Can someone please check my solution to this qualifying exam problem? Thanks!!
For each of the following, either construct a holomorphic function $f$ in the unit disk $D=\{z\in\mathbb{C}|\,|z|<1\}$ with the stated properties, or show that no such function exists.
1.For each sequence $(a_n)\subset D$ such that $\lim_{n\rightarrow\infty}|a_n|=1$, it follows that $\lim_{n\rightarrow\infty}|f(a_n)|+\infty$.
- $|f'(0)|=2$, $|f(z)|\leq 1$ for all $z$ such that $|z|=\frac{1}{2}$, and $\Big|f\Big(\frac{3}{4}\Big)\Big|=\frac{5}{3}.$
3.$|f(z)|\leq 1$ for all $z\in D$, $f\Big(1-\frac{1}{n^2}\Big)=0$ for all $n\in\mathbb{Z}^+$, and $f$ is not identically zero.
My solution:
1.If such $f$ exists, first we claim the set of zeros of $D$ is finite. For otherwise, let $(z_n)$ be an infinite sequence consisting of distinct zeros of $f$ in $D$ such that $(z_n)$ converges to $z\in\mathbb{C}$. If $z\in D$, then $f(z)=0$, which shows that $f$ is identically zero. If $z\notin D$, then $|z_n|\rightarrow 1$ but $f(z_n)\rightarrow 0$. Therefore, the set of zeros of $f$ is finite. Let $\xi_1,...,\xi_n$ be the zeros of $f$ repeated according to multiplicities, let $$g=\prod_{i=1}^n\frac{\xi_n-z}{1-\overline{\xi_n}z},$$ then $\frac{g}{f}$ is nonzero and holomorphic on $D$ and $\Big(\frac{g}{f}\Big)(a_n)\rightarrow 0$ if $(a_n)\subset D$ is a sequence such that $|a_n|\rightarrow 1$. Let $h:\overline{D}\rightarrow\mathbb{C}$ be defined by $$h(z)=\frac{g}{f}(z)$$ if $|z|<1$ and $$h(z)=0$$ if $|z|=1$, then $h$ is holomorphic on $D$ and continuous on $\overline{D}$. However $h=0$ on $\partial D$, so $h=0$ on $\overline{D}$ by the maximum modulus principle. Contradiction!
2.If such $f$ exists, define $g:D(0,2)\rightarrow\mathbb{C}$ by $g(z)=f\Big(\frac{z}{2}\Big)$, then $|g'(0)|=1$, $|g(z)|\leq 1$ for all $z$ such that $|z|=1$ and $\Big|g\Big(\frac{3}{2}\Big)\Big|=\frac{5}{3}$. Cauchy integral formula says $$|g'(0)|=\bigg|\frac{1}{2\pi i}\int_{\partial D(0,1)}\frac{g(z)}{z}\,dz\bigg|,$$ which implies that $|g(z)|=1$ if $|z|=1$. Furthermore, same as before, $g$ only has finitely many zeros in $D(0,1)$. Let $(z_1,...,z_n)$ be the zeros of $g$ in $D(0,1)$, repeated according to multiplicities. Let $$h=\prod_{i=1}^n\frac{z_i-z}{1-\overline{z_i}z},$$ if $n>0$, and $h=1$ if $n=0$; then $\Big|\frac{g}{h}(z)\Big|=1$ and $\Big|\frac{h}{g}(z)\Big|=1$ if $|z|=1$. Apply the maximum modulus principle to $\frac{f}{g}$ and $\frac{g}{h}$, we see that there exists a $c\in\mathbb{C}$ with $|c|=1$, and $g=ch$. This shows that $n>0$. Since $g$ has an analytic continuation to $D(0,2)$, we have $|z_j|\leq\frac{1}{2}$ for each $j$. Furthermore, we can compute that $$1=|g'(0)|=\bigg|\sum_{j=1}^n (1-|z_j|^2)z_1...z_{j-1}z_{j+1}...z_n \bigg|\leq n \frac{1}{2^{n-1}}.$$ However, the equality cannot be reached if $n>1$, so $n=1$. This shows that $1=1-|z_1|^2$, hence $z_1=0$. Therefore $g=cz$ for some constant $c\in\mathbb{C}$ with $|c|=1$. Therefore, it is impossible that $\Big|f\Big(\frac{3}{4}\Big)\Big|=\frac{5}{3}.$
3.Let $$f(z)=\prod_{n=1}^\infty \frac{\bigg(1-\frac{1}{n^2}\bigg)-z}{1-\bigg(1-\frac{1}{n^2}\bigg)z}.$$ Since $\sum_{n=1}^\infty\frac{1}{n^2}<\infty$, the Blaschke product converges.
