I want to prove that there is no function $f$ that:
Is analytic in $\{ |z|<2 \}$.
$f(0) = 1, f(1) = f(-1) = 0$
$|f(z)| \le |z| +1$ for all $z$.
It seems logical to think about using the maximum modulus principle somehow. However, there are two problems. I don't know that on the boundary, for example on $|z| < 1$, the function doesn't get a maximum since I am given only two points. Moreover, I can't use $\frac {1}{f}$ since the inequality is on the wrong side. I don't even know that $\frac {1}{f}$ is analytic.
Actually, I don't see how the information that is given helps me here.
Help would be appreciated
$f(1) = f(-1) = 0$ implies that $$ g(z) = \frac{f(z)}{z^2-1} $$ has only removable singularities and can be extended to a holomorphic function in $\{ |z|<2 \}$. For $|z| \le r$ with $1 < r < 2$ we have $$ |g(z)| \le \max_{|w|=r} |g(w)| \le \frac{r+1}{r^2-1} $$ and for $r \to 2$ it follows that $$ |g(z)| \le \frac{2+1}{2^2-1} = 1 $$ for all $z$. Since $|g(0)| = |f(0)| = 1$, the maximum modulus principle implies that $g$ is constant, and consequently $$ f(z) = 1- z^2 $$ for all $z$.
But then the condition $|f(z)| \le |z| +1$ is violated, e.g. for $z = iy$ with $1 < y < 2$: $$ |f(iy) | = 1 + y^2 > 1 + y = |iy| +1 \, . $$