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 $$\|x-x_0\|<\delta_0 \quad \Rightarrow \quad \|f(x)-f(x_0)\|<M\|x-x_0\|.$$

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 1


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.

  • $\begingroup$ 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? $\endgroup$ Mar 21, 2019 at 18:46
  • $\begingroup$ 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. $\endgroup$ Mar 21, 2019 at 19:00
  • $\begingroup$ Thanks for the clarification! Also thanks for the YouTube links! Will definitely look at them $\endgroup$ Mar 22, 2019 at 10:02

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