# Vector-valued continuously differentiable function of several variables is locally Lipschitz

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 $$0. Now note that $$\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 $$\|df_{(t,y)}\|_{operator} for all $$(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.

• 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 – peek-a-boo May 6 '20 at 9:21
• 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) – peek-a-boo May 6 '20 at 9:26
• @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. – DerivativesGuy May 6 '20 at 9:35
• 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 $(t,y) \in N, \lVert df_{(t,y)} \rVert < C$' – peek-a-boo May 6 '20 at 14:31
• I have updated the proof and also added something regarding the convexity. – DerivativesGuy May 6 '20 at 15:14