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!


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.

  • $\begingroup$ Your definition of local lipschitz on $D$ seems incorrect; if anything, what you describe sounds like "Locally Lipschitz in the $y$ variable". Locally Lipschitz on $D$ means the inequality should read: for all $(t_1, y_1), (t_2, y_2) \in N(t_0,y_0)$, we require $ \lVert f(t_1, y_1) - f(t_2, y_2) \rVert \leq L(t_0, y_0) \lVert(t_1 - t_2, y_1 - y_2) \rVert$. Anyway, the theorem you're asking about is a classic theorem of differential calculus. Here's an answer I wrote a while back with a bunch of references and a hint for how to prove it: math.stackexchange.com/a/3242940/568204 $\endgroup$ – peek-a-boo May 6 '20 at 9:21
  • $\begingroup$ As a side remark: if you're using the standard Euclidean norm on $\Bbb{R}^n$ and $\Bbb{R}^m$, then the $\sqrt{n}$ term is unncecessary in the mean-value inequality. Also, if you take a look at the book by Henri Cartan referenced in my link, you'll also find a much stronger version of the mean-value inequality (whose proof is just as easy/tough as the version you stated) $\endgroup$ – peek-a-boo May 6 '20 at 9:26
  • $\begingroup$ @peek-a-boo Yes, I agree that the condition is different from what one would usually understand as Lipschitz continuity. I think that's why the author says a "Lipschitz condition". I will check out the thread. Regarding the $\sqrt{n}$, it doesn't really matter. Of course you can then just set $D=\sqrt{n}C$. I have simply copied it from my notes and this what comes out of the proof which I have omitted for brevity. $\endgroup$ – DerivativesGuy May 6 '20 at 9:35
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    $\begingroup$ Yup, that's exactly the idea of the proof: pick a small ball with compact closure lying in $D$, then use compactness, and $C^1$-ness to get an upper bound on operator norm then finally use convexity of the ball to apply mean-value theorem. The only real nitpick I have is that in your proof, you never (explicitly) mention what $C$ is. Of course that's an easy fix simply say something like 'by compactness of $N$, there is a $C>0$ such that for all $(t,y) \in N, \lVert df_{(t,y)} \rVert < C$' $\endgroup$ – peek-a-boo May 6 '20 at 14:31
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    $\begingroup$ I have updated the proof and also added something regarding the convexity. $\endgroup$ – DerivativesGuy May 6 '20 at 15:14

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