Ngô Quốc Anh

June 9, 2010

Invariance under fractional linear transformations

Filed under: Giải Tích 5, Giải Tích 6 (MA5205) — Tags: — Ngô Quốc Anh @ 5:02

I just read the following result due to Loss-Sloane published in J. Funct. Anal. this year [here]. This is just a lemma in their paper that I found very interesting.

Let $f$ be any function in $C_0^\infty(\mathbb R \setminus \{0\})$. Consider the inversion $x \mapsto \frac{1}{x}$ and set $\displaystyle g(x) = {\left| x \right|^{\alpha - 1}}f\left( {\frac{1}{x}} \right)$.

Then $g \in C_0^\infty(\mathbb R)$ and $\displaystyle\iint\limits_{{\mathbb{R}^2}} {\frac{{{{\left| {g(x) - g(y)} \right|}^2}}}{{{{\left| {x - y} \right|}^{\alpha + 1}}}}dxdy} = \iint\limits_{{\mathbb{R}^2}} {\frac{{{{\left| {f(x) - f(y)} \right|}^2}}}{{{{\left| {x - y} \right|}^{\alpha + 1}}}}dxdy}$.

Proof. For fixed $\varepsilon$ consider the regions $\displaystyle {R_1}: = \left\{ {(x,y) \in {\mathbb{R}^2}:\left| {\frac{x}{y}} \right| > 1 + \varepsilon } \right\}$

and likewise, $\displaystyle {R_2}: = \left\{ {(x,y) \in {\mathbb{R}^2}:\left| {\frac{y}{x}} \right| > 1 + \varepsilon } \right\}$.

By changing variables $x \mapsto \frac{1}{x}$ and $y \mapsto \frac{1}{y}$ we find that $\displaystyle\begin{gathered} \iint\limits_{{R_1} \cup {R_2}} {\frac{{{{\left| {f(x) - f(y)} \right|}^2}}}{{{{\left| {x - y} \right|}^{\alpha + 1}}}}dxdy} \hfill \\ \qquad= \iint\limits_{{R_1} \cup {R_2}} {\frac{{{{\left| {f\left( {\frac{1}{x}} \right) - f\left( {\frac{1}{y}} \right)} \right|}^2}}}{{{{\left| {x - y} \right|}^{\alpha + 1}}}}{{\left| x \right|}^{\alpha - 1}}{{\left| y \right|}^{\alpha - 1}}dxdy} \hfill \\ \qquad= \iint\limits_{{R_1} \cup {R_2}} {\frac{{{{\left| {g(x) - g(y)} \right|}^2}}}{{{{\left| {x - y} \right|}^{\alpha + 1}}}}dxdy} + \hfill \\\qquad\qquad \iint\limits_{{R_1} \cup {R_2}} {\frac{{{{\left| {f\left( {\frac{1}{x}} \right)} \right|}^2}\left( {{{\left| x \right|}^{\alpha - 1}}{{\left| y \right|}^{\alpha - 1}} - {{\left| x \right|}^{2(\alpha - 1)}}} \right) + {{\left| {f\left( {\frac{1}{y}} \right)} \right|}^2}\left( {{{\left| x \right|}^{\alpha - 1}}{{\left| y \right|}^{\alpha - 1}} - {{\left| y \right|}^{2(\alpha - 1)}}} \right)}}{{{{\left| {x - y} \right|}^{\alpha + 1}}}}dxdy} \hfill \\ \end{gathered}$

which, by symmetry under exchange of $x$ and $y$, $\displaystyle = \iint\limits_{{R_1} \cup {R_2}} {\frac{{{{\left| {g(x) - g(y)} \right|}^2}}}{{{{\left| {x - y} \right|}^{\alpha + 1}}}}dxdy} + 2\iint\limits_{{R_1} \cup {R_2}} {\frac{{{{\left| {f\left( {\frac{1}{x}} \right)} \right|}^2}\left( {{{\left| x \right|}^{\alpha - 1}}{{\left| y \right|}^{\alpha - 1}} - {{\left| x \right|}^{2(\alpha - 1)}}} \right)}}{{{{\left| {x - y} \right|}^{\alpha + 1}}}}dxdy}$.

The second integral can be written as $\displaystyle\int\limits_\mathbb{R} {{{\left| {f\left( {\frac{1}{x}} \right)} \right|}^2}{{\left| x \right|}^{\alpha - 2}}dx} \int\limits_{\left\{ {\left| s \right| > 1 + \varepsilon } \right\} \cup \left\{ {\frac{1}{{\left| s \right|}} > 1 + \varepsilon } \right\}} {\frac{{{{\left| s \right|}^{\alpha - 1}} - 1}}{{{{\left| {1 - s} \right|}^{1 + \alpha }}}}ds}$.

We have $\displaystyle\int\limits_{\left\{ {\left| s \right| > 1 + \varepsilon } \right\} \cup \left\{ {\frac{1}{{\left| s \right|}} > 1 + \varepsilon } \right\}} {\frac{{{{\left| s \right|}^{\alpha - 1}} - 1}}{{{{\left| {1 - s} \right|}^{1 + \alpha }}}}ds} = \int\limits_{\left\{ {\left| s \right| > 1 + \varepsilon } \right\}} {\frac{{{{\left| s \right|}^{\alpha - 1}} - 1}}{{{{\left| {1 - s} \right|}^{1 + \alpha }}}}ds} + \int\limits_{\left\{ {\frac{1}{{\left| s \right|}} > 1 + \varepsilon } \right\}} {\frac{{{{\left| s \right|}^{\alpha - 1}} - 1}}{{{{\left| {1 - s} \right|}^{1 + \alpha }}}}ds}$

and by changing the variable $s \mapsto \frac{1}{s}$ in the last integral we find that this sum vanishes. Letting $\varepsilon \to 0$ yields the desired equality.