We are given that $f$ is analytic in a bounded domain $D$ and continuous on the boundary $\partial D$. Also we can assume that $|f(z)| \leq M$ for $z \in \partial D$ for $z \in \mathbb{C}$. I'm unsure what it means to be continuous on the boundary $\partial D$, but here is what I have thought up so far: Since $f$ is analytic in the bounded domain $D$ we have that $f$ is continuously differentiable at some point $z \in D$. Since $f$ is continuously differentiable at $z$ we have that $f'$ is also continuous at $z$. As $f$ is continuous on the boundary $\partial D$ we have by the definition of $\epsilon - \delta$ continuity that for every $\epsilon &gt;0$ there exists a $\delta &gt;0$ such that $|z-z_0|&lt;\delta \implies |f(z)-f(z_0)|&lt;\epsilon$. We are also allowed to assume that $|f(z)| \leq M$ for $z \in \partial D$ for $z \in \mathbb{C}$. I'm not sure where to head with this problem, or if I have set up what was given correctly. I think since $|f(z)-f(z_0)|&lt;\epsilon$ we can rewrite this as $||f(z)|-|f(z_0)|| \leq \epsilon$ and so $|f(z)| \geq -\epsilon + |f(z_0)|$ where we then have that $M \geq -\epsilon +|f(z_0)|$. But, I don't think this leads in any particular direction.
123093
Suppose  that $V$ is the bounded open subset of the plane and $f\in C(\bar V)\cap H(V)$. Show that if $M\ge 0$ and $|f(z)|\le M$ for all $z\in\partial V$ then $|f(z)|\le M$ for all $z\in V$.  I was trying to apply maximum modulus principle or open mapping theorem but the open set $V$ is not given to be connected and since then i have no idea how to do it.  Any type of help wil be appreciated. Thanks in advance.
80328
Is function $f:\mathbb C-\{0\}\rightarrow\mathbb C$ prescribed by  $z\rightarrow \large{\frac{1}{z}}$ by definition discontinuous at $0$?  Personally I would say: "no". In my view a function can only be (dis)continuous at $z$ if $z$ belongs to its domain.  But I have heard other sounds, that made me curious.   This question was inspired by comments/answers on  question.
21389
Is the following statement true?  Suppose, $ f:D\to \mathbb C $ is an analytic function where $ D $ is the unit disc of radius $ 1 $ around $0 $. Suppose, $ f $ is analytic on the boundary of $ D $ as well. Then prove that, if $ f $ is one-one on the boundary of $ D $, then $ f $ is one-one on $ D $.  P.S. I think I got a solution. But I was wondering if this problem has a trivial solution.  Here is my solution, since nobody has posted a solution yet:-  Suppose, $ \gamma $ is the boundary of $ D $(assuming $ \gamma(t)=e^{2\pi it} $). Then $ f\circ \gamma $ is a simple continuously differentiable curve. So by Jordan curve theorem, $ f\circ \gamma $ divides $ \mathbb C $ into 2 path connected regions $ A_1, A_2 $ such that if $ a\in A_1 $, then $\int_{f\circ \gamma} \frac {1}{z-a} dz=2\pi i$ and if $ a\in A_2 $, then $\int_{f\circ \gamma} \frac {1}{z-a} dz=0$. So, $ f(z)=a $ for some $ a\in A_1,z\in D $ implies $ \int_{\gamma} \frac {f'(z)}{f(z)-a} dz=2\pi i $. So $ f(z)=a $ has only one solution in $ D $. Similarly if $ a\in A_2 $, then $\int_{\gamma} \frac {f'(z)}{f(z)-a}=0 $, so $ f(z)=a $ has no solution in $ D $. If $ f(z)=a $ for some $ z\in D, a\in f\circ \gamma $, then by open mapping theorem, $ f(z)=a' $ for some $ z\in D, a'\in A_2 $, which is not possible. So $ f $ is one-one in $ D $.
32767
Suppose that $f:\Bbb R\to \Bbb R$ is continuous and that $f(x)\to 0$ as $x\to \pm\infty$. Prove that $f$ is uniformly continuous.   Since $f(x)\to 0$ as $x\to \pm\infty$ we have $|f(x)|&lt;\epsilon $ whenever $x&lt;-K$ and $x&gt;K$ where $K&gt;0$.  Also $f$ is uniformly continuous on $[-K,K]$ as the domain is compact.  Hence $f$ is uniformly continuous in $(-\infty,-K)\cup [-K,K]\cup (K,\infty)$.  But how to conclude that $f$ is uniformly continuous on $\Bbb R$ from above as I may chose $x\in (-\infty,-K);y\in [-K,K]$ ,how to conclude that $|f(x)-f(y)|&lt;\epsilon $ from above for a given $\epsilon$?  Please suggest the edits required.
79979
A problem that I have been thinking: For what kind of functions $f:[a,b]\to \mathbb R$ is  $x\mapsto\sup_{t\in [a,x]}f(t)$ continuous? This is not true for bounded functions, take for example $f(x)=0$ in $[0,1]$ and $f(x)=1$ in $(1,2]$. It is however trivially true for continuous monotone functions. Assuming mere continuity for $f$ I can show right continuity of $\max_{t\in[a,x]}f(t)$:   For $\epsilon$ and some $\delta&gt;0$,  $$x\in [x_0-\delta,x_0+\delta]\implies f(x_0)-\epsilon&lt;f(x)&lt;f(x_0)+\epsilon$$ If $x\in [x_0,x_0+\delta]$ then $$\max_{t\in [a,x]}f(t)=\max\left\{\max_{t\in [a,x_0]}f(t),\max_{t\in [x_0,x]}f(t)\right\}$$ If $max_{t\in [a,x_0]}f(t)\ge \max_{t\in [x_0,x]}f(t)$ then $$\left|\max_{t\in [a,x]}f(t)-\max_{t\in [a,x_0]}f(t)\right|=0&lt; \epsilon$$ Otherwise, $$\left|\max_{t\in [a,x]}f(t)-\max_{t\in [a,x_0]}f(t)\right|\le \max_{t\in [x_0,x]}f(t)-\max_{t\in [a,x_0]}f(t)&lt;f(x_0)+\epsilon-\max_{t\in [a,x_0]}f(t)\le \epsilon$$ which proves right continuity. When I tried to prove left continuity I was stuck on showing $$\max_{t\in [x,x_0]}f(t)-\max_{t\in [a,x]}f(t)&lt;\epsilon$$ Can this be circumvented?
84752
I'm doing an exercise that asks me to prove that $f$ is continuous using a $\epsilon$-$\delta$ proof. I have that $$   f(x) = \begin{cases} x\cdot \sin \frac1x,&amp;x\neq 0 \\ 0,&amp;x = 0 \end{cases} $$ I've already managed to show this property for $x=0$. How can I show it for $x \ne 0$, also using a $\epsilon$-$\delta$ proof?  Thank you very much.
27340
Suppose that $f$ is analytic on a domain $D$, which contains a simple closed curve $\gamma$ and the inside of $\gamma$. If $|f|$ is constant on $\gamma$, then I want to prove that either $f$ is constant or $f$ has a zero inside $\gamma$  Here is my take: if $f$ is constant, i dont see a reason why $|f|$ wouldnt be constant :)  if $f$ is not constant, then the max/min modulus principle applies ... meaning $|f|$ can not have any local max/min on D now i am lost at this point ...
15812
Prove that if $f(z)$ is analytic and $f'(z)$ is continuous on a closed curve $\gamma$, then $\int_\gamma\overline{f(z)}f'(z)dz$ is purely imaginary.  I'm not so sure where to start. Maybe parametrize $z$ by $z(t)$, so that the integral becomes $\int_a^b\overline{f(z(t))}f'(z(t))z'(t)dt$. Why will this be purely imaginary?
98381
Let $f$ be analytic in open unit disk, we need to show there exist $\{z_n\}$ with $|z_n|&lt;1$ and $|z_n|\rightarrow 1$ then $f(z_n)$ is bounded.  could any one give me Hints for this one?
27594
If $U$ is a bounded domain in $\mathbb R^n$ whose boundary is smooth, and $f$ is a smooth function on $U$ whose partial derivatives of all orders have a continuous extensions to $\bar U$. For an arbitrary domain $V \supseteq \bar U$, is there a smooth function $\tilde f$ on $V$ extending $f$?
29274
Let $f: \mathbb{C} \to  \mathbb{C}$ be a continuous function such that $f^2$ and $f^3$ are both analytic. Prove that $f$ is also analytic.   Some ideas: At $z_0$ where $f^2$ is not $0$ , then $f^3$ and $f^2$ are analytic so $f = \frac{f^3}{f^2}$ is analytic at $z_0$ but at $z_0$ where $f^2$ is $0$, I'm not able to show that $f$ is analytic.
94221
I'm reading through a development of the maximum modulous principle, but I am stuck verifying a remark:  $$\text{"it is enough to show that $|f|$ is constant, from which we may conclude that $f$ is.''}$$  So I am trying to prove it as a lemma:  Let $f$ be analytic on a domain $D$.  If $|f|$ is constant then so is $f$.  I tried using the fact that $|f|$ constant implies $|f|$ is analytic.  From here this means that $Re(|f(x,y)|)$ is harmonic.  I wrote out the consequence to this using Laplace's Equation, hoping to force the partial derivatives of $f$ to vanish, but it didn't seem to go anywhere.  Any suggestions?
86893
Let $f: \mathbb C \to \mathbb C$ be an analytic function, such that for all $z\in\mathbb C$, $$| \operatorname{Re}(f(z))\operatorname{Im}(f(z)) | \le 1.$$   I have to show that $f$ is constant.   I don't know how to apply Liouville's theorem to the question, but I can tell it is bounded, thus the theorem should apply. Any help would be appreciated.
40237
Suppose $f(x)$ is continuous on the closed interval $[a,b]$. Define $m(x)=\max_{a\leq s\leq x}\, f(s)$, $a\leq x\leq b$. Is $m(x)$ continuous necessarily? Thank you.
83544
If $f(z)$ is continuous in the neighborhood of the point $z=a$, then $$\lim_{r\to 0^+} \int_{\vert z-a\vert =r} \frac{f(z) dz}{z-a}=2\pi i\,f(a)\tag{1}$$  In my desperate intents of proving the above exercise, I tried to prove a 'particular' case of it, which is when $z=0$.   So if $z=0$ and as $f$ is continuous on a neighborhood of zero, given $\epsilon&gt;0$, there exists $\delta&gt;0$ such that $|f(re^{i \theta})-f(0)|\le \epsilon$ for $r\in[0,\delta)$ and $\theta\in\mathbb R$. Hence,  $\left\lvert\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta\right| &lt;2\pi\epsilon \Rightarrow \limsup_{r\to0}\left\lvert\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta\right| \le2\pi\epsilon.$  If $\epsilon\to0$, then $\limsup_{r\to0}\left\lvert\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta\right|=0 \Rightarrow \lim_{r\to0}\int_{0}^{2 \pi}(f(re^{i \theta})-f(0))d\theta=0.$  Is this proof correct?  How can I make the proof for $(1)$ following the 'particular proof'? Can someone help me?
68605
Assume $f$ has a domain of $[a,b]$. Is it possible that $f$ is differentiable on the closed interval $[a,b]$, or must the maximal domain for $f'$ be $(a,b)$?
103954
If $f$ Continuous bounded on $\mathbb{R}$，and $$\lim_{h\to0}\sup_{x\in\mathbb{R}}\left|f(x+h)-2f(x)+f(x-h)\right|=0.$$ prove that $f$ uniformly continuous on $\mathbb{R}$. I try to find a $\delta&gt;0$ to make it arbitrary $x_1,x_2\in \mathbb{R}$and$\left|x_1-x_2\right|&lt;\delta$have$$\left|f(x_1)-f(x_2)\right|&lt;\varepsilon.$$but I don't know how to use this complex limit expression.I hope you can give me some tips, thank you!