From cb0b714c78258c8d2c014384c08898f9eea16d94 Mon Sep 17 00:00:00 2001 From: Floris van Doorn Date: Wed, 3 Apr 2024 17:53:51 +0200 Subject: [PATCH] a few missing braces --- blueprint/src/chapter/main.tex | 6 +++--- docs/_layouts/default.html | 2 +- 2 files changed, 4 insertions(+), 4 deletions(-) diff --git a/blueprint/src/chapter/main.tex b/blueprint/src/chapter/main.tex index bb8aac9f..9236ae71 100644 --- a/blueprint/src/chapter/main.tex +++ b/blueprint/src/chapter/main.tex @@ -4948,11 +4948,11 @@ \section{Separated Trees} \eqref{T*Hölder1} \le \frac{2^{151a^3}}{\mu(B(c(\fp), 4D^{s(\fp)}))} \left(\frac{\rho(y,y')}{D^{s(\fp)}}\right)^{\tau'} \int_{E(\fp)}|g(x)| \, \mathrm{d}\mu(x)\,. $$ - If $y,y' \notin B(c(\fp), 5D^{s(\fp)})$, then $T_\fp^*g(y) = T_\fp^*g(y') = 0$, by \eqref{eq Ks supp t} and the triangle inequality. Then \eqref{T*Hölder2} holds. + If $y,y' \notin B(c(\fp), 5D^{s(\fp)})$, then $T_{\fp}^*g(y) = T_{\fp}^*g(y') = 0$, by \eqref{eq Ks supp t} and the triangle inequality. Then \eqref{T*Hölder2} holds. Finally, if $y \in B(c(\fp), 5D^{s(\fp)})$ and $y' \notin B(c(\fp), 5D^{s(\fp)})$, then $$ - |e(Q(\fu)(y)) T_{\fp}^* g(y) - e(Q(\fu)(y')) T_{\fp}^* g(y')| = |T_\fp^* g(y)| + |e(Q(\fu)(y)) T_{\fp}^* g(y) - e(Q(\fu)(y')) T_{\fp}^* g(y')| = |T_{\fp}^* g(y)| $$ $$ \le \int_{E(\fp)} |K_{s(\fp)}(x,y)| |g(x)| \, \mathrm{d}\mu(x)\,. @@ -4981,7 +4981,7 @@ \section{Separated Trees} $$ Plugging this in and using $a \ge 4$, we get $$ - |T_\fp^* g(y)| \le \frac{2^{103a^3}}{\mu(B(c(\fp), 4D^{s(\fp)}))} \left(\frac{\rho(y,y')}{D^{s(\fp)}}\right)^{\tau'} \int_{E(\fp)} |g(x)| \, \mathrm{d}\mu(y)\,, + |T_{\fp}^* g(y)| \le \frac{2^{103a^3}}{\mu(B(c(\fp), 4D^{s(\fp)}))} \left(\frac{\rho(y,y')}{D^{s(\fp)}}\right)^{\tau'} \int_{E(\fp)} |g(x)| \, \mathrm{d}\mu(y)\,, $$ which completes the proof of the lemma. \end{proof} diff --git a/docs/_layouts/default.html b/docs/_layouts/default.html index 45a7f74c..467937da 100644 --- a/docs/_layouts/default.html +++ b/docs/_layouts/default.html @@ -45,7 +45,7 @@

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