5

$f,g$ are continuous on closed unit disk but analytic on open unit disk and $f(z)=g(z)$ on $|z|=1$, we need to show $f\equiv g$

so $h(z)=f(z)-g(z)$ has zero set $S^1$ which is analytic on open unit disk $D$ and continous on compact unit disk so has maximum attained at boundary which is $0$ so $\max|h(z)|=0$ so $h(z)\equiv 0\Rightarrow f(z)\equiv g(z)$ on compact unit disk. am I right?

José Carlos Santos
  • 427,504
Myshkin
  • 35,974
  • 27
  • 154
  • 332
  • Unfortunately I did not understand how the set of zeroes is relevant to this proof. Could you please clarify? – sequence Mar 17 '17 at 00:25

2 Answers2

2

You are essentially correct, but I don't think you should say that $h(z)$ has the zero set $S^1$, since after all, your goal is to show $f\equiv g$. Instead, you might say that $S^1$ is a subset of the zero set, then by the maximum modulus principle, the maximum is attained on the boundary, so $h\equiv0$ on the closed disk.

Clayton
  • 24,751
  • Unfortunately I did not understand how the set of zeroes is relevant to this proof. Could you please clarify? – sequence Mar 17 '17 at 00:25
1

Let $\epsilon > 0$ By compactness and uniform continuity, you can choose $r$ with $0<r<1$ and so that $|f - g| < \epsilon$ on the circle of radius $r$ centered at zero. By the maximum modulus theorem, $|f - g| < \epsilon$ on the interior of this circle. This should close the ring for you.

ncmathsadist
  • 49,383