Holomorphic images of disks

Consider a disk in the complex plane. If one applies an affine-linear map to this disk, one obtains

For maps that are merely holomorphic instead of affine-linear, one has some variants of this assertion, which I am recording here mostly for my own reference:

Theorem 1 (Holomorphic images of disks) Let be a disk in the complex plane, and be a holomorphic function with .

• (i) (Open mapping theorem or inverse function theorem) contains a disk for some . (In fact there is even a holomorphic right inverse of from to .)
• (ii) (Bloch theorem) contains a disk for some absolute constant and some . (In fact there is even a holomorphic right inverse of from to .)
• (iii) (Koebe quarter theorem) If is injective, then contains the disk .
• (iv) If is a polynomial of degree , then contains the disk .
• (v) If one has a bound of the form for all and some , then contains the disk for some absolute constant . (In fact there is holomorphic right inverse of from to .)

Parts (i), (ii), (iii) of this theorem are standard, as indicated by the given links. I found part (iv) as (a consequence of) Theorem 2 of this paper of Degot, who remarks that it “seems not already known in spite of its simplicity”; an equivalent form of this result also appears in Lemma 4 of this paper of Miller. The proof is simple:

Proof: (Proof of (iv)) Let , then we have a lower bound for the log-derivative of at :

(with the convention that the left-hand side is infinite when ). But by the fundamental theorem of algebra we have

where are the roots of the polynomial (counting multiplicity). By the pigeonhole principle, there must therefore exist a root of such that

and hence . Thus contains , and the claim follows.

The constant in (iv) is completely sharp: if and is non-zero then contains the disk

but avoids the origin, thus does not contain any disk of the form . This example also shows that despite parts (ii), (iii) of the theorem, one cannot hope for a general inclusion of the form

for an absolute constant .

Part (v) is implicit in the standard proof of Bloch’s theorem (part (ii)), and is easy to establish:

Proof: (Proof of (v)) From the Cauchy inequalities one has for , hence by Taylor’s theorem with remainder for . By Rouche’s theorem, this implies that the function has a unique zero in for any , if is a sufficiently small absolute constant. The claim follows.

Note that part (v) implies part (i). A standard point picking argument also lets one deduce part (ii) from part (v):

Proof: (Proof of (ii)) By shrinking slightly if necessary we may assume that extends analytically to the closure of the disk . Let be the constant in (v) with ; we will prove (iii) with replaced by . If we have for all then we are done by (v), so we may assume without loss of generality that there is such that . If for all then by (v) we have

and we are again done. Hence we may assume without loss of generality that there is such that . Iterating this procedure in the obvious fashion we either are done, or obtain a Cauchy sequence in such that goes to infinity as , which contradicts the analytic nature of (and hence continuous nature of ) on the closure of . This gives the claim.

Here is another classical result stated by Alexander (and then proven by Kakeya and by Szego, but also implied to a classical theorem of Grace and Heawood) that is broadly compatible with parts (iii), (iv) of the above theorem:

Proposition 2 Let be a disk in the complex plane, and be a polynomial of degree with for all . Then is injective on .

The radius is best possible, for the polynomial has non-vanishing on , but one has , and lie on the boundary of .

If one narrows slightly to then one can quickly prove this proposition as follows. Suppose for contradiction that there exist distinct with , thus if we let be the line segment contour from to then . However, by assumption we may factor where all the lie outside of . Elementary trigonometry then tells us that the argument of only varies by less than as traverses , hence the argument of only varies by less than . Thus takes values in an open half-plane avoiding the origin and so it is not possible for to vanish.

To recover the best constant of requires some effort. By taking contrapositives and applying an affine rescaling and some trigonometry, the proposition can be deduced from the following result, known variously as the Grace-Heawood theorem or the complex Rolle theorem.

Proposition 3 (Grace-Heawood theorem) Let be a polynomial of degree such that . Then contains a zero in the closure of .

This is in turn implied by a remarkable and powerful theorem of Grace (which we shall prove shortly). Given two polynomials of degree at most , define the apolar form by

Theorem 4 (Grace’s theorem) Let be a circle or line in , dividing into two open connected regions . Let be two polynomials of degree at most , with all the zeroes of lying in and all the zeroes of lying in . Then .

(Contrapositively: if , then the zeroes of cannot be separated from the zeroes of by a circle or line.)

Indeed, a brief calculation reveals the identity

where is the degree polynomial

The zeroes of are for , so the Grace-Heawood theorem follows by applying Grace’s theorem with equal to the boundary of .

The same method of proof gives the following nice consequence:

Theorem 5 (Perpendicular bisector theorem) Let be a polynomial such that for some distinct . Then the zeroes of cannot all lie on one side of the perpendicular bisector of . For instance, if , then the zeroes of cannot all lie in the halfplane or the halfplane .

I’d be interested in seeing a proof of this latter theorem that did not proceed via Grace’s theorem.

Now we give a proof of Grace’s theorem. The case can be established by direct computation, so suppose inductively that and that the claim has already been established for . Given the involvement of circles and lines it is natural to suspect that a Möbius transformation symmetry is involved. This is indeed the case and can be made precise as follows. Let denote the vector space of polynomials of degree at most , then the apolar form is a bilinear form . Each translation on the complex plane induces a corresponding map on , mapping each polynomial to its shift . We claim that the apolar form is invariant with respect to these translations:

Taking derivatives in , it suffices to establish the skew-adjointness relation

but this is clear from the alternating form of (1).

Next, we see that the inversion map also induces a corresponding map on , mapping each polynomial to its inversion . From (1) we see that this map also (projectively) preserves the apolar form:

More generally, the group of Möbius transformations on the Riemann sphere acts projectively on , with each Möbius transformation mapping each to , where is the unique (up to constants) rational function that maps this a map from to (its divisor is ). Since the Möbius transformations are generated by translations and inversion, we see that the action of Möbius transformations projectively preserves the apolar form; also, we see this action of on also moves the zeroes of each by (viewing polynomials of degree less than in as having zeroes at infinity). In particular, the hypotheses and conclusions of Grace’s theorem are preserved by this Möbius action. We can then apply such a transformation to move one of the zeroes of to infinity (thus making a polynomial of degree ), so that must now be a circle, with the zeroes of inside the circle and the remaining zeroes of outside the circle. But then

By the Gauss-Lucas theorem, the zeroes of are also inside . The claim now follows from the induction hypothesis.

The closed graph theorem in various categories

Given a function between two sets , we can form the graph

which is a subset of the Cartesian product .

There are a number of “closed graph theorems” in mathematics which relate the regularity properties of the function with the closure properties of the graph , assuming some “completeness” properties of the domain and range . The most famous of these is the closed graph theorem from functional analysis, which I phrase as follows:

Theorem 1 (Closed graph theorem (functional analysis)) Let be complete normed vector spaces over the reals (i.e. Banach spaces). Then a function is a continuous linear transformation if and only if the graph is both linearly closed (i.e. it is a linear subspace of ) and topologically closed (i.e. closed in the product topology of ).

I like to think of this theorem as linking together qualitative and quantitative notions of regularity preservation properties of an operator ; see this blog post for further discussion.

The theorem is equivalent to the assertion that any continuous linear bijection from one Banach space to another is necessarily an isomorphism in the sense that the inverse map is also continuous and linear. Indeed, to see that this claim implies the closed graph theorem, one applies it to the projection from to , which is a continuous linear bijection; conversely, to deduce this claim from the closed graph theorem, observe that the graph of the inverse is the reflection of the graph of . As such, the closed graph theorem is a corollary of the open mapping theorem, which asserts that any continuous linear surjection from one Banach space to another is open. (Conversely, one can deduce the open mapping theorem from the closed graph theorem by quotienting out the kernel of the continuous surjection to get a bijection.)

It turns out that there is a closed graph theorem (or equivalent reformulations of that theorem, such as an assertion that bijective morphisms between sufficiently “complete” objects are necessarily isomorphisms, or as an open mapping theorem) in many other categories in mathematics as well. Here are some easy ones:

Theorem 2 (Closed graph theorem (linear algebra)) Let be vector spaces over a field . Then a function is a linear transformation if and only if the graph is linearly closed.

Theorem 3 (Closed graph theorem (group theory)) Let be groups. Then a function is a group homomorphism if and only if the graph is closed under the group operations (i.e. it is a subgroup of ).

Theorem 4 (Closed graph theorem (order theory)) Let be totally ordered sets. Then a function is monotone increasing if and only if the graph is totally ordered (using the product order on ).

Remark 1 Similar results to the above three theorems (with similarly easy proofs) hold for other algebraic structures, such as rings (using the usual product of rings), modules, algebras, or Lie algebras, groupoids, or even categories (a map between categories is a functor iff its graph is again a category). (ADDED IN VIEW OF COMMENTS: further examples include affine spaces and -sets (sets with an action of a given group ).) There are also various approximate versions of this theorem that are useful in arithmetic combinatorics, that relate the property of a map being an “approximate homomorphism” in some sense with its graph being an “approximate group” in some sense. This is particularly useful for this subfield of mathematics because there are currently more theorems about approximate groups than about approximate homomorphisms, so that one can profitably use closed graph theorems to transfer results about the former to results about the latter.

A slightly more sophisticated result in the same vein:

Theorem 5 (Closed graph theorem (point set topology)) Let be compact Hausdorff spaces. Then a function is continuous if and only if the graph is topologically closed.

Indeed, the “only if” direction is easy, while for the “if” direction, note that if is a closed subset of , then it is compact Hausdorff, and the projection map from to is then a bijective continuous map between compact Hausdorff spaces, which is then closed, thus open, and hence a homeomorphism, giving the claim.

Note that the compactness hypothesis is necessary: for instance, the function defined by for and for is a function which has a closed graph, but is discontinuous.

A similar result (but relying on a much deeper theorem) is available in algebraic geometry, as I learned after asking this MathOverflow question:

Theorem 6 (Closed graph theorem (algebraic geometry)) Let be normal projective varieties over an algebraically closed field of characteristic zero. Then a function is a regular map if and only if the graph is Zariski-closed.

Proof: (Sketch) For the only if direction, note that the map is a regular map from the projective variety to the projective variety and is thus a projective morphism, hence is proper. In particular, the image of under this map is Zariski-closed.

Conversely, if is Zariski-closed, then it is also a projective variety, and the projection is a projective morphism from to , which is clearly quasi-finite; by the characteristic zero hypothesis, it is also separated. Applying (Grothendieck’s form of) Zariski’s main theorem, this projection is the composition of an open immersion and a finite map. As projective varieties are complete, the open immersion is an isomorphism, and so the projection from to is finite. Being injective and separable, the degree of this finite map must be one, and hence and are isomorphic, hence (by normality of ) is contained in (the image of) , which makes the map from to regular, which makes regular.

The counterexample of the map given by for and demonstrates why the projective hypothesis is necessary. The necessity of the normality condition (or more precisely, a weak normality condition) is demonstrated by (the projective version of) the map from the cusipdal curve to . (If one restricts attention to smooth varieties, though, normality becomes automatic.) The necessity of characteristic zero is demonstrated by (the projective version of) the inverse of the Frobenius map on a field of characteristic .

There are also a number of closed graph theorems for topological groups, of which the following is typical (see Exercise 3 of these previous blog notes):

Theorem 7 (Closed graph theorem (topological group theory)) Let be -compact, locally compact Hausdorff groups. Then a function is a continuous homomorphism if and only if the graph is both group-theoretically closed and topologically closed.

The hypotheses of being -compact, locally compact, and Hausdorff can be relaxed somewhat, but I doubt that they can be eliminated entirely (though I do not have a ready counterexample for this).

In several complex variables, it is a classical theorem (see e.g. Lemma 4 of this blog post) that a holomorphic function from a domain in to is locally injective if and only if it is a local diffeomorphism (i.e. its derivative is everywhere non-singular). This leads to a closed graph theorem for complex manifolds:

Theorem 8 (Closed graph theorem (complex manifolds)) Let be complex manifolds. Then a function is holomorphic if and only if the graph is a complex manifold (using the complex structure inherited from ) of the same dimension as .

Indeed, one applies the previous observation to the projection from to . The dimension requirement is needed, as can be seen from the example of the map defined by for and .

(ADDED LATER:) There is a real analogue to the above theorem:

Theorem 9 (Closed graph theorem (real manifolds)) Let be real manifolds. Then a function is continuous if and only if the graph is a real manifold of the same dimension as .

This theorem can be proven by applying invariance of domain (discussed in this previous post) to the projection of to , to show that it is open if has the same dimension as .

Note though that the analogous claim for smooth real manifolds fails: the function defined by has a smooth graph, but is not itself smooth.

(ADDED YET LATER:) Here is an easy closed graph theorem in the symplectic category:

Theorem 10 (Closed graph theorem (symplectic geometry)) Let and be smooth symplectic manifolds of the same dimension. Then a smooth map is a symplectic morphism (i.e. ) if and only if the graph is a Lagrangian submanifold of with the symplectic form .

In view of the symplectic rigidity phenomenon, it is likely that the smoothness hypotheses on can be relaxed substantially, but I will not try to formulate such a result here.

There are presumably many further examples of closed graph theorems (or closely related theorems, such as criteria for inverting a morphism, or open mapping type theorems) throughout mathematics; I would be interested to know of further examples.

245B, Notes 9: The Baire category theorem and its Banach space consequences

The notion of what it means for a subset E of a space X to be “small” varies from context to context.  For instance, in measure theory, when is a measure space, one useful notion of a “small” set is that of a null set: a set E of measure zero (or at least contained in a set of measure zero).  By countable additivity, countable unions of null sets are null.  Taking contrapositives, we obtain

Lemma 1. (Pigeonhole principle for measure spaces) Let be an at most countable sequence of measurable subsets of a measure space X.  If has positive measure, then at least one of the has positive measure.

Now suppose that X was a Euclidean space with Lebesgue measure m.  The Lebesgue differentiation theorem easily implies that having positive measure is equivalent to being “dense” in certain balls:

Proposition 1. Let be a measurable subset of .  Then the following are equivalent:

1. E has positive measure.
2. For any , there exists a ball B such that .

Thus one can think of a null set as a set which is “nowhere dense” in some measure-theoretic sense.

It turns out that there are analogues of these results when the measure space   is replaced instead by a complete metric space .  Here, the appropriate notion of a “small” set is not a null set, but rather that of a nowhere dense set: a set E which is not dense in any ball, or equivalently a set whose closure has empty interior.  (A good example of a nowhere dense set would be a proper subspace, or smooth submanifold, of , or a Cantor set; on the other hand, the rationals are a dense subset of and thus clearly not nowhere dense.)   We then have the following important result:

Theorem 1. (Baire category theorem). Let be an at most countable sequence of subsets of a complete metric space X.  If contains a ball B, then at least one of the is dense in a sub-ball B’ of B (and in particular is not nowhere dense).  To put it in the contrapositive: the countable union of nowhere dense sets cannot contain a ball.

Exercise 1. Show that the Baire category theorem is equivalent to the claim that in a complete metric space, the countable intersection of open dense sets remain dense. 

Exercise 2. Using the Baire category theorem, show that any non-empty complete metric space without isolated points is uncountable.  (In particular, this shows that Baire category theorem can fail for incomplete metric spaces such as the rationals .) 

To quickly illustrate an application of the Baire category theorem, observe that it implies that one cannot cover a finite-dimensional real or complex vector space by a countable number of proper subspaces.  One can of course also establish this fact by using Lebesgue measure on this space.  However, the advantage of the Baire category approach is that it also works well in infinite dimensional complete normed vector spaces, i.e. Banach spaces, whereas the measure-theoretic approach runs into significant difficulties in infinite dimensions.  This leads to three fundamental equivalences between the qualitative theory of continuous linear operators on Banach spaces (e.g. finiteness, surjectivity, etc.) to the quantitative theory (i.e. estimates):

1. The uniform boundedness principle, that equates the qualitative boundedness (or convergence) of a family of continuous operators with their quantitative boundedness.
2. The open mapping theorem, that equates the qualitative solvability of a linear problem Lu = f with the quantitative solvability.
3. The closed graph theorem, that equates the qualitative regularity of a (weakly continuous) operator T with the quantitative regularity of that operator.

Strictly speaking, these theorems are not used much directly in practice, because one usually works in the reverse direction (i.e. first proving quantitative bounds, and then deriving qualitative corollaries); but the above three theorems help explain why we usually approach qualitative problems in functional analysis via their quantitative counterparts.