My lecture notes on ODEs state the following in the section on unique solutions:
Definition: Let $I \subset \mathbb{R}$ be an interval and let $D \subseteq I \times \mathbb{R}^d$ be open. Then $f : D \to \mathbb{R}^d$ satisfies a local Lipschitz condition on $D$ if $\forall (t_0,y_0) \in D$ there exists a neighbourhood $N(t_0,y_0) \subseteq D$ of $(t_0,y_0)$ and a constant $L(t_0, y_0) \geq 0$ such that
$\|f(t,y_1) − f(t,y_2)\| \leq L(t_0,y_0) \|y_1 − y_2\|$ for all $(t,y_1),(t,y_2) \in N(t_0,y_0)$.
Using the mean value theorem we can show that any function continuousy differentiable on $D$ satisfies a local Lipschitz condition by the mean value theorem.
In my lecture notes on real analysis the mean value theorem is stated as follows:
Theorem: Suppose $f : U \to \mathbb{R}^m$ is differentiable with uniformly bounded operator norm $ \|df_x\|_{operator} \leq C$ for some $C \geq 0$ and all $x \in U$. Then, for all $a,b$ such that $\bar{ab} \in U$
$\|f(b)-f(a)\| \leq \sqrt{n}C \|b-a\|$.
Here $\bar{ab}$ is the segment between $a$ and $b$ and the operator norm for a linear map $L: \mathbb{R}^m \to \mathbb{R}^n, L(x)=Ax$ is defined as
$\|L\|_{operator}:=\|A\|_{operator}:=sup\{\|L(x)\|=\|Ax\| : \|x\| \leq 1\}$.
Also note that $U \subseteq \mathbb{R}^n$.
Now I am a bit confused about what it means for a function $f : U \to \mathbb{R}^m$ to be continuously differentiable.
In the single variable real-valued case the derivative is defined through a limit and we can simply define a function $f' : x \to f'(x)$ and call it the derivative. We then say $f$ is continuously differentiable if $f'$ exists and is continuous $\forall x$.
I think in the multivariate vector-valued case we usually consider the partial derivatives and use the same logic, i.e. we say a function $f : U \to \mathbb{R}^m$ is continuously partially differentiable if all partial derivatives exist and are continuous.
But I don't see how I can apply the mean value theorem given this assumption. For this I would need some notion of continuity for the total differential $df_x$, but this is not a real number as in the one-dimension case but a linear map $df_x : \mathbb{R}^n \to \mathbb{R}^m$ instead. Do we then say that $f$ is continuously differentiable if the function $df: x \to df_x$ is continuous?
Appreciate it if someone could clarify the definition for me and how we can then show that it implies that $f$ satisfies a local Lipschitz condition.
Thanks very much!
Edit:
Found another post that contains an answer to my question. However, would be great if someone could show me why $df: x \to df_x$ is continuous if and only if all partial derivatives are continuous.
Edit 2:
Proof idea:
Since $D$ is open $\exists B_{\epsilon}(t_0,y_0) \subseteq D$. So $\bar{B}_r(t_0,y_0) \subset B_{\epsilon}(t_0,y_0)$ for $0<r<\epsilon$. Now note that $\bar{B}_r(t_0,y_0)$ is closed by definition and bounded since $\bar{B}_r(t_0,y_0) \subset B_{\epsilon}(t_0,y_0)$. So $\bar{B}_r(t_0,y_0)$ is compact.
Since $f$ is continuously differentiable on $N=\bar{B}_r(t_0,y_0)$, we have that $\|df_{(t,y)}\|_{operator}<C$ for all $(t,y) \in N$ by compactness of $N$. So by the mean value theorem
$\|f(t_1,y_1)-f(t_2,y_2)\| \leq \sqrt{n}C\|(t_1,y_1)-(t_2,y_2)\|$ which shows that $f$ is Lipschitz continuous on $B_r(t_0,y_0)$ with Lipschitz constant $K=\sqrt{n}C$. In particular, this holds for $t_1=t_2$ which proves the claim.
Note that we have used the convexity of $\bar{B}_r(t_0,y_0)$, see for example here (the proof is for the open ball, but the argument is the same.