2

Problem from Bak,Newmann -Complex Analysis:

Show that if $f$ is a continuous real valued function and $|f|\leq 1$. Then $|\int _{|z|=1} f|\leq 4$

What I did is :taking $C:z(t)=\cos t+i\sin t$ as the parametrized curve.

$|\int _Cf(z)|\leq \int _C |f|<$Length of the curve $C$ since $|f|\leq 1 $ and Length of $C=1\implies |\int _Cf(z)|\leq 1$ ..

But I dont understand why the author has given the upper bound so large as $4$. Am I wrong or the author has been a bit casual in the question.

Please help.

Learnmore
  • 31,062
  • 1
    You have the wrong length of the curve, the length of a circle with radius $r$ is $2\pi r$, so with $r = 1$ the ML inequality gives you a bound of $2\pi > 4$. – Daniel Fischer Jan 06 '16 at 11:35
  • Can you confirm that the problem really is as you have stated. As indicated by @DanielFischer, the statement is clearly false. – mickep Jan 06 '16 at 12:21
  • @mickep I haven't indicated that the statement is false (it may be true, the bound of $2\pi$ from the ML inequality is not sharp), just that the OP's argument is incorrect, since they used the wrong length. – Daniel Fischer Jan 06 '16 at 12:24
  • @DanielFischer As I read it, $f(z)=1$ is a counterexample to the statment in the yellow box. (Sorry if I put wrong words in your mouth.) – mickep Jan 06 '16 at 12:25
  • @mickep The OP omitted the differential/measure with respect to which one integrates. I'm pretty sure that should be $dz$ and not $\lvert dz\rvert$ (since for the latter the assertion indeed is clearly wrong). And $\int_{\lvert z\rvert = 1} 1,dz = 0$. – Daniel Fischer Jan 06 '16 at 12:29
  • @DanielFischer This is more or less why I requested confirmation from OP about the formulation of the problem. I should have made it more clear. – mickep Jan 06 '16 at 12:31
  • yes thank you @DanielFischer; the length is $2\pi$;but one more question can you please state why the hypothesis $f$ a real valued function needed? – Learnmore Jan 06 '16 at 13:11
  • @mickep;I have quoted the exact problem it is given in Page 55 Problem 6 – Learnmore Jan 06 '16 at 13:12
  • @Amartya Because without that restriction, the absolute value of the integral can be $2\pi$, e.g. $\int_{\lvert z\rvert = 1} \overline{z},dz = 2\pi i$. I would however be surprised if the $dz$ was also omitted in the book, since it is essential. – Daniel Fischer Jan 06 '16 at 13:15

0 Answers0