Differentiable Convex Function with multiple global minimum.

Question

Does there exist Differentiable Convex Function with multiple global minimum (except for constant function)?

Answer 1

Here is an example convex, differentiable, and non-constant function $f:\mathbb{R}\rightarrow\mathbb{R}$ $$f(x) = \left\{ \begin{array}{ll} (x-1)^4 &\mbox{ if $x \geq 1$} \\ 0 & \mbox{ if $x \in [0, 1)$} \\ x^4 & \mbox{ if $x<0$} \end{array}\right. $$ This has an infinite number of global mins, and is consistent with the David Ullrich answer in that indeed we have $f'(x)=0$ over the entire interval $x\in[0,1]$.

Answer 2

No, not unless you count piecewise-functions which include a constant function like $y = 0$ for $-2<x<2$, $x^2-4$ else.

Answer 3

No. If $f$ is convex then $f'$ is non-decreasing, so we cannot have $f'=0$ at more than one point (unless $f'=0$ on an entire interval).

Answer 4

Unless the interval where the minima lie is constant, then no. Convexity implies that all local optima are global optima. If there were a interval where the values were not constant, where multiple local optima were global optima, it would violate the properties of convexity.

Try drawing it out for added intuition!