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this is more of a broader question.

Say I have an analytic complex function $f$ that is defined on the open unit circle, but I know that the limit of $f$ when $|z|$ approaches 1 is 0. Can you define a function $g$ that satisfies:

for $|z| \lt 1$, $g(z) = f(z)$

for $|z| = 1$ $g(z) = 0$

And would that function remain analytic?

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  • $\begingroup$ Just a thought: If $g$ were harmonic you would get $g=0$. Do you think that you can construct non-zero $f$ such that $g$ is harmonic? $\endgroup$ – Yanko Jan 26 at 14:46
  • $\begingroup$ Well, the question actually came from a broader question which was "If $f$, which is defined on a square $(0,1)^2$, satisfies $|f| \le Re(z)$, prove $f = 0$ $\endgroup$ – Guy Schwartzberg Jan 26 at 15:03
  • $\begingroup$ Don't your hypotheses imply that $f$ is identically zero in the open disk? $\endgroup$ – Andreas Blass Jan 26 at 18:19
  • $\begingroup$ Well yes, that's what I need to prove, and that's how I am trying to prove it using the identity theorem $\endgroup$ – Guy Schwartzberg Jan 27 at 10:16

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