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Let $f:\mathbb R^m\to \mathbb R^n$ be a differentiable function such that $\lim_{|x|\to \infty}f'(x)\cdot x=0$. I would to know how to prove this function $g:\mathbb R^m\to \mathbb R^n$ defined by $g(x)=f(2x)-f(x)$ is bounded.

My only guess was using the definition of the derivative:

$$g(x)=f(2x)-f(x)=f'(x)\cdot x+r(x)$$

I took the norm, but I didn't make any progress:

$$|g(x)=|f(2x)-f(x)|\le |f'(x)\cdot x+r(x)|\le\ldots ?$$

I need help how to continue these reasoning. Am I going in the wrong direction?

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Set $$ g(t,x)=f(tx). $$ Then $$ \frac{\partial}{\partial t}g(t,x)=f'(tx)\cdot x, $$ and hence $$ f(2x)-f(x)=\int_1^2 \frac{\partial}{\partial t}g(t,x)\,dt $$ Thus $$ |f(2x)-f(x)|\le \int_1^2 |g_t(t,x)|\,dt=\int_1^2 |\,f'(tx)\cdot x|\,dt=\int_1^2 \frac{1}{t}|\,f'(tx)\cdot tx|\,dt\le \sup_{y\in\mathbb R^m} |\,f'(y)\cdot y|. $$ Note that as $\lim_{|x|\to\infty}f'(x)\cdot x=0$, then $\sup_{y\in\mathbb R^m} |\,f'(y)\cdot y|<\infty$.

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    How do you guarantee that $g_t(t,x)$ is integrable? – 2018-01-25
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Nothing guarantees that the derivative of the curve defined by @Yiorgos is integrable and this makes his solution questionable. I give an alternative solution below.

For each $x\in \Bbb R^m$, define the curve \begin{align*} \begin{array}{cccc} F_x:&\Bbb R&\to&\Bbb R^n\\ &t&\mapsto&|x|f(tx) \end{array}. \end{align*}

For each $x$ fixed, $F_x$ is differentiable, since it is the composition $t\mapsto tx\mapsto f(tx)\mapsto|x|f(tx)$. Furtheremore, $F_x'(t)=|x|f'(tx)\cdot x$, by the chain rule.

Since $\displaystyle \lim_{|x|\to \infty}f'(x)\cdot x=0$, there exists $M>0$ such that $|x|>M\implies |f'(x)\cdot x|<1$. Of course, $g(x)=f(2x)-f(x)$ is bounded on $B[0;M]$ (because it is continuous and the closed ball is compact).

It remains to show that $g$ is bounded on $\Bbb R^m-B[0;M]$. Let $x\in \Bbb R^m-B[0;M]$, i.e., $|x|>M$. For $t\in [1,2]$, we have $|tx|=|t||x|\geq |x|>M$ and, therefore, $|f'(tx)\cdot tx|<1$, for all $t\in [1,2]$. Then we have $|F_x'(t)|=||x|f'(tx)\cdot x|=\frac{1}{t}|x||f'(tx)\cdot tx|\leq |x|$, for all $t\in [1,2]$. By the Mean Value Inequality applied to $F_x$ over $[1,2]$, we get \begin{align*} |x||g(x)|&=|x||f(2x)-f(x)|=|F_x(2)-F_x(1)|\leq |x||2-1|=|x|,\end{align*} and this implies that $|g(x)|\leq 1$, if $|x|>M$.

This proves that $g$ is bounded on $\Bbb R^m$.