Proving or Disproving the Existence of a Function on the Unit Disc 
Let $f$ be holomorphic in the unit disc and continuous on its closure. Prove or disprove that there exists such $f$ so that $f(e^{i\theta})=e^{-i\theta}$ for $0<\theta<\frac{\pi}{4}$.

I believe that there is no such function since on the boundary of the disc it acts as the map $z\mapsto \bar{z}$, which is not holomorphic. But I don't see how to prove this rigorously, nor do I see what known theorems could be applied. I am familiar with analytic continuation, but I am not sure if that applies since $f$ need not be holomorphic on the boundary. Any hints or insights would be greatly appreciated.
 A: Consider the function $g(z)=f(z)$ if $|z|\leq1$, $g(z)=1/z$ if $z=r e^{i\theta}$, $0<\theta<\pi/4$, $r\geq 1$ (say). It is continuous and holomorphic except perhaps on the arc $e^{i\theta}$, $0<\theta<\pi/4$. By Morera theorem it is then holomorphic everywhere (where defined), and so it must coincide with $1/z$, which is however not holomorphic at $z=0$. So such $f$ doesn't exist.
A: Such holomorphic function could not exist. Before delving into the proof, I think it is important to see the underlying key idea (lemma, if you will)

Lemma: If $f$ is a holomorphic function on the unit disk and continuous on its boundary ($S^1$); furthermore $f$ is identically zero on some arc (say, the first quarter of the upper-half arc) of $S^1$, then $f$ is also identically zero on the closure of the unit disk.

Knowing this, the problem is more or less easy now. Indeed, define $g(z):=zf(z)-1$. Then certainly, $g$ enjoys all the properties of $f$, i.e., $g$ is a holomorphic function on the unit disk and continuous on $S^1$. Furthermore for every $z=e^{i\theta},$ whenever $0<\theta<\pi/4$, $g(e^{i\theta})=e^{i\theta}e^{-i\theta}-1=0.$ In other words, $g$ is identically zero on the first quarter of the upper-half arc of $S^1$. So the above lemma tells us that $g\equiv 0$. Therefore, $zf(z)=1,$ for all $z\in \bar D(0,1)$. Now there are many things that will go wrong here if this is true. For example here is one way to see this: 
Note that $f$ can't have any zero on $D(0,1)$; or else, say, $z_0$ is a zero, then $0=z_0f(z_0)=1$ (!). Thus $z\mapsto 1/f(z)$ is a well-defined holomorphic map on the unit disk that is also continuous on the boundary. The Maximum modulus principle tells us that $z \mapsto 1/f(z)$ achieves its maximum somewhere on the boundary. But since $zf(z)=1$, $|1/f(z)|=1$ on the boundary. Therefore, $|1/f(z)|\leq 1$ on $D(0,1)$. Or $|f(z)|\geq 1$ on $D(0,1)$. There is nothing to prevent us from repeating the same argument to find that $|f(z)|\leq 1$ on $D(0,1)$. So $|f(z)|\equiv 1$ on $D(0,1)$. 
Let $z_0=re^{i\theta}$ be a point inside the disk. Then $r=|z_0|=|z_0f(z_0)|=1$. But this is clearly non-sense because $r<1$. So we are done.
Now back to the proof of the lemma, to see this is true, consider $$h(z)=f(z)f(e^{i\pi/4}z)f(e^{i2\pi/4})f(e^{i3\pi/4})f(e^{i4\pi/4})f(-z)f(-e^{i\pi/4}z)f(-e^{i2\pi/4})f(-e^{i3\pi/4})f(-e^{i4\pi/4}).$$ Note that $h$ now is a holomorphic function on the unit disk, continuous on the boundary, and also identically zero on the boundary. So $h\equiv 0$ on the the whole disk. Then the uniqueness of analytic continuation (or principle of isolated zero, either one is fine) tells us that $f$ must also be identically zero on the disk (and its closure too!).
Hope this helps!
