Holomorphically Convex Hull a Subset of the convex hull of This comes from Hörmander's "An Introduction to Complex Analysis in Several Variables".
We defined the $A(\Omega)$-hull (analytic functions in an open set $\Omega$). $\hat{K}$ of a compact set $K\subset\Omega$ by $\hat{K}=\{z;z\in\Omega, |f(z)|\leq\sup_K |f| \operatorname{for every } f\in A(\Omega) \}$.
The book says, if we consider $f(z)=e^{az}$ for every complex number $a$, we obtain $\hat{K}\subseteq \operatorname{convex hull of }K$. 
I do not get how he concluded this result. I do not know how to turn a $\hat{z}\in\hat{K}$ into a linear combination of elements $z\in K$ using the exp function. Are there specific $a$ I need to choose? Can I construct this?
Also, he says "Furthermore, it is clear that $\overset{*}{K}=\hat{K}$ ". I'm assuming that the $K$ with the weird star mark on top represents the convex hull? Even then, I do not understand the reverse inclusion.
 A: The exponential function grows in module as the exponential of the real part. Therefore, the set of all $z$ such that $|exp(az)|\leq \sup_K |exp(a\times\cdot)|$ is a half space containing $K$, and meeting $K$. You get all such half spaces, if you vary $a$ in $\mathbb{C}$ or even on the unit circle. Their intersection is the convex hull of $K$ by some famous theorem on convex sets (Krein-Milman?).
So by restricting yourself to the exponential functions you get the convex hull of $K$. The hull you're interested in is a subset of that set.
I don't know what $K^*$ stands for, but it won't be the convex hull in general. for instance, if you take $\Omega=\mathbb{C}\setminus\lbrace 0\rbrace$ and $K=$ the unit circle, and $f(z)=z, ~g(z)=\frac{1}{z}$, you see that the hull you're interested in is just $K$ itself. The convex hull may not even be a subset of $\Omega$.
A: Proof of the theorem that convex sets are holomorphically convex, in the case of  several complex variables (as suggested by O. Eroshkin).
L. Hörmander states on page 8 of "Introduction to Complex Analysis in Several Variables", without details,  "... if we consider $f\left(z\right)  =e^{az}$ for every complex number $a,$ we obtain $\widehat{K}%
\subset~$convex hull of $K$". The symbol $\widehat{K}$ denotes the holomorphically convex hull of a compact set $K$ with respect to an open subset $\Omega$ of $\mathbb{C}$. As usual, $z=x+iy$ where $x,y~\mathbb{\in }\mathbb{R}.$ Also on page 37 the same statement is made for $\mathbb{C}^{n}$, also without much in the way of details. This note provides a proof in the case $\mathbb{C}^{n}$.
Theorem. 
If $K$ is a compact convex set in $\mathbb{R}^{n}$, and if $w\not \in K$, then
there is a hyperplane in $\mathbb{R}^{n}$ that does not intersect $K$.
Proof. I denote by $\left\vert x\right\vert $ the length of the vector $x$ in
$\mathbb{R}^{n}$ and by $\left\langle w,x\right\rangle $ the Euclidean inner
product. The squared distance between two points $x$ and $y$ in $\mathbb{R}%
^{n}$ is given by $\left\vert x-y\right\vert ^{2}=\left\vert x\right\vert
^{2}+\left\vert y\right\vert ^{2}-2\left\langle x,y\right\rangle $.
\vspace{0in}Letting $w$ be fixed point in $\mathbb{R}^{n}$, the set of $x$
that satisfies the equation $\left\langle w,x\right\rangle =\left\vert
w\right\vert ^{2}$ is a hyperplane that contains $w$ and lies a distance
$\left\vert w\right\vert $ from the origin.
Given $w\not \in K$, because $K$ is compact there is a nearest point in $K$ to $w$ and there is no loss in generality by assuming that this point is $0$.
Assume for contradiction that the hyperplane $\left\langle w,x\right\rangle
=\left\vert w\right\vert ^{2}$ has a nonempty intersection with $K$ and $b$ is
a point in this intersection. Using the distance formula and the hyperplane
condition that $\left\langle w,x\right\rangle =\left\vert w\right\vert ^{2}$
we get $\left\vert w-b\right\vert ^{2}=\left\vert b\right\vert
^{2}-\left\vert w\right\vert ^{2}$ and this implies that $\left\vert
b\right\vert \geq\left\vert w\right\vert $. If $\left\vert b\right\vert
=\left\vert w\right\vert $, then the midpoint between the closest point $0$
and $b$ is $\frac{1}{2}b$ which by convexity is in $K$ and has length
$\frac{1}{2}\left\vert w\right\vert $. This contradicts that $\left\vert
w\right\vert $ is the distance between $w$ and $K$.
In the remaining case $\left\vert b\right\vert >\left\vert
w\right\vert $ and letting $p=\frac{\left\vert w\right\vert ^{2}}{\left\vert
b\right\vert ^{2}}b$, by convexity, $p\in K$, and $\left\vert w-p\right\vert
^{2}=\left\vert w\right\vert ^{2}+\frac{\left\vert w\right\vert ^{4}%
}{\left\vert b\right\vert ^{4}}\left\vert b\right\vert ^{2}-2\left\langle
w,\frac{\left\vert w\right\vert ^{2}}{\left\vert b\right\vert ^{2}%
}b\right\rangle $. Recalling that $b$ is in the hyperplane, we have
$\left\langle w,b\right\rangle =\left\vert w\right\vert ^{2}$, and hence
$\left\vert w-p\right\vert ^{2}=\left\vert w\right\vert ^{2}-\frac{\left\vert
w\right\vert ^{4}}{\left\vert b\right\vert ^{2}}=\frac{\left\vert w\right\vert
^{2}}{\left\vert b\right\vert ^{2}}\left(  \left\vert b\right\vert
^{2}-\left\vert w\right\vert ^{2}\right)  <\left\vert w\right\vert ^{2}.$
This contradicts that $w$ is the closest point to $K$. ■
For the next result we identify $\mathbb{C}^{n}$ with $\mathbb{R}^{2n}$ in the usual way: $\left(  z_{1},\ldots,z_{n}\right)  =\left(  \operatorname{Re}% \left(  z_{1}\right)  ,\operatorname{Im}\left(  z_{1}\right)  ,\ldots
,\operatorname{Re}\left(  z_{n}\right)  ,\operatorname{Im}\left(
z_{n}\right)  \right)  $. Given $w\in\mathbb{C}^{n},$ define $a\in
\mathbb{C}^{n}$ by $a_{j}=\operatorname{Re}\left(  w_{j}\right)
-i\operatorname{Im}\left(  w_{j}\right)  $. Then it is routine to show that
the inner product $\left\langle w,z\right\rangle $ in $\mathbb{R}^{2n}$ equals
$\operatorname{Re}\left(  \sum\limits_{j=1}^{n}a_{j}z_{j}\right)  $.
Theorem. 
Let $K$ be a compact set in an open set $\Omega\subset$ $\mathbb{C}^{n}.$ Then the convex hull of $K$ is holomorphically convex in $\Omega$.}
Proof. Let $w\in\mathbb{C}^{n}$ be a point that is in $\Omega$ but not in the the convex hull $L$ of $K$. Then there exists a  hyperplane $H$ with equation $\left\langle w,x\right\rangle =\left\vert w\right\vert ^{2}$ that contains $w$ but does not intersect $L$.  Define $H^{+}$ to be the set where $\left\langle w,x\right\rangle >\left\vert w\right\vert ^{2}$, and call it the positive half space defined by $H$. The negative half space $H^{-}$ is defined in a similar way.
Each half space is open and connected. Because $L$ is connected and does not intersect $H$, it must lie in one or the other half space.
If $L$ lies in the negative half space, then $\left\langle w,z\right\rangle -\left\vert w\right\vert ^{2}<0$ for $z$ in $L$. Because this function is continuous and negative on the compact set $L$, then it achieves a maximum value $N<0$. If we define $a\in$ $\mathbb{C}^{n}$ as in the previous comments, then the function $f\left(  z\right)  =\exp\left(  \left\langle
w,z\right\rangle -\left\vert w\right\vert ^{2}\right)  =\exp\left(\sum\limits_{j=1}^{n}a_{j}z_{j}-\left\vert w\right\vert ^{2}\right)  $ is holomorphic on $\mathbb{C}^{n}$, and therefore also on $\Omega$. Because $\left\vert f\left(  z\right)  \right\vert =\exp\left(  \operatorname{Re} \left(  \left\langle w,z\right\rangle -\left\vert w\right\vert ^{2}\right)
\right)  $, then $\left\vert f\left(  w\right)  \right\vert =1$, and because $N$ is negative, $\left\vert f\left(  z\right)  \right\vert \leq e^{N}<\left\vert f\left(  w\right)  \right\vert =1$. This shows that $L$ is holomorphically convex in $\Omega$.
A similar proof works if $L\ $is in the positive half space. ■
A: L. Hörmander states on page 8 of "Introduction to Complex Analysis in Several Variables":
"… if we consider $f\left(  z\right)  =e^{az}$ for every
complex number $a,$ we obtain $\widehat{K}\subset~$convex hull of
$K$". The symbol $\widehat{K}$ denotes the holomorphically
convex hull of a compact set $K$ with respect to an open subset $\Omega$ of
$\mathbb{C}$. As usual, $z=x+iy$ where $x,y~\mathbb{\in}\mathbb{R}.$
Several attempts to prove this can be found on the internet, but all that I
have seen are incomplete or invalid. What follows is a correct proof which is
based on the following theorem that I prove later. 
Theorem 1. A compact set $K\subset\mathbb{C}$ is convex if and only if $K$ satisfies the
condition: $K~\text{is connected, and}$, for all w not in K, there is a line through w that does not intersect K.
Theorem 2. $\widehat{K}\subset convex~hull\left(  K\right)  $
Proof. Let $w$ be a point that is in $\Omega$ but not in the the
convex hull $H$ of $K$. Theorem 1 applies to $H$ so there is a line through
$w~$that does not intersect $H$. Because rotations and translations are
holomorphic, without loss of generality we can assume that $w=0$ and that the
separation line is the $y$ - axis. We do the proof in the case that $H$ lies
to the left of the $y$ - axis, a similar proof works if $H$ is to the right.
Define $f\left(  z\right)  =e^{-z}$ in which case $\left\vert f\left(
z\right)  \right\vert =e^{x}$. Let $d$ be the (positive) distance of $H$ from
the $y$ - axis, so $z\in H$ implies that $x\leq-d$. Then $\left\vert f\left(
z\right)  \right\vert =e^{x}\leq e^{-d}<1=f\left(  0\right)  $. This shows
that $0\not \in \widehat{K}$.
Having shown that if $w\not \in convex~hull\left(  K\right)  $, then
$w\not \in \widehat{K}$ it follows that  $\widehat{K}\subset convex~hull\left(  K\right)  $.
If $H\ $lies to the right of the $y$ -- axis then defining  $f\left(
z\right)  =e^{-z}$ leads to the same result. $\blacksquare $
Comments


*

*John C asked for a proof of this result, March 29, 2011, 4:48. The same
day at 5:09 Choi responded with the good idea of separating $K$ by a line in
the plane. This is  the correct idea, but Choi did not give details. John C's
follow up at 6:16 is not correct. He used compactness and admitted that he did
not use convexity. By theorem 1 this approach cannot be correct. 

*On March 29, 2011, at 5:18 and later at 6:24 Bégassat gives a hint
involving the Krein Milman theorem. Bégassat  suggests use of the fact that
on each half space there is a function of the form $\exp\left(  az\right)  $
such that $\left\vert f\left(  z\right)  \right\vert \leq\sup\limits_{K}%
\left\vert f\right\vert $. But this inequality goes in the wrong direction and
what is needed is  for each $z\not \in $ the convex hull, at least one
function such that  $\left\vert f\left(  z\right)  \right\vert >\sup
\limits_{K}\left\vert f\right\vert .$

*On March 6, 2012, 13:04, Berk asked the same question on StackExchange.
The response by Makholm states "We can prove by geometric
reasoning in $\mathbb{R}^{2}$  that the convex hull of a compact set $K$ can
be characterized as exactly the intersection of all closed balls that contain
K." This is incorrect, the convex hull is the intersection
of all convex sets containing $K$, not just balls, and therefore the suggested
function $f\left(  x\right)  $ does not do the job.

*Wong, March 6, 2012, 13:41 suggests it is true that the convex hull of a
compact set $K$ in $\mathbb{C}$ is equal to all $z$ such that $\left\vert
L\left(  z\right)  \right\vert \leq\sup\limits_{K}\left\vert L\right\vert $
where $L$ runs through the family of complex affine functions $f\left(
z\right)  =az+b$. I don't know if this is correct (does anyone, and if so please give
a reference), but if so it would lead to the desired result, but not in the
spirit of Hörmander's hint.
Omitted proofs.
Lemma. If $K$ is a compact convex set in $\mathbb{C}$, then for every $w\not \in K$,
$\exists$ line through $w$ that does not intersect $K$ (informally, there is a
line through $w$ that separates $K$: $K$ is on one side of the line)
Proof. By compactness, there is a point $a\in K$ that is closest to $w$.
Construct the line $l$ through $w$ and perpendicular to the line $aw$. Assume
for contradiction that there were a point $b\in K\cap l$. Then the line $ab$
is contained in $K$ by convexity. Let $m$ be the line constructed by dropping
a perpendicular from $w$ to $ab,$ intersecting at $d$ which is contained in
$K$ because it is between $a$ and $b,$ both of which are in the convex set
$K$. Then the distance between $d$ and $w$ is less than the distance between
$a$ and $w$, because the triangle $awd$ is a right triangle so its hypotenuse
$aw$ is larger than the side $ad$. This contradicts the fact that $a$ was the
closest point in $K$ to $w$. $\blacksquare $
Theorem 1:  A compact set $K\subset\mathbb{C}$ is convex if and only if $K$ satisfies the
condition: $K~\text{is connected, and}$, for all w not in K, there is a line through w that does not intersect K.
Proof. Suppose that $K$ is convex and compact. By convexity $K$ is pathwise
connected and therefore connected so by the lemma if $w\not \in K$, the
remainder of the condition is satisfied.
Conversely, assume that the condition is satisfied and assume
for contradiction that there were $a,b\in K$ such that the line segment
$\left[  a,b\right]  $ between $a$ and $b$ contained a point $w$ not in $K.$
Then there is a line through $w~$that does not intersect $K$. This line
divides the plane into two connected open sets. More precisely, there are two
open half planes whose union contains $K$. This violates the condition that
$K$ is connected. $\blacksquare $
