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g (x) \overset{x \to \infty}{\to} \infty

Implies

g^{'} (x) \leq g^{1 + ε} (x)

posted 3y ago by TextKit‭ · edited 3y ago by TextKit‭

Answer

#3: Post edited by $user avatar$ TextKit‭ · 2022-05-14T23:09:16Z (almost 3 years ago)

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## My questions
1. Can you cite or scan the question from the source? I am leery, because the claim appears false.
2. 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 $h \leq 1$ everywhere, and $h = 0
~~$o u t s i d e t h e i n t e r v a l$ [-0.25,0.25] $. L e t$ 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 $H (n) = 2 n$ .
Let $g (x) = x + \int_{x}^{0} H (t) d t$ . Then certainly $g$ is $C^{1}$ , increasing, and tends to infinity.
I constructed $h \leq 1$ , and $0$ outside a finite interval. Thus $\int_{x}^{0} a h (a t) = \int_{x / a}^{0} h (u) d u \leq C$ for some constant C. Then $g (x) \leq x + C (x + 1)$ , since at most $x + 1$ of the rescaled $h's
~~eq 0$.~~
So $g^{1 + ϵ}$ grows at most polynomially. But we just constructed $g$ so that $g^{'} \geq H$ has an exponentially growing SUBsequence!

## My questions
1. Can you cite or scan the question from the source? I am leery, because the claim appears false.
2. 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 $h \leq 1$ everywhere, and $h = 0
$ outside the finite interval $[- 0.25, 0.25]$ . Let $H (x) = h (x) + 2 h (2 (x - 1)) + 4 h (4 (x - 2)) + 8 h ((x - 3) / 8) + . . .$
Then $H (x)$ converges to a continuous function, because the supports of each summand are disjoint. Observe that $H (n) = 2 n$ .
Let $g (x) = x + \int_{x}^{0} H (t) d t$ . Then certainly $g$ is $C^{1}$ , increasing, and tends to infinity.
I constructed $h \leq 1$ , and $h = 0$. Thus $\int_{x}^{0} a h (a t) = \int_{x / a}^{0} h (u) d u \leq C$ for some constant C. Then $g (x) \leq x + C (x + 1)$ , since at most $x + 1$ of the rescaled $h's
eq 0$.
So $g^{1 + ϵ}$ grows at most polynomially. But we just constructed $g$ so that $g^{'} \geq H$ has an exponentially growing SUBsequence!

#2: Post edited by $user avatar$ TextKit‭ · 2022-05-14T23:08:05Z (almost 3 years ago)

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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 $h \leq 1$ everywhere, and $h = 0
$o u t s i d e t h e i n t e r v a l$ [-0.25,0.25] $. L e t$ 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 $H (n) = 2 n$ .
Let $g (x) = x + \int_{x}^{0} H (t) d t$ . Then certainly $g$ is $C^{1}$ , increasing, and tends to infinity.
I constructed $h \leq 1$ , and $0$ outside a finite interval. Thus $\int_{x}^{0} a h (a t) = \int_{x / a}^{0} h (u) d u \leq C$ for some constant C. Then $g (x) \leq x + C (x + 1)$ , since at most $x + 1$ of the rescaled $h^{'} s \neq 0$ .
So $g^{1 + ϵ}$ grows at most polynomially. But we just constructed $g$ so that $g^{'} \geq H$ has an exponentially growing SUBsequence!

## My questions
1. Can you cite or scan the question from the source? I am leery, because the claim appears false.
2. 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 $h \leq 1$ everywhere, and $h = 0
$o u t s i d e t h e i n t e r v a l$ [-0.25,0.25] $. L e t$ 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 $H (n) = 2 n$ .
Let $g (x) = x + \int_{x}^{0} H (t) d t$ . Then certainly $g$ is $C^{1}$ , increasing, and tends to infinity.
I constructed $h \leq 1$ , and $0$ outside a finite interval. Thus $\int_{x}^{0} a h (a t) = \int_{x / a}^{0} h (u) d u \leq C$ for some constant C. Then $g (x) \leq x + C (x + 1)$ , since at most $x + 1$ of the rescaled $h^{'} s \neq 0$ .
So $g^{1 + ϵ}$ grows at most polynomially. But we just constructed $g$ so that $g^{'} \geq H$ has an exponentially growing SUBsequence!

#1: Initial revision by $user avatar$ TextKit‭ · 2022-05-14T23:07:32Z (almost 3 years ago)

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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!

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