$f:D \rightarrow \mathbb C$ holomorphic, $D$ a convex set, $Re(f'(z))&gt;0$. Prove that $f$ is injective.  What I tried: Assume that $f$ is not injective. So there are $a,b \in D$ so that $f(a)=f(b)$.   So we get that:  $\int_a^bf'(z)dz=f(a)-f(b)=0 $  How do I use the fact that $Re(f'(z))&gt;0$? Or that's not the direction at all?  Thank you.
75384
Suppose that $f$ is differentiable on a convex domain (open and connected) D and Re($f')&gt;0$ in D. How can we prove that f is injective on D?  I found some answers using the mean value theorem for holomorphic functions, but I am not familiar with that theorem. Actually I want to use the mean value theorem for integrals, which tells us that the value of an differentiable function $f$ at the center $z_{0}$ of the disk $|z-z_{0}|&lt;r$ is equal to the integral mean of its values on the boundary of that disk: $f(z_{0})=\int^{2\pi}_{0}{f(z_{0}+re^{it})dt}$.  Can you help me proving it by the mean value theorem for integrals?
46080
We want to prove the following:     If $f$ is a holomorphic function on the unit disc $\mathbb{D}$ s.t. $f(z) \neq 0$ for $z \in \mathbb{D}$, then there is a holomorphic function $g$ on $\mathbb{D}$ such that $f(z)=e^{g(z)}$ for all $z \in \mathbb{D}$.   For such a $g$, we have that $\frac{f'(z)}{f(z)}=g'(z)$ on $\mathbb{D}$ since $f$ has non-zero derivative.   Perhaps then we define $g(z)=\int_\gamma \frac{f'(z)}{f(z)}dz$ where $\gamma$ is a path in $\mathbb{D}$ from $0$ to $z$.  Does this work?
4030
Maybe I am not good at looking for the right questions but I haven't seen this task anywhere so I hope it is no duplicate.  I have to prove the following statement:     Let $V$ be a finite dimensional vector space and $f:V \rightarrow V$ a linear map. Show that $f$ is injective $\Longleftrightarrow$ $f$ is surjective.   The following is already known:  $(i)$ $\ker f$ and $\def\Im{\operatorname{Im}}\Im f$ are linear subspaces $(ii)$ There exists an isomorphism $V/\ker f\rightarrow \Im f$ $(iii)$ If $U\subset V$ is a linear subspace then $\dim V/U=\dim V-\dim U$ $(iv)$ $f$ is injective $\Longleftrightarrow$ $\ker f=\{0\}$  So my approach is the following:  $"\Longrightarrow "$ Let $f:V\rightarrow V$ be injective. Then due to $(iv)$ $\dim\ker f=0$. Hence due to $(ii)$ and $(iii)$ $\dim V/\ker f=\dim V=\dim\Im f$. Since $\dim V=\dim\Im f$ it implies that $f$ is surjective.  $"\Longleftarrow "$ Let $f$ be surjective. Then $\dim\Im f=\dim V$. Additionally, because of $(ii)$ it is $\dim\Im f=\dim V/\ker f$. Thus $\dim V=\dim V/\ker f=\dim V-\dim\ker f$. This means that $\dim\ker f=0$ and hence $\ker f=\{0\}$ and thus $f:V\rightarrow V$ is injective.  For me it all makes sense but maybe I missed something.  Thank you in advance.
111761
Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be an holomorphic function such that $f'(z_0) \neq 0$ for some $z_0 \in \mathbb{C}$.Prove that there is $r&gt;0$ such that, if $|z-z_0|&lt;r$ and $z \neq z_0 $, then $f(z) \neq f(z_0)$.  It seems very trivial, but I can't solve it...
87244
If $f:\mathbb{D}\rightarrow\mathbb{D}$ is holomorphic then prove that   $$\frac{|f(0)|-|z|}{1 + |f(0)||z|} \leq|f(z)| \leq\frac{|f(0)| + |z|}{1 - |f(0)||z|} $$   I have been wracking my brain for hours and am out of ideas.  The hypothesis seems to suggest an application of the Schwarz lemma but I cannot find a suitable function - I've tried functions of the form $$\frac{f(z) - f(0)}{\mbox{scale factor}}$$ and none have worked.  I also tried working backwards to obtain an inequality and got $$-|z|\leq|f(z)||1\pm f(0)z| - |f(0)|\leq |z|\;\;\;\mbox{(*) }$$ If I could prove $$|f(z)[1\pm f(0)z] - f(0)|\leq 1$$ then I could obtain (*), but with so few conditions on $f$ I have no idea how to proceed.
64505
Let $a&gt;0$ and $f: \mathbb R \rightarrow \mathbb R$ with a property $\forall x,y \in \mathbb R;$ such that $0\;&lt;\;y\;-\;x\;&lt;\;a$ $$f(x)\;&lt;f(y).$$ Prove if the statements are true or false:  $1)$ $f$ must be injective.  $2)$ $f$ must be surjective.  I tried to use a quarter of a circle with radius $r=a$ or a right-angled triangle and translate it. When I moved it, I saw new values of $f$. I concluded the maximum value in the previous interval can't appear again in another interval when the distance between one point in the first interval and one point in the second is less than $a$. I also considered using an absolute value function, but I haven't done anything. I think, but I haven't proven it, that the function must be bijective.
122625
We consider $T:\mathbb{C}\to\mathbb{C}$ defined by $T(z)=\lambda z+\mu\overline{z}$, where $\lambda ,\mu\in\mathbb{C}$. I want to prove that if $T$ is injective, then $\lambda\cdot\overline{\lambda}\neq\mu\cdot\overline{\mu}$.  Clearly, if $\mu =0$, then $\lambda\neq 0$ because $T$ is injective.  Now, if $\lambda\cdot\overline{\lambda}=\mu\cdot\overline{\mu}$ and $\mu\neq 0$, then I wanted to find a $z\in\mathbb{C}\setminus\{0\}$ such that $z=-\dfrac{\overline{\lambda}}{\overline{\mu}}\overline{z}$, because we would have that:  $T(z)=-\lambda\dfrac{\overline{\lambda}}{\overline{\mu}}\overline{z}+\mu\overline{z}=\overline{z}(-\lambda\dfrac{\overline{\lambda}}{\overline{\mu}}+\mu)=0$  ('Cause this would prove that $\lambda\cdot\overline{\lambda}=\mu\cdot\overline{\mu}$ implies $T$ is not injective.)  So, the question is, can we always find such a $z\in\mathbb{C}\setminus\{0\}$?
122614
$f$ is an entire function, and it satisfies $f(\mathbb{C}) \subseteq \{z \in \mathbb{C} \mid \operatorname{Im} z &gt; 0\}$. Show that $f$ is constant.   I want to take advantage of the Liouville's Theorem, but I just can't figure out the relationship between its image part with its module.
92203
Function $f(z)$ is holomorphic at least within the unit disk.   If for all arguments $|z|&lt;1$ we have $|f(z)| \le 1$, and there exists a zero of order $n$ at $z=z_0$, then show that $|f(0)| \le |z_0|^n$.  How should I tackle this problem? Can it be generalized if for $|z|&lt;R$ we have $|f(z)| \le M$ ?    So far I only understand that there exists a holomorphic $g(z)$ with $g(z_0) \ne 0$ such that $f(z) = (z-z_0)^n g(z)$,then $|f(z)| = |z-z_0|^n |g(z)| \le 1$ and moreover $|f(0)| = |z_0|^n |g(0)|$, but that doesn't seem useful since apparently I know too little about $g(z)$. In particular why should $|g(0)| \le 1$?
47373
Let $f\in C[0,1]$ be such  that $\forall \phi\in C[0,1]$ with $\int_0^1\phi \, dx=0$,we have $\int_0^1f\phi \, dx=0$.  Then $f$ is constant.  May be someone can help me proof this problem. 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!
43808
There doesnt exist a holomorphic $f: \mathbb C\setminus0 \rightarrow \mathbb C\setminus 0 $, so that $e^{f(z)} = z $ for all $z \in \mathbb C \setminus 0 $  I tried to work with the derivative of f(z) but i didnt manage to get anything out of it.
20383
Let $A \subset \mathbb{R}^n$ be closed and $d_A(x) = \inf_{a \in A} \|x-a\|$. I need to show that $d_A$ is convex if and only if $A$ is convex.  I have no good ideas in either direction. Can somebody please give me a hint?  For the direction ($\Longleftarrow$), my attempt was:  Let $a_1 := \operatorname*{argmin}_{a \in A} \|x-a\|, a_2 := \operatorname*{argmin}_{a\in A} \|y-a\|$, then:  \begin{align*} d_A(x) &amp;= \inf_{a \in A} \|\lambda x + (1 - \lambda)y - a\| = \inf \|\lambda(x-a) + (1-\lambda) (y-a)\|\\ &amp;\leq \lambda \|x-a_1\| + (1-\lambda)\|y-a_2\| = \lambda d_A(x) + (1-\lambda)d_A(y) \end{align*}  However, I'm not sure whether the inequality in the third step is actually correct.
61631
How to prove the real value function $f$ is convex on $[a,d]$ if $f$ is convex on both $[a,c]$ and $[b,d]$? ($a&lt;b&lt;c&lt;d$).    I try to show that: for arbitrary $x\in(a,b), z\in(c,d)$, and for all $y\in (x,z)$ the inequality   $\dfrac{f(y)-f(x)}{y-x}\leqslant \dfrac{f(z)-f(y)}{z-y}$  holds. Then $f$ is convex on $[a,d]$.   But it seems to need some inequality tricks, I can't finish it.
45536
Let $f \in \mathbf R[x]$  and for every $k \in \mathbf Z$, $f(k)\in \mathbf Z$. Does this imply that $f \in \mathbf Z[x]$?   I tried to find a counter example, but my intuition tells me it's true and I don't know how to prove the statement either.
5700
Prove that if $T''$ is injective then $T$ it's injective, but the converse it's not true.   I proved the first part " $T''$ injective implies $T$ injective".  It's easy to prove that $T''$ is injective if and only if the image of $T'$ is dense on $X'$. So If I prove that the $cl({Ran(T')})$ it's a proper subset of $X'$ then I'm done. But even this it's difficult.
18951
If I have $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f$ is onto and $f\circ f\circ f = f$, how can I prove that $f$ is bijective? I know that I only have to prove that it is 1-to-1 because I'm given the fact that it's onto, but how can I use the fact that $f\circ f\circ f=f$ to prove that it is 1-to-1?  Thanks!