# Lipschitz continuous and Jacobian matrix

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.

• Consider the inequality $|f(x) - f(y)| \leq L|x-y|$. We can rewrite it as $\frac{|f(x)-f(y)|}{|x-y|} \leq L$. Is there any equation/formula from calculus you recall that you can relate to the left-hand side of this inequality? In particular, can you think of a formula that involves a derivative?
– kkc
Mar 22 '19 at 19:17
• The result is indeed trivial for $n=m=1$ by using the definition of derivative for one implication and by using the mean value theorem for the other one. The question is about extending this result to multidimension. Mar 25 '19 at 8:33

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: $$\frac{|f(x)-f(x+tv)|}{t}, by the compactness of the shpere one may choose a common $$\epsilon$$ which gives : $$\frac{|f(x)-f(z)|}{|x-z|} 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 : $$|J_{f}(x)|≤L$$.