Assume that $f_n,f:D(0,1)\rightarrow C$ are holomorphic. and $f_n \rightarrow f$ uniformly on every compact set. we need to show that $f'_n \rightarrow f'$ uniformly on every compact set.  I kind of have no idea how to approach this one, its from a test I attended and I'm going for a second go at it and I want to understand the first one, first.
120844
Let $\Omega$ be an open subset in $\mathbb{C}$. Let $\{f_n\}$ be a sequence of holomorphic functions on $\Omega$ such that $f_n\to f$ pointwise and converges uniformly on any compact subset $K\subseteq \Omega$. Then by Cauchy Theorem and Morera Theorem, $f$ is holomorphic. Let $f_n$ and $f'$ be the derivatives of $f_n$ and $f$ respectively. Prove that $f_n'\to f'$ uniformly on any compact subset $K\subseteq \Omega$.   How to prove?
71420
Suppose $f$ is holomorphic in the unit disk $D=\{z: |z|&lt;1\}$ and $|f(z)|\leq |f(z^2)|$ on $D$. Show that $f$ is a constant function.   My Try:  I wanted to use Strong Maximum Principle, but failed to prove that $f$ is bounded on $D$. I want to try this problem in my own. Can anybody please give me just a hint?
120910
Let $f$ be a holomorphic function in $H={\{z\in\mathbb{C}}:\Re(z)&gt;0\}$ such that it is continuous in $cl(H)$, $|f(z)|\leq1$ $\forall z\in \delta H$ and $\lim_{|z|\to\infty}f(z)=0 $. Prove that $|f(z)|\leq1$ $\forall z\in H$. I thought about taking the bounded set $H\cap D(0,R), R&gt;0$. This way we could use the maximum modulus principle to prove it but I do not really know how to relate it with the limit since I would need $|f(z)|\leq1$ $\forall z\in H\cap D(0,R)$, and I do not have it on $D(0,R)$.
97532
Define $(p_n)_n$ recursively by $p_0(x) = 0$ and $p_{n+1}(x) = p_n(x) + \frac{1}{2}(x - p_n(x)^2)$.  Prove that $p_n(x) \rightarrow \sqrt{x}$ uniformly for $0 \leq x \leq 1$ as $n \rightarrow \infty$.  I'm not sure where to go with this.  I can show inductively that $\forall n,\  0 \leq p_n(x) \leq 1$ for $0 \leq x \leq 1$, and I think that $p_n(x)$ is increasing with $n$, but I have no idea whether this helps me.  I know that I need to show that $\forall \epsilon&gt;0: \exists N \in \mathbb{N}: \forall n \geq N: \forall x\in [0,1]: \big|p_n(x) - \sqrt{x}\big| &lt; \epsilon$ Edit:  I just had a thought.  I'm given a version of Dini's theorem that reads  Let $X$ be a compact metric space, $(f_n)_n$ a sequence of real-valued continuous functions on $X$, $f$ a continuous function on $X$ such that: (i) $f_n \rightarrow f$ pointwise on $X$, and (ii) $f_n(x) \geq f_{n+1}(x) $ for all $x \in X, n \in \mathbb{N}$. Then: $f_n \rightarrow f$ uniformly on $X$.  Obviously, (ii) is not true for my sequence, but what if I took $f_n(x) = -p_n(x)$ and $f(x) = - \sqrt{x}$?  Then I could prove (ii) to be true, I know that $[0,1]$ is compact, and $f_n(x)$ is real-valued.  The only part I would need to prove is (i), which is where I get confused.  Correct me if I'm wrong, but after that I could say that since $(-p_n)_n \rightarrow -\sqrt{x}$, then $(p_n)_n \rightarrow \sqrt{x}$.  Any advice?
50916
Let $f: D(0,1)\to \mathbb C$ be a holomorphic function. How to show that there exists a sequence $\{z_n\}$ in $D(0,1)$ such that $|z_n| \to 1$ and $\exists M&gt;0$ such that $|f(z_n)|&lt;M,\forall n \ge 1$ ?   My try: If not, then $\lim_{|z|\to 1} |f(z)|=\infty$. So in particular, $f$ has finitely many zeroes in $D(0,1)$. Also, $1/f$ is meromorphic in $D(0,1)$ with $\lim _{|z|\to 1}\dfrac {1}{|f(z)|}=0$. I am not sure where to go from here. Please help.
38173
Suppose function $f$ is holomorphic on $\mathbb C-\{0\}$ and satisfies $$|f(z)| \le \sqrt {|z|} + \frac{1}{\sqrt {|z|}}.$$ Prove that $f$ is a constant function. I think it's related to Laurent series representation and ML inequality but I have no idea how to prove it. Can anyone give me some hints? Thanks in advance!
62345
I came across this question whilst doing some research into complex analysis, and I just can't see what to do!  Let $f(z)$ be a holomorphic function on $\mathbb{C}$. Show that $\overline{f(\overline{z})}$ is holomorphic, whilst $f(\overline{z})$ is holomorphic if and only if $f(z)$ is constant.  I know that holomorphic means that the function is differentiable everywhere, and I need to apply the Cauchy-Riemann equations somehow, but I'm not sure how to approach this.
28987
I need to prove that if $f: (0,1) \rightarrow \mathbb{R}$ is Uniformly continuous then it is  bounded.   Thank you.
86990
In Stein's complex analysis, there is one theorem 5.3 of Chapter 2 which says that if holomorphic functions  $f_n$ converges uniformly to $f$ on any compact subset of an open set $\Omega$, then so does $f'_n$.   In his proof, he said "without loss of generality we can assume $f_n$ converges uniformly to $f$ on all of $\Omega$". So I think what he meant was that from the conditions of the theorem, we can get $f_n$ converges uniformly to $f$ on all of $\Omega$.  However, I don't know how to prove this assumption. Any help would be welcome.
43676
I am asked to prove this theorem:  If $f:U \rightarrow C$ is holomorphic in $U$ and invertible, $P\in U$ and if $D(P,r)$ is a sufficently small disc about P, then   $$f^{-1}(w) = \frac{1}{2\pi i} \oint_{\partial D(P,r)}{\frac{sf'(s)}{f(s)-w}}ds$$     The book says to "imitate the proof of the argument principle" but I am not seeing the connection.
116437
I know I can find an $f_n$ that converges to f uniformly, but can we use this to argue $f_n$ is continuous? Is there a theorem I can state?
96552
Does there exist a holomorphic function $f$ defined over $D = \{ z : |z| &lt; 1 \}$  such that $|f| \rightarrow \infty$ when $|z| \rightarrow 1$?  My approach:  If such an $f$ exists, then for a given $\varepsilon &gt; 0$, there exists an $M &gt; 0$ such that $|f(z)|&gt;M$ if $1-\varepsilon &lt; |z| &lt; 1$  Since $f$ has no zeros in that set, I can consider $g = 1/f$, I would like to apply Maximum Modulus Principle and get a contradiction, but actually I am not sure I can.
106718
Let $f$ be a continuous function on the closed unit disk and holomorphic inside it, such that $f(z) = 0$ for all $ z = \exp (it) $ with $0 ≤ t ≤ \pi/2 $. Prove that $ f(z) = 0 $ for all $ z \in \overline{D}(0,1) $.  I have worked this exercise for absurdity, but it has not reached any conclusion. I would like to know a more sophisticated test, perhaps by a direct method.
788
Let $f$ be holomorphic in a domain $U$, then $f$ has at most countably zeros.  How can I prove this statement? I have seen it in Sheng Gong  explicitly, but without any indications.
46926
$f$ is analytic function on open unit disk, and $|f(z)|\le 1-|z|\forall z\in D$, we need to show $f\equiv 0$, just a hint please.
44605
Let $f:[0,1]\rightarrow [0,1]$ be a homeomorphism. Show that , there exists a sequence of polynomials  $$(P_n(x))_n$$ such that $P_n(x)$ converge uniformly to $f$ on $[0,1]$ and every $P_n(x)$ is a homeomorphism from $[0,1]$ to itself.   I think that if we put additional condition that f be $C^1$ then $f'$ will be non-negative and continuous. Moreover, Weierstrass theorem says that there will be exist non-negative sequence of polynomials $(Q_n(x))_n$ such that it converges uniformly to $f'$ on $[0.1]$. Now it seems that $$P_n(x)=\int_{\mathbb{0}}^{x}Q_n(t)\:dt$$ works .  Is this true ? Is there other ways?   Thanks
105695
Let $\Omega$ be a bounded open subset of $\mathbb{C}$ and $f:\Omega\rightarrow\Omega$ be holomorphic in $\Omega$. Prove that if there exists a point $z_0$ in $\Omega$ such that $$f(z_0)=z_0~~~~\text{and}~~~~f'(z_0)=1$$ then $f$ is linear.  Please give some hints. Thanks in advance!