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Fixed typo in isometry proof.
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Bokuan Li
2026-03-16 21:10:21 -04:00
parent b591904469
commit ae69a73fba
1 changed files with 1 additions and 1 deletions
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src/measure/radon/c0.tex
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@@ -62,7 +62,7 @@
so so
\[ \[
\abs{\int \phi d\mu} \le \sum_{j = 1}^n |\mu(A_j)| + 2n\eps \abs{\int \phi d\mu} \ge \sum_{j = 1}^n |\mu(A_j)| - 2n\eps
\] \]
As such a $\phi$ exists for all $\eps > 0$, $\norm{I_\mu}_{C_0(X; \complex)} \ge \sum_{j = 1}^n |\mu(A_j)|$. Since this holds for all such partitions, $\norm{I_\mu}_{C_0(X; \complex)} \ge |\mu|(X)$. Therefore the map $\mu \mapsto I_\mu$ is isometric. As such a $\phi$ exists for all $\eps > 0$, $\norm{I_\mu}_{C_0(X; \complex)} \ge \sum_{j = 1}^n |\mu(A_j)|$. Since this holds for all such partitions, $\norm{I_\mu}_{C_0(X; \complex)} \ge |\mu|(X)$. Therefore the map $\mu \mapsto I_\mu$ is isometric.