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hermanni

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

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- Thread starter hermanni
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- #1

hermanni

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

- #2

tiny-tim

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tell us what

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

- #3

DonAntonio

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

A step function is non-decreasing, convex (or concave) and we can make it bounded, but won't be continuous. For example

[tex]f(x)=0\,\,,\,if\,\, x\in [0,1)\,,\,f(x)=1\,\,,\,if\,\, x\geq 1\,[/tex]

DonAntonio

- #4

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How is a step function convex??

- #5

DonAntonio

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How is a step function convex??

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

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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

I don't see how. Take the function

[tex]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.[/tex]

This is the function you mean right??

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

[tex]f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)[/tex]

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

[tex]f(tx+(1-t)y)=1[/tex]

while

[tex]tf(x)+(1-t)f(y)=3/4[/tex]

and this does not satisfy the inequality.

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

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- #7

DonAntonio

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I don't see how. Take the function

[tex]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.[/tex]

This is the function you mean right??

No, my function's defined on the convex non-negative ray [itex]\,[0,\infty)\,[/itex]

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

[tex]f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)[/tex]

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

[tex]f(tx+(1-t)y)=1[/tex]

while

[tex]tf(x)+(1-t)f(y)=3/4[/tex]

and this does not satisfy the inequality

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

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:

[tex]f(0)=0\,,\,f(x)=1\,\,,\,\forall\,x>0[/tex]

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

DonAntonio

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- #8

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

- #9

DonAntonio

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

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

DonAntonio

- #10

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That's not nondecreasing

- #11

pwsnafu

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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.

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