Is this function always differentiable?

Question

I need some help with the following exercise:

Assume $A \subset \mathbb{R}$ is a compact set. Is the following function always differentiable?

$f: \mathbb{R}^{n+1} \rightarrow \mathbb{R}, \quad f(x_0, ... x_n) := \sup\limits_{x \in A} \prod\limits_{j=0}^{n} (x-x_j)$

My idea was to show this directly for all partial derivatives:

$\lim\limits_{h \rightarrow 0} \frac{f(x_0,..,x_j+h,..,x_n)-f(x_0,...,x_n)}{h}= \lim\limits_{h \rightarrow 0} \frac{\sup\limits_{x \in A} ((x-x_j-h)\prod\limits_{i\in\{0,...,n\}\setminus\{j\}} (x-x_i))- \sup\limits_{x \in A} (\prod\limits_{i=0}^{n} (x-x_i))}{h}=\lim\limits_{h \rightarrow 0} \frac{\sup\limits_{x \in A} (\prod\limits_{i=0}^{n} (x-x_i)-h(\prod\limits_{i\in\{0,...,n\}\setminus\{j\}}(x-x_i))- \sup\limits_{x \in A} (\prod\limits_{i=0}^{n} (x-x_i))}{h}$

At this point I think it should be possible to show that this limit is finite in general or to search for a example for a non differential case should be easier. Can someone give me a hint how to proceed?

Answer 1

Hint: Consider the situation where $A=[0,1]$. Is $g(x) = f(x,x) = \sup_{y\in A} (x-y)^2$ differentiable everywhere?

Answer 2

Let's test the claim of differentiability with a simple example . . .

Let $n=1$, and let $A=\{-1,1\}$.

Then $f:\mathbb{R}^2\to R$ is given by $$f(x,y)=\max\{(-1-x)(-1-y),(1-x)(1-y)\}$$ and since \begin{align*} (-1-x)(-1-y)&=1+x+y+xy\\[4pt] (1-x)(1-y)&=1-x-y+xy\\[4pt] \end{align*} it follows that $$ f(x,y)= \begin{cases} 1+x+y+xy&&\text{if}\;x+y \ge 0\\[4pt] 1-x-y+xy&&\text{if}\;x+y < 0\\[4pt] \end{cases} $$ Now consider the question of whether the partial derivative of $f$ with respect to $x$ exists at the point $(0,0)$.

First, letting $h$ approach $0$ from the right, we get \begin{align*} \lim_{h\to0^+}\frac{f(h,0)-f(0,0)}{h} &=\lim_{h\to0^+}\frac{(1+h)-1}{h}\\[4pt] &=\lim_{h\to0^+}\frac{h}{h}\\[4pt] &=1\\[4pt] \end{align*} Next, letting $h$ approach $0$ from the left, we get \begin{align*} \lim_{h\to0^-}\frac{f(h,0)-f(0,0)}{h} &=\lim_{h\to0^-}\frac{(1-h)-1}{h}\\[4pt] &=\lim_{h\to0^-}\frac{-h}{h}\\[4pt] &=-1\\[4pt] \end{align*} It follows that $f_x(0,0)$ does not exist.

As a consequence, $f$ is not differentiable.