Holomorphic function on unit disk. Let $F\in C(\bar{D}(0,1))$ and holomorphic on $D(0,1)$. Suppose that $\mid F(z)\mid \leq 1$ when $\mid z\mid=1$. Prove that $\mid F(z)\mid \leq 1$ for $z \in \bar{D}(0,1)$. 
I need to show that the points inside the circle have a modulus less than or equal to 1. My assumption is that I am to manipulate absolute values on the Cauchy Integral formula. So far I have for any $\alpha \in D(0,1), integrating around unit circle:
$$F(\alpha)=\frac{1}{2\pi i} \int_{\gamma} \frac{F(\zeta)}{\zeta-\alpha}d\zeta$$
Clearly then: 
$$\mid F(\alpha)\mid =\mid \frac{1}{2\pi i} \int_{\gamma} \frac{F(\zeta)}{\zeta-\alpha}d\zeta \mid\leq\frac{1}{2\pi}\int_{\gamma}\frac{\mid F(\zeta)\mid}{\mid \zeta-\alpha \mid}d\zeta \leq \frac{1}{2\pi}\int_{\gamma}\frac{1}{\mid \zeta-\alpha \mid}d\zeta$$
Obviously there is some use of the fact that the length of what we are integrating over is $2\pi$.
 I can throw is that $\mid\zeta-\alpha \mid\geq\mid\zeta|-\mid\alpha\mid=1-\mid\alpha\mid$ on the concerned disk, but otherwise I am stuck.
Thanks in advance!
 A: That is a consequence of the “maximum modulus principle,“ and you can prove it using the Cauchy Integral Formula (compare https://proofwiki.org/wiki/Maximum_Modulus_Principle). The “trick” is to apply the CIF not to an arbitrary point, but to the point where $|F|$ attains its maximal value:
As a continuous function, $|F|$ attains it maximum on the closed disk $\bar D(0,1)$ at some point $\alpha$. Now assume that
$$
 |F(\alpha)| = \max \{ |f(z)| : z \in \bar D(0,1) \} > 1 \, .
$$
It is clear that $\alpha$ lies in the open disk $D(0, 1)$. If  $r > 0$ is sufficiently small then $D(\alpha,r) \subset D(0, 1)$ and we can apply the Cauchy Integral Formula to $\alpha$ and the circle $|z - \alpha| = r$.
Conclude that  $|F(z)| = |F(\alpha)|$ on the circle $|z - \alpha| = r$ and consequently, $|F(z)| = |F(\alpha)|$  for $|z-\alpha| \le r$, i.e. that $|F|$ is constant in a neighborhood of $\alpha$. 
Then use (for example) the Cauchy-Riemann equations to show that $F$ is constant in a neighborhood of $\alpha$.
Finally it follows from the identity principle that $F$ is constant in $D(0, 1)$ and because of the continuity, constant on $\bar D(0, 1)$. This is a contradiction because $|F(\alpha)| > 1$ but $|F(z)| \le 1$ on the boundary.
