Show that there does not exist any holomorphic function on the open unit disk and continuous on the closed unit disk with the given property. [duplicate]

Question

Let $\mathbb D : = \left \{z \in \mathbb C\ :\ \left \lvert z \right \rvert < 1 \right \}.$ Prove that there is no continuous function $f : \overline {\mathbb D} \longrightarrow \mathbb C$ such that $f$ is holomorphic on $\mathbb D$ and $f(z) = \overline z$ for all $z \in \partial \mathbb D.$

My Attempt $:$ By Cauchy Integral Formula we get $f^{(k)} (0) = 0$ for any $k \geq 0.$ But since $f(z) = \sum\limits_{k = 0}^{\infty} \frac {f^{(k)} (0)} {k!} z^k$ we have $f(z) \equiv 0$ on $\mathbb D$ and this cannot happen since $f$ is continuous on $\overline {\mathbb D}$ and $f(z) = \overline z$ for all $z \in \partial \mathbb D.$

Could anyone please have a look at my solution? Thanks in advance.

Answer 1

I don't know whether you can just apply Cauchy integral formula on the boundary. The version that I know does not allow it.

However, the function $g(z) := zf(z)$ for all $z \in \overline{\mathbb{D}}$ fulfils the following for all $z \in \partial\mathbb{D}$: $$ g(z) = z f(z) = z\bar z = \lvert z \rvert^2 = 1 $$ As a holomorphic function that is real on the boundary, it has to be constant. Therefore $zf(z) = 0%$ for all $z \in \mathbb{D}$ as $g(0)=0$. From this we conclude that $f=0$. This contradicts the boundary condition.

Answer 2

By Cauchy-Goursat, $\int_{B_r(0)} f(z) dz =0$ for every radius $r<1$. Convince yourself that by continuity, this entails $\int_{B_1(0)} f(z) dz =0$ as well. (This is the nontrivial part, you need that the smaller circles uniformly approach the big one, and actual epsilon-estimates. Cf. Cauchy's integral theorem and domain boundaries).

But it's very well known that $\int_{B_1(0)} \bar z dz = \int_{B_1(0)} z^{-1} dz =2\pi i\neq 0$.