# Thread: 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.

6. Hrm, let me think about this. You might be right, or I'm missing another condition.

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

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