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. ■
