Consider a function $f:\mathbb{R}^n\longrightarrow\mathbb{R}^m$ with partial derivatives everywhere so that the Jacobian matrix is well-defined. Let $L>0$ be a real number. Is it true that:
$$|f(x)-f(y)|\leq L|x-y|,\forall x,y \Longleftrightarrow |J_f(x)|_2\leq L,\forall x$$
where $|\cdot|$ denotes the euclidean vector norm and $|\cdot|_2$ the spectral matrix norm.
[Math] Lipschitz continuous and Jacobian matrix
jacobianlipschitz-functionsmultivariable-calculus
Best Answer
Fix $\delta > 0$, let's take $x \in \mathbb{R}^n$ where the derivatives exist, then prove that $f$ is $(L+\delta)$-lipschitz. for that let $v$ of norm $1$ and consider the quantity : $$ \frac{|f(x)-f(x+tv)|}{t}$$ Where $t\in\mathbb{R}$, we know that the limit equals : $J_{f}(x)(v)$ namely the derivative in the direction of $v$, so there exist an $\epsilon_{v}$ such that: $\frac{|f(x)-f(x+tv)|}{t}<L+\delta, \text{for} |t|<\epsilon_{v}$, by the compactness of the shpere one may choose a common $\epsilon$ which gives : $$ \frac{|f(x)-f(z)|}{|x-z|}<L+\delta, \text{if} |x-z|<\epsilon. $$ Now take the segment which link $x$ to $y$, cover it by balls where $f$ is $(L+\delta)$-Lipschitz on each ball, and use the triangle inequality on the centers of the balls and the intersections ( we can manage to have finite number of balls, each ball is centered on a point on the segment and intersect only two other balls, with intersection points on the segment, except for the ball centered on $x$ and the one centered on $y$). It suffices now to take $\delta$ to $0$ to get the result.
Now for the other direction, we see that $J_{f}(x)(v) = \lim \frac{|f(x)-f(x+tv)|}{t}≤L$, hence : $|J_{f}(x)|≤L$.
PS: there is a surprising theorem on Lipschitz functions ( also locally Lipschitz functions ) which says that they are differentiable almost everywhere, it's called Rademacher's theorem, note also that differentiable is more strong than only admitting directional derivatives.