Proving Convexity of Bounded Function F

[talolard]

Homework Statement

Hey, the original question is not in english, so I am translating. So just to make sure I'm understood, i take convex to mean that the graph of the function is below the tangent.

The question:
Let F be a convex function and F is bounded from above by some number C, prove that F is static (again, my translation, by static I mean that for every x F(X)=a)

The Attempt at a Solution

I don't think I'm close, but I am stumped, some mild hin tto point me in the right direction would be great.
we will write F as a taylor expansion:
f(x_0)+f'(x_0)(x-x_0) > f(x_0)+f'(x_0)(x-x_0) + \frac {f''(c)(x-x_0)^2}{2} <a \iff
<br /> 0 &gt; \frac {f&#039;&#039;(c)(x-x_0)^2}{2} &lt;a - f(x_0)+f&#039;(x_0)(x-x_0) \iff
<br /> 0 &gt;\frac {f&#039;&#039;(c)}{2} &lt; \frac {a}{(x-x_0)^2} - \frac {f(x_0)+f&#039;(x_0)}{x-x_0}

 

[snipez90]

I don't think Taylor's theorem is required here. The other flaw with your approach is that you are assuming some value for a beforehand in trying to show the function is constant. I think a better approach is to show that f(x) = f(y) for every real x,y.

I think it's a lot easier if you consider two points x and y and suppose x < y. If f(y) > f(x), what can you say about the behavior of f as x approaches infinity? Similarly, what if f(y) < f(x)? Remember in working with these two cases, you are trying to obtain a contradiction if your goal is to show the function is constant. But there is really only one statement in the hypothesis you can directly contradict.

 

[Tinyboss]

The definition I learned of convex is that the graph lies above the tangent. I don't think the problem works with your definition: for example, f(x)=-x² is bounded above by 0, and lies below its tangent.

 

[snipez90]

The function doesn't need to be strictly convex.

 

[talolard]

I don't think Taylor's theorem is required here. The other flaw with your approach is that you are assuming some value for a beforehand in trying to show the function is constant. I think a better approach is to show that f(x) = f(y) for every real x,y.

I think it's a lot easier if you consider two points x and y and suppose x < y. If f(y) > f(x), what can you say about the behavior of f as x approaches infinity? Similarly, what if f(y) < f(x)? Remember in working with these two cases, you are trying to obtain a contradiction if your goal is to show the function is constant. But there is really only one statement in the hypothesis you can directly contradict.

Ok, it took me a few days to work this over and I'm still not there. Here's what I have so far:

Since F is convex then every point has one sided deriviatives such that
x&gt;y f_{-}&#039;(y) \leq f_{+}&#039;(y) \leq f_{-}&#039;(x) \leq f_{+}&#039;(x)
which means that the function is monotonic increasing.
let x>y then lim_{x-&gt; \infty} f(x) = c because the function is monotonic increasing and bounded from above. But this means that
lim_{x-&gt; \infty} \frac {f(x) -f(y)}{x-y}= lim_{x-&gt; \infty} \frac {c -f(y)}{x-y} =0 Which means that there exists a point y<"a"<x where f&#039;(a)=0 but this means that for any x<a f&#039;(x)=0

Now take y>x>a. Then lim_{x-&gt; \infty} f(x) = c and f(y)=c so for all x after a certain point f&#039;(x)=0 then all points before that point must also have f&#039;(x)=0 because of the monotonity of f. Then f is constant.

Is that correct?
Thanks