Differentiability implies Lipschitz continuity (multivariable)

Question

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?

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.