I'm reading Ordinary Differential Equations by Andersson and Böiers. There is a Lemma regarding Lipschitz continuity which I have some questions about. $\pmb f$ is a vector-valued function, and $\pmb x,\pmb y$ vectors in $\mathbb{R}^n$ that depend on $t$.
Lemma. Assume that $\Omega\subseteq \mathbb{R}\times\mathbb{R}^n$ 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 \begin{equation}|\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 \end{equation}
I do not understand exactly how the convexity is being used. That $\Omega$ is convex means that $\pmb{y}+s(\pmb{x}-\pmb{y})\in\Omega$ for $s\in [0,1]$. Geometrically, what does it mean to integrate the function $\frac{\mathrm{d}}{\mathrm{d}s}\pmb{f}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))$ over $[0,1]$? Is $\frac{\mathrm{d}}{\mathrm{d}s}\pmb{f}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))$ well-defined?
I'd be very grateful if someone could clarify how we obtain $(2)$ from the chain rule. I'm familiar with $$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s},\tag4$$ where $z = z(x(t,s), y(t,s) )$, but here it seems like we have a vector valued function which depends on $s$ in a slightly unusual way.
Small detail maybe, but I'm a bit confused about the notation $f$ in $\max_i\sup_{\Omega}\left|\frac{\partial f}{\partial x_i}\right|$. The authors have been using $\pmb f$ to denote the function. Would it be incorrect to replace $f$ by $\pmb f$ here?
The proof should mention that we fix $t$, and thus $(t,\pmb x)$ and $(t,\pmb y)$ are two fixed points in $\Omega$.
The function $\frac{\mathrm{d}}{\mathrm{d}s} \pmb{f}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))$ is well-defined since, as shown in 2, it can be evaluated using the chain rule, and the derivatives exist and are bounded on $\Omega$ since $\pmb f$ is continuously differentiable in a neighborhood of $\overline{\Omega}$. Convexity in this proof is necessary since the function is being integrated over the line connecting $(t,\pmb x)$ and $(t,\pmb y)$.
The function can be viewed as a composition of $\pmb f$ with $\pmb{g}_t(s)=(t,\pmb{y}+s(\pmb{x}-\pmb{y}))$. The subscript denotes that $t$ is held fixed. We could omit the subscript $t$ and only write $\pmb{g}(s)$. The derivative of $D\pmb{f}\circ \pmb{g}(s)$ is given by the chain rule; $D\pmb{f}(\pmb{g}(s))D\pmb{g}(s)$, where \begin{align} D\pmb{f}(\pmb{g}(s))&=\begin{bmatrix} \frac{\partial f_1}{\partial t}\Bigr|_{\pmb{g}(s)} &\frac{\partial f_1}{\partial z_1}\Bigr|_{\pmb{g}(s)} & \dots & \frac{\partial f_1}{\partial z_n}\Bigr|_{\pmb{g}(s)} \\ \vdots &\vdots & \ddots & \vdots \\ \frac{\partial f_n}{\partial t}\Bigr|_{\pmb{g}(s)} & \frac{\partial f_n}{\partial z_1}\Bigr|_{\pmb{g}(s)} & \dots & \frac{\partial f_n}{\partial z_n}\Bigr|_{\pmb{g}(s)}, \end{bmatrix}, \\ D\pmb{g}(s)&=\begin{bmatrix} 0 \\(x_1-y_1) \\ \vdots \\ (x_n-y_n) \end{bmatrix}. \end{align} So \begin{align} D\pmb{f}(\pmb{g}(s))D\pmb{g}(s)&=\begin{bmatrix} \sum_{j=1}^{n}\frac{\partial f_1}{\partial z_j}\Bigr|_{\pmb{g}(s)}(x_j-y_j) \\ \vdots \\ \sum_{j=1}^{n}\frac{\partial f_n}{\partial z_j}\Bigr|_{\pmb{g}(s)}(x_j-y_j) \end{bmatrix}\\ &=\sum_{j=1}^n \frac{\partial \pmb{f}}{\partial z_j}(t,\pmb{y}+s(\pmb{x}-\pmb{y})) (x_j-y_j). \end{align}
$f$ is most likely a typo here for $\pmb{f}$. $|\cdot|$ is the Euclidean norm. For this norm, we have $|x_i-y_i|\leq |\pmb x-\pmb y|$. Moreover, using $\left|\int_a^b \pmb{f}(s)\mathrm{d}s\right|\le\int_a^b\left|\pmb{f}(s)\right|\mathrm{d}s$ and the triangle inequality, we have \begin{align} |\pmb{f}(t,\pmb{x})-\pmb{f}(t,\pmb{y})|&\leq \int_0^1\sum_{i=1}^n\left|\frac{\partial \pmb{f}}{\partial z_i}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))\right|\left|(x_i-y_i) \right| \mathrm{d}s \\ &\leq \int_0^1 nK\left|\pmb x-\pmb y \right|\mathrm{d}s \\ &=nK|\pmb{x}-\pmb{y}| \end{align}
Note, a function $f$ is bounded on a set $A$ if $\sup_A |f|$ is finite. A continuous function is bounded on a compact set, therefor $K$ exists.