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Let $K$ be an arbitrary compact subset of domain $\Omega$.

Why is the holomorphically convex hull of $K$ is contained in the convex hull of $K$?

Holomorphically convex hull of $K$ is defined as $\hat{K}_\Omega= \{z \in \Omega: |f(z)| \leq \sup_K |f|, \forall f\in A(\Omega)\}$.

I know that convex hull of $K$ is the smallest convex set that contains $K$. How can we express this formally in this context?

Thank you!

3 Answers 3

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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 $K$.

In the complex case, consider now a point $w$ outside the convex hull of $K$. Then there must be a closed ball $\overline{B_r}(z_0)$ that contins $K$ but does not contain $w$. Now apply your definition of holomorphically convex hull to $f(z) = z-z_0$.

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    @HenningMakholm Oh, this is trivial! I was confused with the definition. Thank you both for your help!2012-03-06
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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

  1. 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.

  2. 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 $\left\vert f\left( z\right) \right\vert >\sup \limits_{K}\left\vert f\right\vert .$

  3. 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.

  4. 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 $

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    It is unclear what your comments are referring to. Is that a post in MSE?2018-07-31
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Proof of the theorem that convex sets are holomorphically convex, in the case of several complex variables.

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 $. 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. ■