0

Let $A=[1,2]\cup [3,4]\subset \Bbb{R}$. $x\in \Bbb R$, let $f(x)=\text{inf}\{|x-y|: y\in A\}$. Then the function $f(x)$ is continuous but not differentiable at finitely many point.

Thanks!

Unknown
  • 3,073
  • yes more or less the graph is something like \_/\_/ and the places where you can’t differentiate are the points where the gradient changes – Calvin Khor Nov 20 '20 at 14:14

1 Answers1

0

Let $A=[1,2]\cup [3,4]\subset \Bbb{R}$. $x\in \Bbb R$, let $f(x)=\text{inf}\{|x-y|: y\in A\}$. Then the function $f(x)$ is continuous but not differentiable at finitely many point.

Since the distance function is uniformly continuous therefore $f(x)$ is continuous.

It remains to show that $f(x)$ is not differentiable at finitely many points.

distance from any point(less than 1) to the set $A$ and then distance from any point to the set $A$(greater than 3). We have two slopes from the left most and the right most. Then slope continues from 2 to 3. Then we have sharp edge between 2 and 3 and also sharp edge at 1,2,3,4 and between 2 and 3. For a function differentiable at some point the slope has to be continuous.

Unknown
  • 3,073