Ngô Quốc Anh

May 1, 2010

A useful identity in a book due to L. Ahlfors

Filed under: Các Bài Tập Nhỏ, Giải Tích 5, Nghiên Cứu Khoa Học — Tags: — Ngô Quốc Anh @ 3:12

Let $\mathbf{x},\mathbf{y}$ be points in $\mathbb R^n$. If we denote by $\mathbf{x}^\sharp$ the reflection point of $\mathbf{x}$ with respect to the unit ball, i.e. $\displaystyle \mathbf{x}^\sharp = \frac{\mathbf{x}}{|\mathbf{x}|^2}$

we then have the following well-known identity $\displaystyle |\mathbf{x}|\left| {{\mathbf{x}^\sharp } - \mathbf{y}} \right| = |\mathbf{y}|\left| {{\mathbf{y}^\sharp } - \mathbf{x}} \right|$.

The proof of the above identity comes from the fact that $\displaystyle |\mathbf{x}|\left| {\frac{\mathbf{x}}{{|\mathbf{x}{|^2}}} - \mathbf{y}} \right| = \sqrt {1 + |\mathbf{x}{|^2}|\mathbf{y}|^2 - 2\mathbf{x} \cdot \mathbf{y}} = |\mathbf{y}|\left| {\frac{\mathbf{y}}{{|\mathbf{y}|^2}} - \mathbf{x}} \right|$.

Indeed, by squaring both sides of $\displaystyle |\mathbf{x}|\left| {\frac{\mathbf{x}}{{|\mathbf{x}{|^2}}} - \mathbf{y}} \right| = \sqrt {1 + |\mathbf{x}|^2|\mathbf{y}|^2 - 2\mathbf{x} \cdot \mathbf{y}}$

we arrive at $\displaystyle |\mathbf{x}|^2\left( {\frac{{|\mathbf{x}|^2}}{{|\mathbf{x}|^4}} - 2\frac{{\mathbf{x} \cdot \mathbf{y}}}{{|\mathbf{x}|^2}} + |\mathbf{y}|^2} \right) = 1 + |\mathbf{x}|^2|\mathbf{y}|^2 - 2\mathbf{x} \cdot \mathbf{y}$

which is obviously true. Similarly, the last identity also holds. If we replace $\mathbf{y}$ by $-\mathbf{y}$ we also have $\displaystyle |\mathbf{x}|\left| {{\mathbf{x}^\sharp }+ \mathbf{y}} \right| = |\mathbf{y}|\left| {{\mathbf{y}^\sharp } + \mathbf{x}} \right|$.

Generally, if we consider the reflection point of $\mathbf{x}$ over a ball $B_r(0)$, i.e. $\displaystyle \mathbf{x}^\sharp = \frac{r^2\mathbf{x}}{|\mathbf{x}|^2}$

we still have the fact $\displaystyle |\mathbf{x}|\left| {{\mathbf{x}^\sharp } - \mathbf{y}} \right| = |\mathbf{y}|\left| {{\mathbf{y}^\sharp } - \mathbf{x}} \right|$.

Indeed, one gets $\displaystyle |\mathbf{x}|\left| {\frac{{{r^2}\mathbf{x}}}{{|\mathbf{x}{|^2}}} - \mathbf{y}} \right| = {r^2}|\mathbf{x}|\left| {\frac{\mathbf{x}}{{|\mathbf{x}{|^2}}} - \frac{\mathbf{y}}{{{r^2}}}} \right| = {r^2}\left| {\frac{\mathbf{y}}{{{r^2}}}} \right|\left| {\frac{{\frac{\mathbf{y}}{{{r^2}}}}}{{{{\left| {\frac{\mathbf{y}}{{{r^2}}}} \right|}^2}}} - \mathbf{x}} \right| = \left| \mathbf{y} \right|\left| {\frac{{{r^2}\mathbf{y}}}{{|\mathbf{y}{|^2}}} - \mathbf{x}} \right|$.

Similarly, $\displaystyle |\mathbf{x}|\left| {{\mathbf{x}^\sharp } + y} \right| = |y|\left| {{y^\sharp } + \mathbf{x}} \right|$.

Such identity is very useful. For example, in $\mathbb R^n$ ( $n\geqslant 3$) the following holds $\displaystyle\iint\limits_{{\mathbf{x}} = r} {\frac{{d{\sigma _{\mathbf{x}}}}}{{{{\left| {{\mathbf{x}} - {\mathbf{y}}} \right|}^{n - 2}}}}} = \min \left\{ {\frac{1}{{|{\mathbf{y}}|^{n - 2}}},\frac{1}{r^{n - 2}}} \right\}$.

This type of formula has been considered before when $n=3$ here. For a general case, Lieb and Loss introduced another method in their book published by AMS in 2001. Here we introduce a completely new proof. At first, if $|\mathbf{y}|>r$ by the potential theory, one easily gets $\displaystyle\iint\limits_{{\mathbf{x}} = r} {\frac{{d{\sigma _{\mathbf{x}}}}}{{{{\left| {{\mathbf{x}} - {\mathbf{y}}} \right|}^{n - 2}}}}} = \frac{1}{{|{\mathbf{y}}|^{n - 2}}}$.

If $|\mathbf{y}|, one needs to make use of the reflection point of $\mathbf{y}$ and the above identity to go back to the first case. The point here is $|\mathbf{y}^\sharp|>r$. The integral is obviously continuous as a function of $\mathbf{y}$. The above argument is due to professor X.X.W.

1. Dear Ngo,
Thanks so much for your useful post. But I don’t know how to apply the potential theory in order to obtain the formula in the case $|\mathbf{y}| >r$. Could you please explain it in more detail?
Thanks a lot! Best regards!

Comment by Blacknblanc — June 4, 2011 @ 23:59

• This is a long story. You may think in the following way: assuming $\Omega$ is bounded and $n \geqslant 3$, function $\displaystyle \Gamma(\mathbf{x}-\mathbf{y})=\frac{1}{n(2-n)\omega_n}\frac{1}{|\mathbf{x}-\mathbf{y}|^{n-2}}$

is actually the fundamental solution of Laplacian where $\omega_n$ is the volume of unit ball in $\mathbb R^n$. For an integrable function $f$, we can define the Newtonian potential of $f$ given by $\displaystyle w(\mathbf{x})=\int_\Omega \Gamma(\mathbf{x}-\mathbf{y})f(\mathbf{y})d\mathbf{y}.$

There is a theorem saying that $\displaystyle \Delta w(\mathbf{y})=f(\mathbf{y}).$

In other words, $\displaystyle f(\mathbf{y})=\int_\Omega \Gamma(\mathbf{x}-\mathbf{y})\Delta f(\mathbf{x})d\mathbf{x}.$

Furthermore, if $f$ is harmonic, then we immediately have $\displaystyle f(\mathbf{y})=\int_{\partial\Omega} \left( f(\mathbf{x})\frac{\partial \Gamma}{\partial \nu}(\mathbf{x}-\mathbf{y})-\Gamma(\mathbf{x}-\mathbf{y})\frac{\partial u}{\partial \nu}(\mathbf{x})\right)d\sigma_\mathbf{x}.$

Notice that function $|\mathbf{x}|^{2-n}$ itself is harmonic provided $|\mathbf{x}|>0$.

It would be great if you could post your proof here for everyone.

Comment by Ngô Quốc Anh — June 5, 2011 @ 0:12

2. Thanks for your detailed explanation. The rest of the proof is simple by applying the mean value theorem $\displaystyle \Gamma(y)=\frac{1}{n\omega_n r^{n-1}}\int_{\partial B_r}^{}\Gamma(x-y)d\sigma_x .$

And I think that you have forgotten the constant $n \omega_n r^{n-1}$ in the formula.

Comment by Blacknblanc — June 5, 2011 @ 5:48

• No, you cannot apply the mean value theorem here. In view of the mean value theorem, the ball $B_r$ is centered at $\mathbf{y}$ with radius $r$. In my original formula, point $\mathbf{y}$ is outside of the ball centered at $\mathbf{x}$ with radius $r$, as such, $|\mathbf{y}|^{2-n}$ is harmonic within this ball.

Similarly, when $\mathbf{y} you cannot use the potential theory since function $|\mathbf{y}|^{2-n}$ has a singularity within the ball. As such, you have to consider its reflection point to bring the point out of the ball.

My suggestion is that you have to calculate all partial derivatives and do estimate. Therefore, there is no extra coefficient as you said.

Comment by Ngô Quốc Anh — June 5, 2011 @ 5:56

• You wrote that “In my original formula, point $\mathbf{y}$ is outside of the ball centered at $\mathbf{x}$ with radius $r$ ” —> I think that point $\mathbf{y}$ is outside of the ball $B_r(0)$ centered at the origin $0$ with radius $r$.

Next, you misunderstood my idea. In fact I consider a point $\mathbf{y}$ outside of $B_r(0)$ and set $u_\mathbf{y}(\mathbf{x})=\Gamma(\mathbf{x}-\mathbf{y})$. Then $u_\mathbf{y}$ is harmonic in a domain $\Omega$ such that $B_r(0) \subset \Omega$ and $\mathbf{y} \notin \Omega$. Due to the mean value theorem, we obtain $\displaystyle u_\mathbf{y}(0)=\frac{1}{n\omega_n r^{n-1}}\int_{\partial B_r(0)}^{}u_\mathbf{y}(\mathbf{x})d\sigma_\mathbf{x},$

which is the desired formula in the case $|\mathbf{y}|>r$. So I still think that there must be the constant $n \omega_n r^{n-1}$ in the formula.

In the end,I have a (maybe naive) question: what happens if $|\mathbf{y}|=r$?

Comment by Blacknblanc — June 5, 2011 @ 15:48

• Oh yes, you was right. So integrals mean the average integrals. So let me rewrite the correct formula $\displaystyle\frac{1}{|B_r(0)|}\iint\limits_{{\mathbf{x}} = r} {\frac{{d{\sigma _{\mathbf{x}}}}}{{{{\left| {{\mathbf{x}} - {\mathbf{y}}} \right|}^{n - 2}}}}} = \min \left\{ {\frac{1}{{|{\mathbf{y}}|^{n - 2}}},\frac{1}{r^{n - 2}}} \right\}$

where $|B_r(0)|$ is the volume of the ball considering. For the case $|\mathbf{y}|=r$, you may prove the continuity of $u_\mathbf{y}(0)$ up to the boundary, I mean $y \to \partial B_r(0)$, to seal the case.

Comment by Ngô Quốc Anh — June 5, 2011 @ 17:43

3. Dear Ngo,
Could you please tell me in which page of the book of Lieb and Loss I can find another method to prove this indentity.
Thanks a lots.

Comment by blacknblanc — June 16, 2011 @ 17:24

• Page 249. Check it out.

Comment by Ngô Quốc Anh — June 16, 2011 @ 18:54

4. Blog đẹp thế QA.

Comment by Thuan — June 21, 2011 @ 10:23

• Dạ, viết lúc rảnh rỗi thôi thầy Thuận ạ.

Comment by Ngô Quốc Anh — June 21, 2011 @ 14:47

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