Bounded Gradient implies Lipschitz proof with the mean value theorem

Question

Let $f:\mathbb{R}^n \to \mathbb{R}$ with $|| \nabla f(x)|| \leq M$ (say it is the Euclidean norm), then f is Lipschitz.

I have seen proofs that do this for the case where $f:\mathbb{R} \to \mathbb{R}$ by applying the mean value theorem. I am wondering if there is a proof available that shows how the mean value theorem is applied to the problem for a function from $f:\mathbb{R}^n \to \mathbb{R}$?

Answer 1

I don't understand where is your problem. This is exactcly the same proof in higher-dimension.

By the mean value theorem we have :

$$\| f(x) - f(y) \| \leq \sup_{x \in \mathbb{R}^n} \| \nabla f(x) \| \|x -y \| \leq M \| x - y\|$$

Answer 2

Here's an approach using the fundamental theorem of calculus and an integral estimate in lieu of the mean value theorem:

Let

$x, y \in \Bbb R^n; \tag 1$

let

$\gamma:[0, 1] \to \Bbb R^n \tag 2$

be given by

$\gamma(t) = x + t(y - x); \tag 3$

then $\gamma(t)$ is a line segment 'twixt

$\gamma(0) = x \; \text{and} \; \gamma(1) = y; \tag 4$

then

$f(y) - f(x) = f(\gamma(1)) - f(\gamma(0))$ $= \displaystyle \int_0^1 \dfrac{df(\gamma(t))}{dt} \; dt = \int_0^1 \nabla f(\gamma(t)) \cdot \dot \gamma(t) \; dt = \int_0^1 \nabla f(\gamma(t)) \cdot (y - x) \; dt \tag 5$

therefore,

$\vert f(y) - f(x) \vert = \left \vert \displaystyle \int_0^1 \nabla f(\gamma(t)) \cdot (y - x) \; dt \right \vert \le \displaystyle \int_0^1 \vert \nabla f(\gamma(t)) \vert \vert y - x \vert \; dt \le \vert y - x \vert \int_0^1 M \; dt = M\vert y - x \vert, \tag 6$

which shows that $f(x)$ is in fact globally Lipschitz continuous with Lipschitz constant $M$. $OE\Delta$.

It will be observed that there is more than one similarity 'twixt this and the MVT approach; both are based on "one-dimensionalizing" the problem by restriction to a path joining $x$ and $y$, and both exploit the global bound $\vert \nabla f(x) \vert \le M$ to obtain the global Lipschitz constant $M$. Formally, by way of the mean value theorem we would write

$f(y) - f(x) = f(\gamma(1)) - f(\gamma(0)) = (f(\gamma(r))'(1 - 0) = (f(\gamma(r))', 0 < r < 1; \tag 7$

and note that

$f(\gamma(r))' = \nabla f(\gamma(r)) \cdot \dot \gamma(r) = \nabla f(\gamma(r)) \cdot (y - x); \tag 8$

if we combine (7) and (8) and take norms, the desired result is obtained.