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I'm having problems with this question:

Let $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$ be a differentiable function such that $\lim_{|x|\rightarrow\infty}(Df)_x\cdot x=0$. Then the function $g:\mathbb{R^m}\rightarrow\mathbb{R}^n$ defined by $g(x)=f(2x)-f(x)$ is limited.

I have to prove this result.

What I have tried so far: If that limit is so, it means that for every $\epsilon >0$ exists a $\delta >0$ such that, if $| x|>\delta $ then $|(Df)_x\cdot x|<\epsilon$. In particular, $M|x|\leq\epsilon$ where $M=\sup\{||(Df)_x||:x\in\mathbb{R}^m\}$.

$\textbf{I don't know if my last statement is true!}$

Assuming it is true, let's suppose that $g$ is not limited. That is, for every $\epsilon >0$ there exists a $x\in\mathbb{R}^m$ such that $|g(x)|>\epsilon$. By definition of $g$, we have:

$$|f(2x)-f(x)|>\epsilon$$

By the Mean Value Inequality, we have:

$$M|2x-x|=M|x|\geq |f(2x)-f(x)|>\epsilon$$

and this gives us $M|x|>\epsilon$ and $M|x|\leq \epsilon$, which is a contradiction.

Can someone tell me if this is correct? Thanks in advance.

  • 0
    The statement $M|x| \le \varepsilon $ is not true (or cannot be derived from what you know) I guess it would only be true if $f$ is constant. First of all, you can derive an inequality only for large $x$, secondly you only have knowlegde about $Df$ in radial directions.2012-04-05

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