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Rudin ask:

If $g$ is a positive function on $(0,1)$ such that $g(x)\rightarrow \infty$ as $x\rightarrow 0$. Then there is a convex function $h$ on $(0,1)$ such that $h\le g$ and $h(x)\rightarrow \infty$ as $x\rightarrow 0$. True or False? Is the problem changed if $(0,1)$ is replaced by $(0,\infty)$ and $x\rightarrow 0$ replaced by $x\rightarrow \infty$?

I thought about using piece-wise linear functions $f_{n}$ to approximate $g$, and use convex functions $h_{ni}$ to approximate $f_{n}$. The selecting $h_{nn}$ we would be able to approximate $g$. But this missed a point; $g$ can be arbitrally strange. For example if we work with a basis for $\mathbb{R}/\mathbb{Q}$, then $g$ can even be dense in $\mathbb{R}$ in any subset of $(0,1)$. So a piecewise approximation is impossible. But is an approximation by convex functions still possible?

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  • $\begingroup$ I think this is impossible. $\endgroup$ – Bombyx mori Dec 25 '12 at 13:24
  • $\begingroup$ $g$ cannot have a dense range on every subinterval because it tends to $\infty $ . Begin by defining $f(x)$ to be the infimum of $g$ on $ (0,x)$ - this is a monotone function. $\endgroup$ – user53153 Dec 25 '12 at 15:17
  • $\begingroup$ This is a duplicate of math.stackexchange.com/questions/50549/… $\endgroup$ – Bombyx mori Dec 25 '12 at 15:39
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Following Pavel M's suggestion, let us define $f(x)=\inf_{(0,x)}g(x)$. Then $f(x)$ is monotonely decreasing and $f(x)<g(x)$ for all $x$. Further, $f(x)\rightarrow \infty$ as $x\rightarrow 0$. So it suffice to find a sequence of convex functions $f_{n}$ approximating $f$.

Since $f$ is monotonely decreasing, during the interval $[a,b]$ $f(x)$'s value has to be between $f(a)$ and $f(b)$. So it is possible to make a convex function such that $f_{n}(x)\le f(b)-\frac{1}{2^{n}}$ on $[a,b]$.

We now define $f_{n}$ by a step by step process, with $f_{1}$ defined on $[1/4,3/4]$, $f_{2}$ defined on $[1/8,1/4]\cup [1/4, 3/4]\cup [3/4,7/8]$ and so on, with a convex "bump" function between neighboring intervals. Eventually we defined $\lim_{n\rightarrow \infty} f_{n}$ on the whole $(0,1)$. By our construction $f_{n}$ is convex for all $n$, and $f_{n}\le f\le g$, $\lim_{x\rightarrow 0}f_{n}(x)=\infty$ as desired.

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    $\begingroup$ Sorry, this solution does not look complete. It's not clear how $f_n$ are patched together to form a single convex function. "Convex bump between neighboring intervals" is much too vague. Note that you were not asked to "approximate" $g$, just to have growth near zero. I would let $h=\sup h_n$ where each $h_n$ is convex and piecewise linear of the form $h_n(x)=\max(M_n-k_n x,0)$. Then adjust $M_n,k_n$ so that $h_n\le f$ (one could work directly with $g$, too) and $M_n\to \infty$. $\endgroup$ – user53153 Dec 25 '12 at 19:40
  • $\begingroup$ I think we can define the bump function step by step when we glue two linear functions, and then induct on to any of them. I will read your comment carefully. $\endgroup$ – Bombyx mori Dec 26 '12 at 7:48

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