I &\leq \sum_{i=1}^n \sup_s \|\mathbf{F}_i\||x_i - y_i| \\ &\leq \left(\max_i \sup_s \|\mathbf{F}_i\|\right) \sum_{i=1}^n |x_i - y_i|.\end{split}$$ Now $$\begin{split} \sum_{i=1}^n |x_i - y_i| &= \left|(1, \dots, 1) \cdot (|x_1-y_1|, \dots, |x_n - y_n|)\right| \\ & \leq \| (1, \dots, 1) \| \|(|x_1-y_1|, \dots, |x_n - y_n|)\| \\ & = \sqrt{n} \|\mathbf{x} - \mathbf{y}\| \\ & \leq n \|\mathbf{x} - \mathbf{y}\|.\end{split}$$ so that $$I \leq \left(\max_i \sup_s \|\mathbf{F}_i\|\right) \sum_{i=1}^n |x_i - y_i| \leq \left(\max_i \sup_s \|\mathbf{F}_i\|\right) n \|\mathbf{x} - \mathbf{y}\|.$$

 

[psie]

Thank you for replying.

In the statement of the lemma, the authors stipulate that $\pmb f$ be $C^1$ in a neighborhood of the closure of $\Omega$ (I interpret $\overline{\Omega}$ to be the closure of $\Omega$). Do you by any chance know why this is required? I see why we need the function to be $C^1$, since we are integrating its derivative, but I do not see why this condition needs to be satisfied in the closure of $\Omega$.

 

[pasmith]

To guarantee boundedness of the derivatives we require that they be continuous (hence $C^1$) on a compact domain. This is the generalisation of the theorem that a function continuous on a closed, bounded interval is bounded, but a function continuous on an unclosed bounded interval may not be bounded (eg. $x^{-1}$ on $(0,1]$).

For the Euclidean metric on $\mathbb{R}^m$, a subset is compact if and only if it is closed and bounded. We are given that $\Omega \subset \mathbb{R} \times \mathbb{R}^n \simeq \mathbb{R}^{n+1}$ is bounded, but we are not told that it is closed; its closure, however, is certainly closed and bounded.