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## My questions

Can you scan the question as you were assigned? I am leery, because the claim appears false.

Is g(x) supposed to be convex? 

## Game plan


I shall construct, on top of $f(x)=x$, a function which occasionally jumps up a constant amount over littler and littler intervals. This means I can create a function with the growth rate of $x$, but with as much derivative growth as I want.


## Counter Example

 Let $h(x)$ be any nonnegative continuous function with the property that  $h(0) = 1$ and $h \le 1$ everywhere, and $h = 0
$ outside the interval $[-0.25,0.25]$. Let $H(x)= h(x) + 2h(2(x-1)) + 4h(4(x-2)) + 8h((x-3)/8) + ...$ 

Then $H(x)$ converges to a continuous function, because the supports of each summand are disjoint. Observe that $H(n) = 2n$.

Let $g(x) = x + \int^0_x H(t) \, dt$. Then certainly $g$ is $C^1$, increasing, and tends to infinity. 

I constructed $h \le 1$, and $0$ outside a finite interval. Thus $ \int^0_x  ah(at) = \int^0_{x/a} h(u) du \le C$ for some constant C. Then $g(x) \le x + C(x+1)$, since at most $x+1$ of the rescaled $h's \neq 0$.

So $g^{1+\epsilon}$ grows at most polynomially. But we just constructed $g$ so that $g' \ge H$ has an exponentially growing SUBsequence!