I'm reading Conway's complex analysis book and on page 103/104, he said:
I didn't understand why he meant by "length of T" and why $|g(z)|\le \epsilon/l$ for any $z$ on $T_1$ and because of that $|\int_{T_1}g|\le \epsilon$.
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$\begingroup$ In this proof, it seems to mean the perimeter of triangle. (in any event, if you replace the word "length" by "perimeter" in the proof, the arguments continue to work). $\endgroup$– achille huiNov 7, 2016 at 6:51
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As explained at the bottom of p. 100 in the book, the length of a triangle is the length of its perimeter.
answered Nov 7, 2016 at 7:03
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$\begingroup$ Thank you very much, you're completely right! Do you know which theorem he used when he said $|\int_{T_1}g|\le \epsilon$? $\endgroup$ Nov 7, 2016 at 7:21
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$\begingroup$ That's 1.17(b). But you should also know it as the M-L inequality. $\endgroup$ Nov 7, 2016 at 7:39