Suppose $f$ is continuous and: $$ |f(e^{i \theta})| \leq M $$ Suppose: $$ |\int_{|z|=1}f(z)dz| = 2\pi M $$ I am trying to show that $f(z) = c \bar{z}$ for some constant $c$ with $|c|=M$
122647
The problem is:  Suppose that $f(z)$ is continuous on a domain $D$ that contains the unit circle, and that $f(z)$ satisfies: $$\vert f(e^{i\theta})\vert \leq M \; \forall \theta \, \in [0, 2\pi )$$ and $$\bigg \vert \int _{\vert z \vert=1} f(z) \text{d}z \bigg \vert =2\pi M $$ show that there exists a $c\in \mathbb{C}$ such that $f(z)=c \bar{z}$ for all $\vert z\vert=1$.  I've tried lots of different approaches such as defining a function $g(z)=\frac{f(z)}{\bar{z}}$ and trying to show that $g'(z)=0$ so that $g$ is constant on the circle, or trying to use some bounding theorems for analytic functions to show that $g$ is constant but this requires $g$ to be analytic. Some other methods were using the Cauchy-Riemann equations but I couldn't find any reason why $g$ should be analytic.  Any help would be greatly appreciated, thanks!
28787
Let $f (z) $ be an entire function such that $|f (z)|≤K|z|$, $∀z∈\mathbb{C}$, for some $K&gt;0$. If $f (1) =i$, the value of $f (i) $ is (A)  $ 1 $ (B)$-1$ (C) $i$ (D) $-i$      how can I able to solve this problem?totally stuck.
115100
Let $V$ be an open connected subset of $\mathbb{C}$.  Let $f:\rightarrow \mathbb{C}$ be a function continuous on a contour $C$.  Define $F(z)=\int_C \frac{f(\zeta)}{\zeta - z} d\zeta$ on $V$.  My question is:   How can i show that $F$ is analytic at every point except $C$? How can I show that  $F'(z)=\int_C \frac{f(\zeta)}{(\zeta - z)^2} d\zeta$ where $z$ not in $C$?
49561
Is it true that  $ f(\bar{z})=\overline {f(z)}$, Where z is complex?  I think it holds when $f(z)$ is holomorphic since we have $f(z)=p(x,y)+iq(x,y)=p(z,0)+iq(z,0)$ Any help...
115888
I need help with the following exercise:  Prove that if $f$ and $g$ are analytic functions on a region $G$ such that $\bar{f}g$ is analytic then $f$ is constant or $g\equiv0$  Thanks!
77706
Theorem 2.1 : Suppose that $f$ is an integrable function on the circle with $\hat f(n)=0$ for all $n \in \Bbb Z$. Then $f(\theta_0)=0$ whenever $f$ is continuous at the point $\theta_0$.      Proof : We suppose first that $f$ is real-valued, and argue by contradiction. Assume, without loss of generality, that $f$ is defined on $[-\pi,\pi]$, that $\theta_0=0$, and $f(0) \gt 0$.      Since $f$ is continuous at $0$, we can choose $ 0\lt \delta \le \frac \pi2$, so that $f(\theta) \gt \frac {f(0)}2$ whenever $|\theta| \lt \delta$. Let $$p(\theta)=\epsilon + \cos\theta,$$   Where $\epsilon \gt 0$ is chosen so small that $|p(\theta)| \lt 1 - \frac \epsilon2$, whenever $\delta \le |\theta| \le \pi$. Then, choose a positive $\eta$ with $\eta \lt \delta$, so that $p(\theta) \ge 1 + \frac \epsilon2$, for $|\theta| \lt \eta$. Finally, let $p_k(\theta)=|p(\theta)|^k$, and select $B$ so that $|f(\theta)| \le B$ for all $\theta$. This is possible since $f$ is integrable, hence bounded.      By construction, each $p_k$ is a trigonometric polynomial, and since $\hat f(n)=$ for all $n$, we must have $\int_{-\pi}^{\pi} f(\theta)p_k(\theta)\,d\theta=0$ for all $k$.   I understood the first paragraph clearly. But the rest is not making it's way into my head.   In the beginning of second paragraph, how does the given range works for choosing $\delta$? If the continuity is used to get the range, then how? How can we choose $\epsilon$ so small such that, $|p(\theta)| \lt 1 - \frac \epsilon2$, whenever $\delta \le |\theta| \le \pi$? How can we choose positive $\eta$ with $\eta \lt \delta$, so that $p(\theta) \ge 1+ \frac \epsilon2$, for $|\theta| \lt \eta$. Why do we must have $\int_{-\pi}^{\pi}f(\theta)p_k(\theta)\,d\theta=0$ for all $k$?
79513
Let $f$ be a continuous complex-valued function on the unit interval. For any complex number $z$, define $F(z)=\int _0 ^1 f(t) e^{zt} dt$.  How do I show that $F$ is entire?
92992
Let $S=\{f:\mathbb{R} \to  \mathbb{R}\}$ that satisfies: $\forall f\in S$, $f'$ exists and $f'$ is continuous and $f(0)=f(1)=0$.  Please prove that $\exists \delta :\forall f\in S$ s.t.  $\int_{0}^{1} (f(x))^2dx\leq \delta \int_{0}^{1} (f'(x))^2dx$. Thanks in advance.
16048
mathematicians! I want to ask to all wise people about a problem I met at the quiz to obtain some ideas. The problem is following. It may not be accurate since the problem is dependent on my memory   Let $f : [0, \infty) \rightarrow \mathbb{R}$ and $f$ is differentiable. $|f'(x) |\; \leq\  |f(x)|$ for all $x \in [0, \infty)$. $f(0) = 0$. Then prove that $f(x) = 0$ for all $x \in [0,\infty)$.  At the quiz, I applied the Mean Value Theorem and the Cauchy-Schwarz inequality.   For any $x$, $|f(x)| = |\{f(x) - f(0)\}(x-0)| = |f'(c_{1})(x)|$ for some $c_{1} \in (0,x)$, and then  $$|f'(c_{1})(x)| \leq |f'(c_{1})||x| \leq |f(c_{1})||x|$$ by Cauchy-Schwarz and given assumption. By doing so consecutively, we can obtain the inequality that  $$ |f(x)| \leq |f(c_{n})||x||c_{1}|\ldots |c_{n-1}| $$ for any $x \in [0,\infty)$. As $\{c_{i}\}, \; i\in \mathbb{N}$ is decreasing sequence and $f$ is continuous, $|f(c_{n})|$ can be arbitrary close to $0$ when $c_{n} \rightarrow 0$. Thus the inequality above can be bounded by $\epsilon$.  I think my answer is right, but a bit uncertain. Could you give me a certainty? Thank you very much.
109651
Was reading some notes and it states that $f(z) = \frac{1 - e^{iz}}{z^2}$ can be written as $f(x) = \frac{-iz}{z^2} + E(z)$ where $E(z)$ is bounded as $z \rightarrow 0.$ I don't exactly see why. Help is appreciated.
33619
Well, after spending hours on this problem, I'm still stuck, so I thought I'd turn to you guys. The problem statement is as follows.  Let $f$ be a complex-valued function that is $C^1$ in the disk $|z - z_0| &lt; R$ for some $R&gt;0$, and let $C_r$ be the circle $|z - z_0| = r$. Show that $$\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0).$$  I tried parametrizing the circle in the usual way and differentiating to see if I get something useful, but no luck. I just can't get the LHS to give me something to do with $\frac{\partial f}{\partial \bar{z}}$, so that I could then put the RHS under the integral and show that the difference between the two can be less than arbitrary $\epsilon &gt; 0$ by the continuity of partial derivatives.  Thanks in advance for your help!
113752
Let $f(x)$ be a strictly positive continuous function in $[0,1]$ such that $f^2(t) \leq 1 + 2\int_0^t f(s)\, \mathrm{d}s$. Prove that $f(t)\leq t+1$.    Obviously, if $f(x)$ can be differentiated, it is true. But I can't deal with the case that it is not differentiated.
80576
$f(z)$ is a function such that $|f(z)| &gt; |z|$ for all $z\in\mathbb{C}$  How would you prove the function $\frac{1}{f(z)}$ is bounded. I think you have to use the $|f(z)|\leq M$ but I'm not sure.
21692
This is the problem from the book Notes on Complex Function Theory, Sarason.      Let the complex valued function $f$ defined and $\mathcal{C}^1$ in the disk $|z-z_0|&lt;R.$ For $0&lt;r&lt;R$ let $C_r$ denote the circle $|z-z_0|=r$, with the counterclockwise orientation. Prove that $$ \lim_{r\to 0}\frac{1}{r^2}\int_{C_r}{f(z)}\ dz=2\pi i\ \frac{\partial f}{\partial \bar{z}}(z_0) .$$   I found the answer to my question  But I wanted to do this by basic theory of complex integration. Any help will be appreciated.   Thanks.
97399
Specifically, the question is as follows:     Let $\mathbb{D}$ be the open unit disk in $\mathbb{C}$ and suppose that $f:\bar{\mathbb{D}}\rightarrow\mathbb{C}$ is a continuous function such that both $\Re f$ and $\Im f$ are harmonic.       (a) Show that   $$f(re^{i\theta})=\frac{1}{2\pi}\int_{-\pi}^\pi f(e^{it})P_r(\theta-t)dt$$   for all $re^{i\theta}$ in $\mathbb{D}$.       (b) Show that $f$ is holomorphic in $\mathbb{D}$ if and only if   $$\int_{-\pi}^\pi f(e^{it})e^{int}dt=0$$   for all $n\geq1$.   I believe (a) is a direct consequence of Poisson's representation theorem for harmonic functions, stated as follows:  Let $\mathbb{D}$ be the open unit disk and $f:\partial\mathbb{D}\rightarrow\mathbb{R}$ be continuous. Then there exists a continuous function $u:\bar{\mathbb{D}}\rightarrow\mathbb{R}$ such that $u(z)=f(z)$ if $z\in\partial\mathbb{D}$, $u$ is harmonic on $\mathbb{D}$, and $u(re^{i\theta})=\frac{1}{2\pi}\int_{-\pi}^\pi P_r(\theta-t)f(e^{it})dt$ and $u$ is unique.  It seems to me that this can just be applied to the real and imaginary parts of the function $f$ from the question to derive the desired result, but maybe I'm making a careless mistake?  For (b), it's not obvious to me how part (a) assists in the solution. This is a homework question, so I am looking for assistance with understanding it, but not a full solution. Thanks.
1790
Let $f$ be a real valued and continuous function over $[a,b]$ with $a &lt;b$ such that $f(a)=f(b)$ , My question here is :     Question (claim): How to show that : there  exists  $c \in [a, \frac{a+b}{2} ]$ such that : $f(c)=f(c+\frac{b-a}{2})$.   Note: I have tried intermediate values but I didn't succeed.
115039
Let $f$ be entire and non-constant. Assuming $f$ satisfies the functional equation $f(1-z)=1-f(z)$, can one show that the image of $f$ is $\mathbb{C}$?  The values $f$ takes on the unit disc seems to determine $f$...  Any ideas?