# Complex-diff functions equal at the boundary are identical

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.

• Use the analytic continuation principle – JoseSquare Feb 26 '19 at 20:01

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 $$\overline{\Delta}$$.

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

$$f = g. \tag 5$$ .