# Differentiability implies Lipschitz continuity (multivariable)

I am studying from Marsden: Elementary Classical Analysis ($$2^{\rm{nd}}$$ ed.). I am not able to write down the complete proof of the following theorem (Theorem 6.3.1, page 334). The theorem essentially says that:

Let $$A\subset \mathbb{R}^n$$ be an open set and let $$f:A \to \mathbb{R}^m$$ be a differentiable function. Then $$f$$ is locally Lipschitz, i.e. for each $$x_0\in A$$, there is $$M>0$$ and $$\delta_0>0$$ such that $$\|x-x_0\|<\delta_0 \quad \Rightarrow \quad \|f(x)-f(x_0)\|

I could not locate the proof anywhere, nor was able to generalize the one-variable proof to multivariable (For example: here, here. This one is slightly different, with an extra condition of $$f'$$ being continuous.)

The reason I cannot apply these arguments to multivariable case is all these proofs use MVT.

Can someone help with the proof? OR direct me to a reference / book which has a proof?

## 1 Answer

HINT: Just use the definition of the derivative. If $$f$$ is differentiable at $$x_0$$, then (given $$\epsilon>0$$ there's $$\delta>0$$ so that) $$\|f(x)-f(x_0) - f'(x_0)(x-x_0)\| < \epsilon\|x-x_0\|$$ whenever $$\|x-x_0\|<\delta$$. Taking $$M=\|f'(x_0)\| + \epsilon$$ should do it.

• thats exactly the problem I am facing with generalizing the proofs from these links. In multivariate situation, what is $|f'(x_0)|$? Do you mean the determinant of the Jacobian matrix? – Mike V.D.C. Mar 21 at 18:46
• No, I mean the (operator) norm of the linear map. $\|A\| = \max\limits_{\|x\|=1} \|Ax\|$. (Of course, we're using $\|\cdot\|$ for three different norms here! Sorry about that.) You might find some of my YouTube lectures on multivariable mathematics linked in my profile helpful. – Ted Shifrin Mar 21 at 19:00
• Thanks for the clarification! Also thanks for the YouTube links! Will definitely look at them – Mike V.D.C. Mar 22 at 10:02