# Is this limit equal to 1?

Let $x\in R^{n}$ be fixed ($x\neq 0$) and $r>0$. Evaluate the limit:$$\lim_{r\rightarrow \infty}\frac{V(B(x,r)\cap B(0,r))}{V(B(0,r))}$$

where $V$ stands for volume and $B(x,r)$ is the ball with center x and radius $r$.

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Do you consider euclidian norm? –  Davide Giraudo Sep 27 '12 at 15:09

The volume of $B(x,r)$ is $ar^n$ for some constant $a$ which dependends on $n$, but whose exact value will not concern us.

Let $d=||x||$ be the distance between the two centers $0$ and $x$. If $r>d$ then the triangular inequality implies that $$B_{r-d}(0)\subset B(x,r)\cap B(0,r)$$ Hence $${V(B(x,r)\cap B(0,r))\over V(B(0,r))}>{V(B_{r-d}(0))\over V(B(0,r))}={a(r-d)^n \over ar^n}=\Bigl(1-{d\over r}\Bigr)^n$$ When $r\to\infty$ then the last expression will $\to 1$.

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Don't you need also an upper bound? –  enzotib Sep 27 '12 at 15:39
Yes, but the ratio between the volumes is obviously $<1$, since one set is contained in the other. –  Per Manne Sep 27 '12 at 15:48
You're right, thanks. –  enzotib Sep 27 '12 at 15:50
I really liked your solution Per Manne. Thanks. –  Tomás Sep 27 '12 at 16:00

Yes if we work with euclidian norm. Let $I(r):=V(B(x,r)\cap B(0,r))$. Then $$\left|\frac{I(r)}{|B(0,r)|}-1\right|=\frac 1{|B(0,r)|}\int_{\Bbb R^n}\chi_{\{||x-y|>r\}}\chi_{\{y\leq |r\}}dy.$$ In the set on which we integrate, we have $|x|²-2\langle x,y\rangle+|y|^2\geq r^2\geq |y|^2$ hence $2\langle x,y\rangle\leq |x|^2$. This is a bounded set, hence the integral is uniformly bounded by an universal constant (independent of $r$). By the way, it gives a speed of convergence.

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All Euclidean balls have the same volume, so we may as well use $V(B(\frac{x}{2},r))$ in the denominator. Now draw a picture to see that:

$$V\bigg(B\bigg(\frac{x}{2},r-\frac{|x|}{2}\bigg)\bigg)\leq V(B(x,r)\cap B(0,r))\leq V\bigg(B\bigg(\frac{x}{2},\sqrt{r^2 - \frac{|x|^2}{4}}\bigg)\bigg).$$

(enzotib points out below that it's easier to use a second radius of $r + |x|/2 > \sqrt{r^2 - |x|^2/4}$.)

Since the volume of a ball of radius $r$ is $Cr^n$, $|x|$ is fixed, and the radii of both the first and last balls are order $1$ in $r$, their volumes go as $Cr^n$. So when we divide through by $Cr^n$, we have that both the first and last expression go to $1$, forcing the middle to go to $1$ as desired.

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The including ball can also have, more simply, $R=r+|x|/2$? –  enzotib Sep 27 '12 at 15:42
Oh yes, that's simpler. I drew a picture of two intersecting circles, with centers $0$ and $x$. The midpoint between the centers is $x/2$, and the two radii come from the largest circle centered at $x/2$ contained in the intersection and the smallest circle centered at $x/2$ which contains the intersection. This latter has radius $\sqrt{r^2 - |x|^2/4}$. –  Neal Sep 27 '12 at 15:47
My estimate comes from thinking of the union instead of the intersection, my fault. –  enzotib Sep 27 '12 at 15:51
Not a problem! The union is bigger than the intersection and $r + |x|/2$ still grows with order $1$ in $r$. :) –  Neal Sep 27 '12 at 15:53
Yes. Use zooming $x\mapsto \displaystyle\frac1r\cdot x$. Then volume is shrinked by $\displaystyle\frac1{r^n}$, both in the enumenator and dominator, so the quotient is fix, $0$ stays, $r$ is constantly 1, and $x$ goes to $x/r \to 0$.