# Concaveness and continuity

1. Aug 19, 2012

### hermanni

Hi all,
I have the following question: Suppose f: [0, ∞) $\rightarrow$ℝ and f is concave , nondecreasing and bounded on [ 0, ∞) . Does it follow that f is continuous on [ 0, ∞) ? Thanks in advance, H.

2. Aug 19, 2012

### tiny-tim

hi hermanni!

tell us what you think, and then we'll comment!

(start by writing out the definition of "concave")

3. Aug 19, 2012

### DonAntonio

A step function is non-decreasing, convex (or concave) and we can make it bounded, but won't be continuous. For example
$$f(x)=0\,\,,\,if\,\, x\in [0,1)\,,\,f(x)=1\,\,,\,if\,\, x\geq 1\,$$

DonAntonio

4. Aug 19, 2012

### micromass

How is a step function convex??

5. Aug 19, 2012

### DonAntonio

Very simple: it fulfills the definition of concave upwards (if the steps go up) or concave downwards (steps down), just as a constant

function does as well.

DonAntonio

6. Aug 19, 2012

### micromass

I don't see how. Take the function

$$f:]-1,1[\rightarrow \mathbb{R}:x\rightarrow \left\{\begin{array}{l} 0~\text{if}~x\leq 0\\ 1 ~\text{if}~x>0 \end{array}\right.$$

This is the function you mean right??

According to the definition, this function is convex if for all $x,y\in ]-1,1[$ and $t\in [0,1]$ holds that

$$f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$$

But take x=-1/2, y=1/2 and t=1/4, then

$$f(tx+(1-t)y)=1$$

while

$$tf(x)+(1-t)f(y)=3/4$$

and this does not satisfy the inequality.

Furthermore, it violates the following theorem: http://planetmath.org/ContinuityOfConvexFunctions.html [Broken]

Last edited by a moderator: May 6, 2017
7. Aug 19, 2012

### DonAntonio

No, my function's defined on the convex non-negative ray $\,[0,\infty)\,$
Point taken, but eventhough my example indeed doesn't quite fit in the traditional definition, I will change it slightly and still

will we get a counterexample:

$$f(0)=0\,,\,f(x)=1\,\,,\,\forall\,x>0$$

The above function is convex upwards in $\,[0,\infty)\,$ but not continuous there.

DonAntonio

Last edited by a moderator: May 6, 2017
8. Aug 19, 2012

### micromass

OK, but I think the OP cares more about convex downwards...

9. Aug 19, 2012

### DonAntonio

No problem: interchange zero and one in my last message's definition.

DonAntonio

10. Aug 19, 2012

### micromass

That's not nondecreasing

11. Aug 19, 2012

### pwsnafu

Hint: suppose you already have a function that is continuous, non-decreasing, concave and bounded on the open interval $(0,\infty)$. Then continuity on $[0,\infty)$ is completely determined by $f(0)$.

12. Aug 19, 2012

### haruspex

There is perhaps some confusion about the terms convex and concave. I believe that the standard definition is such that for twice differentiable functions convex would mean the second derivative is never negative. I.e. convexity/concavity is as viewed from 'below'.
This would make DonAntonio's counterexample at post #7 valid. But I suspect the OP was using the terms inversely, so suppose 'convex' was intended. As micromass/pwsnafu point out, the continuity is assured everywhere except at 0 by the convexity alone. The only question is whether the non-decreasing condition makes it continuous at 0+. I believe it does by a simple variation on the proof at the PlanetMath site quoted.