show that function is convex

Question

Let $f:\mathbb{R}\to\overline{\mathbb{R}}$. Show that

$$f\left(x\right)=\begin{cases} +\infty & \mbox{ if }x\in\left(0,\infty\right)\\ 0 & \mbox{ if }x=0\\ -\infty & \mbox{ if }x\in\left(-\infty,0\right) \end{cases}$$

is convex.

How do you define $\infty + (-\infty)$? – Daniel Fischer Oct 05 '13 at 13:13 — Daniel Fischer, Oct 05 '13 at 13:13

score 4 · Accepted Answer · answered Oct 05 '13 at 13:30

4

define $+\infty−\infty=0$

In that case the function is not convex. We have

$$\infty = f\left(\frac13\right) = f \left(\frac23\cdot 1 + \frac13\cdot (-1)\right),$$

but

$$\frac23f(1) + \frac13 f(-1) = \frac23\infty + \frac13(-\infty) = \infty - \infty = 0 < \infty$$

with that definition. For a convex function, we'd have $f(1/3) \leqslant \frac23 f(1) + \frac13 f(-1)$.

answered Oct 05 '13 at 13:30

Daniel Fischer

206,697

what happen if we define $+\infty -\infty =+\infty$ – Muniain Oct 05 '13 at 13:43
1

In that case, we get a convex function. – Daniel Fischer Oct 05 '13 at 13:44

show that function is convex

1 Answers1