Question

Hello,

If $f:\mathbb{R} \to \mathbb{R}$ a differentiable function, it is very easy to find its Lipschitz constant. Is there any way to extend this to functions $f: \mathbb{R} \to \mathbb{R}^n$ (or similar)?

Answer 1

Let $f = (f_1,\ldots,f_n): [a,b] \rightarrow \mathbb{R}^n$ be a continuously differentiable function. (See the comments above for an explanation as to why the hypotheses have been strengthened.)

For $1 \leq i \leq n$, let

$L_i = \max_{x \in [a,b]} |f_i'(x)|$,

so that, by the Mean Value Theorem, for $x,y \in [a,b]$,

$|f_i(x)-f_i(y)| = |f_i'(c)||x-y| \leq L_i |x-y|$.

Then, taking the standard Euclidean norm on $\mathbb{R}^n$,

$|f(x)-f(y)|^2 = \sum_{i=1}^n |f_i(x)-f_i(y)|^2 \leq (\sum_{i=1}^n L_i^2) \ |x-y|^2$,

so

$|f(x)-f(y)| \leq \sqrt{(\sum_{i=1}^n L_i^2)} \ |x-y|$.

Thus we can take

$L = \sqrt{\sum_{i=1}^n L_i^2}$.

Since all norms on $\mathbb{R}^n$ are equivalent -- i.e., differ at most by a multiplicative constant -- the choice of norm on $\mathbb{R}^n$ will change the expression of the Lipschitz constant $L$ in terms of the Lipschitz constants $L_i$ of the components, but not whether $f$ is Lipschitz.

Answer 2

In fact a statement similar to what was described by Pete Clark is true for all normed vector spaces: Let $X$ and $Y$ be normed vector spaces. A (total) differentiable function $f:X\to Y$ is Lipschitz iff its derivative is bounded. Every upper bound for the differential is a Lipschitz constant.

One direction follows from the mean value theorem: $\|f(x)-f(y)\|\leq \|Df(\xi)\|\cdot\|x-y\|$ for some $\xi$ on the straight line from $x$ to $y$.

The other follows immediately from $Df(x)=\lim_{h\to 0}\frac{f(x)-f(x+h)}{\|h\|}$

Answer 3

Another nice way to do this computation (which generally gives more precise information) is to use the formula

$ f(x + h) - f(x) = [ \int_0^1 Df(x + th) dt ] h$

Also, the formula for $Df(x)$ in Hahn's post above is not correct.