If $f'(x)\cdot x$ goes to zero then $f(2x)-f(x)$ is bounded.

by Pedro   Last Updated January 20, 2018 22:20 PM

Let $g:\mathbb R^m\to\mathbb R^n$ be defined by $g(x)=f(2x)-f(x)$ where $f:\mathbb{R}^m\to\mathbb{R}^n$ is a given differentiable function. The problem is to prove that if $\lim_{|x|\to\infty} f'(x)\cdot x=0$ then $g$ is bounded.

My idea is to show that $\lim_{|x|\to\infty}|g(x) |=0$. (If that limit is valid, then there exists $K_1>0$ such that $|x|>K_1\Rightarrow|g(x) |<1$. Furthermore, as $|g|$ is continuous and $B[0,K_1]$ is compact, there exists $K_2>0$ such that $|g(x) |\leq K_2$ for all $x\in B[0,K_1 ]$. So, $|g(x) |\leq \max\{1,K_2\}$ for all $x\in\mathbb R^m$).

For $m=1$, I've done the following.

Let $(x_n )\subset\mathbb R$ be a sequence such that $\lim_{n\to\infty }|x_n |=\infty$. By the Mean Value Theorem, for each $n\in\mathbb N$ there exists $c_n$ between $x_n$ e $2x_n$ such that

$$0\leq|g(x_n)|=|f(2x_n)-f(x_n)|\leq|f'(c_n)\cdot x_n|=\frac{|x_n|}{|c_n|}|f'(c_n)\cdot c_n|\leq 1\cdot|f'(c_n)\cdot c_n|.\tag{1}$$

Notice that $\lim_{n\to\infty}|c_n |=\infty$. So, taking the limit with $n\to\infty$, we get $$\lim_{n\to\infty}|g(x_n )|=0$$ which implies the desired result.

The above solution does not work to arbitrary $m$ because, for $m\neq 1$, $x_n$ and $c_n$ are not scalars and thus the equality in $(1)$ is not valid.

So, I need help to treat the general case.

Thanks.



Answers 2


Here's a suggestion which is similar in spirit to your strategy, but with the fundamental theorem of calculus replacing the mean value theorem.

Let $\vec{v}$ be a unit vector in $\mathbb{R}^m$, and let $t>0$ be a scalar. Then $g(t\vec{v})$ can be computed with the fundamental theorem of calculus as $$g(t\vec{v}) = f(2t\vec{v}) - f(t\vec{v}) = \int_t^{2t}f'(s\vec{v})\cdot \vec{v}\,\,ds = \int_t^{2t}\frac{1}{s}\,[f'(s\vec{v})\cdot(s\vec{v})]\,\,ds.$$ What can you conclude?

froggie
froggie
July 13, 2014 08:06 AM

Since $f'(x)\cdot x\to0$ as $\left\| x\right\|\to\infty$ then by choosing $\| x\|$ large enough (say for $\|x\|>R$) we can guarantee that $\|Df(x)\cdot x\|<1$. So for $x$ in this region we have that since $\mathbb{R}^{m}$ is convex we can apply mean value theorem and get:

$$\| g(x)\|=\| f(2x)-f(x)\|=\left\| \int_{0}^{1}(Df((1+t)x)\cdot x)\,dt\right\|$$

$$=\left\|\int_{0}^{1}\frac{1}{1+t}(Df((1+t)x)((1+t)x))\,dt\right\|\le\int_{0}^{1}\frac{1}{1+t}\,dt=\ln(2)$$

On the other hand $g$ is the difference of two differentiable functions hence on the complement of the previous region ($\| x\|\le R$) $g$ is bounded since this set is compact. A proof of the version of the mean value theorem used can be found here: http://en.wikipedia.org/wiki/Mean_value_theorem.

user71352
user71352
July 13, 2014 08:09 AM

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