# Finding a convex function dominated (point wise) by a given positive function on (0,1)

This problem is actually the Exercise 8 in Chapter 3 of Rudin's Real and Complex analysis book. The problem is as follows:

If $g$ is a positive function on $(0,1)$ such that $g(x) \to \infty$ as $x \to 0$. Does there exist a convex function $h$ on $((0,1)$ such that $h(x) \leq g(x)$ for all $x \in (0,1)$ and $h(x) \to \infty$ as $x \to 0$?

My guess is 'Yes', there does exist such a function but can not prove it. Any help will be appreciated. Thanks in advance!

• I'm thinking of doing something like taking the convex hull of the graph, but I have no idea how to make it rigorous. – Akiva Weinberger Aug 25 '15 at 19:17

## 1 Answer

The supremum of a family of convex functions is again convex. So let

$$h(x) = \sup \{ \varphi(x) : \varphi \text{ is convex and } \varphi(t) \leqslant g(t) \text{ for all } t \in (0,1)\}.$$

Then $h$ is a convex function, and $h(x) \leqslant g(x)$ for all $x\in (0,1)$. Now it remains to see that $\lim\limits_{x\to 0} h(x) = +\infty$.

Let $M > 0$. Since $\lim\limits_{x\to 0} g(x) =+\infty$, there is an $\varepsilon > 0$ such that $x \leqslant \varepsilon \implies g(x) > 2M$. Then

$$\psi(x) = \begin{cases} 2M\bigl(1 - \frac{x}{\varepsilon}\bigr) &, 0 < x < \varepsilon \\ \qquad 0 &, \varepsilon \leqslant x < 1\end{cases}$$

is a convex function on $(0,1)$ with $\psi(x) < g(x)$ for all $x\in (0,1)$ and $\psi(x) > M$ for $x < \frac{\varepsilon}{2}$, hence we have $h(x) > M$ for $x < \frac{\varepsilon}{2}$. Since $M$ was arbitrary, this shows $\lim\limits_{x\to 0} h(x) = +\infty$.

• Is it really possible to prove? I would expect something like $x^{-1}|\sin(x^{-1})|$ to be a problem. – Giovanni Aug 25 '15 at 19:37
• That doesn't have $\lim\limits_{x\to 0} g(x) = +\infty$, it has zeros arbitrarily close to $0$. – Daniel Fischer Aug 25 '15 at 19:43
• @Giovanni The last step is quite simple, one just needs to prove that for every $K > 0$ there is a convex function $f \le g$ with $\lim \limits_{x \to 0} f(x) \ge K$. – Dominik Aug 25 '15 at 19:44
• @Daniel Fischer: right, you are correct! I had a limsup in my mind.. – Giovanni Aug 25 '15 at 19:45
• @Daniel Thanks. I didn't think about considering the collection of convex function and then taking the supremum. Anyway thanks again. – Timon Aug 25 '15 at 20:08