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I want to prove that if $f, g$ are continuous functions in $\bar{\Delta}$ (closure of the unit disc), are complex-diff in $\Delta$ and $f=g$ on the boundary $\delta\Delta$, then they are equal on all $\Delta$.

I know I can't apply Cauchy's Integral Formula, as it is only valid for curves inside the unit disc. Also, as they are holomorphic functions, the maximum is in the boundary, but I don't know how to combine both facts.

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  • $\begingroup$ Use the analytic continuation principle $\endgroup$ – JoseSquare Feb 26 '19 at 20:01
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You can apply maximum principle to the function $f-g$. If applied, we have $$ \max_{z\in\overline \Delta} |f(z)-g(z)|=\max_{z\in \partial \Delta}|f(z)-g(z)|=0, $$ that is $f=g$ on $\overline{\Delta}$.

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Note that $f - g$ is holomorphic in $\Delta$ and continuous on $\bar \Delta$ and that

$f - g = 0 \; \text{on} \; \partial \Delta; \tag 1$

thus

$\vert f - g \vert = 0 \; \text{on} \; \partial \Delta; \tag 2$

since the maximum modulus principle implies that

$\vert f - g \vert \ge 0 \tag 3$

attains its maximum on $\partial \Delta$, we have

$\vert f - g \vert = 0 \tag 4$

everywhere on $\bar \Delta$; thus

$f = g. \tag 5$ .

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