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How can I prove that for any $x$ and $y$ in $\mathbb{R}^n$, $$ \left|\frac{2}{\pi}\arctan(|x|)\frac{x}{|x|}- \frac{2}{\pi}\arctan(|y|)\frac{y}{|y|}\right| < M|x-y|$$ for some $M > 0$.

I tried it using the fact that derivative of $\arctan$ function is always less than equal to $1$.

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It looks like an application of the mean value theorem. Why don't you try differentiating $x \mapsto \frac{x}{|x|}\arctan |x|$? – Siminore Jul 15 '12 at 17:16
Maybe it is worth noting that $\frac{x}{|x|} \arctan|x|=\arctan x$ for $x\in\mathbb{R}$. – Fabian Jul 15 '12 at 17:38
@Fabian except $x = 0$ ;) – qoqosz Jul 16 '12 at 8:20
up vote 2 down vote accepted

Denote $$ f(x)=\frac{2}{\pi}\frac{\arctan|x|}{|x|} x $$ then by mean value theorem for vector valued functions we have $$ |f(x)-f(y)|\leq \left(\sup\limits_{z\in[x,y]}\Vert(Df)(z)\Vert\right)|x-y| $$ where $Df$ is a Jacobi matrix of $f$. It is remains to show that this supremum is always finite and independent of choice of $x,y\in\mathbb{R}^n$.

Note that $$ (Df)(z)_{ij}=\frac{\partial}{\partial z_j}\left(\frac{2}{\pi}\frac{\arctan|z|}{|z|} z_i\right)= \left(\frac{1}{1+|z|^2}-\frac{\arctan|z|}{|z|}\right)\frac{z_i z_j}{|z|^2}+\frac{\arctan|z|}{|z|}\delta_{i,j} $$ This value is bounded by some constant $C_{i,j}$ independent of $z$ because functions $$ \frac{1}{1+|z|^2}\qquad\frac{\arctan|z|}{|z|}\qquad\frac{z_i z_j}{|z|^2} $$ are bounded. Thus all elements of matrix $(Df)(z)$ are bounded by some constant $C=\max C_{i,j}$, and this constant is independent of vector $z$. Hence operator norm $\Vert (Df)(z)\Vert$ of matrix $(Df)(z)$ is also bounded by another constant $M$ independent of vector $z$. Finally, $$ |f(x)-f(y)|\leq \left(\sup\limits_{z\in[x,y]}\Vert(Df)(z)\Vert\right)|x-y|\leq M|x-y| $$

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Thank You , Actually I did not know the mean value theorem for vector valued functions. Can you suggest me some reference for that. – shane Jul 15 '12 at 18:18
@VarunJindal, I've added the reference – Norbert Jul 15 '12 at 18:23

We have $\frac{x}{|x|} \arctan |x| = \arctan x$ for $x\in\mathbb{R}$. Furthermore, $\arctan' x = \frac{1}{1+x^2} \leq 1$.

Starting from $0 < \frac{2}{\pi} \frac{1}{1+x^2} \leq \frac{2}{\pi}$, you get your inequality by integrating from $x$ to $y$.

(and in fact $M= \frac{2}{\pi}$)

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Could you elaborate more you argument. I don't get it, because I don't understand how do you integrate from one point in $\mathbb{R}^n$ to another point in $\mathbb{R}^n$. – Norbert Jul 15 '12 at 18:26

The following argument does not use multivariate calculus:

For $r\geq0$ put $r':=\arctan r$. Looking at the graph of $\arctan$ one immediately sees that $r'\leq r$ and $|r'-s'|\leq|r-s|$.

It suffices to consider the two-dimensional situation. Given the two points $$x=(r,0)\ ,\quad y=s(\cos\theta,\sin\theta)$$ with $r\geq0$, $s\geq0$ we consider the "scaled" points $$x':=(r',0)\ ,\qquad y':=s'(\cos\theta,\sin\theta)\ .$$ In terms of $x'$, $y'$ your "new distance" $\delta(x,y)$ satisfies $$\eqalign{\Bigl({\pi\over2}\delta(x,y)\Bigr)^2=|x'-y'|^2&=r'^2+s'^2-2r's'\cos\theta\cr &=(r'-s')^2+2r's'(1-\cos\theta)\cr &\leq(r-s)^2+2rs(1-\cos\theta)\cr &=|x-y|^2\ .\cr}$$ This proves $$\delta(x,y)\leq{2\over\pi}|x-y|\ .$$ Letting $x=0$, $y\to0$ one easily verifies that this inequality is best possible.

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