# Analytic contiunation

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?

• 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? – Yanko Jan 26 at 14:46
• 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 $f = 0$ – Guy Schwartzberg Jan 26 at 15:03
• Don't your hypotheses imply that $f$ is identically zero in the open disk? – Andreas Blass Jan 26 at 18:19
• Well yes, that's what I need to prove, and that's how I am trying to prove it using the identity theorem – Guy Schwartzberg Jan 27 at 10:16