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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.

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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. –  user20266 Apr 5 '12 at 12:42
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1 Answer

The norms of the differentials $(Df)_x, \;x\in \mathbb{R}^m$ need not be bounded, i. e. that supremum you write could be $\infty.$

We can try to do it following these steps:

(a) Let $f:\mathbb{R}^m\to \mathbb{R}$ be a differentiable function such that $\lim_{||x||\to \infty} Df_x(x)=0$. Then the function $g:\mathbb{R}^m\to \mathbb{R}$ defined by $g(x)=f(2x)-f(x)$ is bounded. Proof: By the Mean Value Theorem, for every $x\in \mathbb{R}^m$ there exists $r(x)\in [1,2]$ with $g(x)=f(2x)-f(x)=(Df)_{r(x)x}(x)$ . Now $(Df)_{r(x)x}(x)=\frac{1}{r(x)}(Df)_{r (x)x}(r(x)x),$ where $\frac{1}{r(x)}\le 1,$ and $\lim_{||x|| \to \infty}(Df)_{r (x)x}(r(x)x)=0.$ (For this last limit, note that $||x||\to \infty$ if and only if $||r(x)x||\to \infty$ because $1\le r(x)\le 2.$) Hence $g(x)=f(2x)-f(x)$ is a continuous function such that $\lim_{||x||\to \infty}g(x)=0.$ Such a function must be bounded. (Proof: Fix $\varepsilon=1;$ there exists $M>0$ such that $|g(x)|\le 1$ if $||x||>M.$ Hence $g$ is bounded outside the ball centered at 0 and with radius $M$. Inside the ball it is bounded too, because any continuous function is bounded on compact sets.)

(b) Let $f:\mathbb{R}^m\to \mathbb{R}^n$ be a differentiable function such that $\lim_{||x||\to \infty} Df_x(x)=0$. Then the function $g:\mathbb{R}^m\to \mathbb{R}^n$ defined by $g(x)=f(2x)-f(x)$ is bounded. Proof: Let $f=(f_1,f_2,\cdots,f_n).$ Note that for every $x\in \mathbb{R}^m,$ the components of $Df_x(x)$ are $\nabla (f_j)_x(x),$ $j=1,\cdots, n.$ Since $\lim_{||x||\to \infty} Df_x(x)=0,$ we can deduce $\lim_{||x||\to \infty}\nabla (f_j)_x(x)=0$ for every $j$. Hence each one of the functions $f_j$ satisfies the hypotheses in (a). We deduce that the functions $g_j:\mathbb{R}^m\to \mathbb{R}$ defined by $g_j(x)=f_j(2x)-f_j(x)$ are bounded. This implies that the function $g:\mathbb{R}^m\to \mathbb{R}^n$ given by $(g_1,g_2,\cdots,g_n)$ is bounded. But it is immediate that $g(x)=f(2x)-f(x).$

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I cannot see why the limit implies that $(Df)_x\cdot x$ is limited. Could you give me a hint? –  Marra Apr 5 '12 at 12:55
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Claro, Gustavo :) I was too brief and maybe not entirely right. I've edited my answer. –  Xabier Domínguez Apr 5 '12 at 13:30
    
Mustn't $f$ be a $C^1$ function to grant the existence of this $\theta$? –  Marra Apr 5 '12 at 14:16
    
I don't think so... en.wikipedia.org/wiki/… –  Xabier Domínguez Apr 5 '12 at 14:23
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@Xabier : the Mean Value Theorem (with equatlity) is false in higher dimension. Take for example $f : R \rightarrow C, x \mapsto exp(ix)$, and look at $x=0$ and $x'=2\pi$ (this example is given in the link you mention). –  user10676 Apr 5 '12 at 15:34
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