# Math Help - Convex Function Proof

1. ## Convex Function Proof

Hey all,

Let f be a positive function on (0, 1) such that $f(x) \to \infty$ as $x\to 0$. Must there exist a convex function g such that $g(x)\le f(x)$ with $g(x) \to \infty$ as $x \to 0$? Prove or give a counter example.

My Progress: Truthfully, I haven't made all that much. My intuition is that g exists. We can make a bunch of assumptions WLOG about f, e.g. that it is monotone; we can also assume that f is a step function or piecewise linear, but so far I haven't been very successful at coming up with anything even in these cases.

2. If the problem is completely stated, examine the function $f(x)=\cot(\pi x).$

3. There exist functions

$
g(x)=f(x)-x^2
$

and not

$
g(x)=f(x)-x^2 \; sin(1/x).
$

4. Whoops, the question isn't stated fully. We require $g(x) \to \infty$ as $x \to 0$ as well. So, we need f to be bounded from below by convex g where g also goes to infinity. As originally stated, obviously $g: x \mapsto 0$ would work.

I don't think either of the prior posts give any help; obviously not the posters fault since the question was incomplete.

5. I still claim my post # 2 is a counterexample. I'm not sure how I could prove it, but I think the function I gave will outrun any convex function to negative infinity as x goes to 1.

Sorry for the sloppiness, I'm not sure if the question is even correctly stated as is. However, I would still like an answer to the question if we require $f \ge 0$ as well.

7. Yes, sorry, the modified wording was still wrong. The functions are positive. I deffinitely had that in mind when I was thinking through everything, however I've been thinking about it on my own for some time and was anxious to get the question up here so I could get an answer

8. If you require $f\ge 0,$ then I think it can be done. Thinking about how, though...

9. One basic idea would be this: Start with a function like $g(x)=(1/x^{n})-M,$ where $n$ is large enough to make $g$ duck under function $f.$ Then make $M$ large enough to get you to the x axis faster than $f$ does. Then glue the zero function onto this much. So you have this:

$g(x)=\begin{cases}(1/x^{n})-M,\quad&0 0,\quad&a\le x<1.\end{cases}.$

Here $a$ is defined by

$\dfrac{1}{a^{n}}-M=0,$ or

$a^{n}=\dfrac{1}{M},$ or

$a=\dfrac{1}{M^{1/n}}.$