User malik younsi - MathOverflow

What can be said about pairs of matrices P,Q that satisfies $(P^{-1})^T \circ P = (Q^{-1})^T \circ Q$ ?

2010-05-27T16:07:49Z

Let $P,Q$ be $n$ by $n$ invertible matrices. Suppose further that $P$ and $Q$ satisfies the following equation :

$$(P^{-1})^T \circ P = (Q^{-1})^T \circ Q$$

where $\circ$ denotes the Hadamard matrix product, which is simply the entrywise product.

Then what can be said about $P$ and $Q$? More precisely, I want to know if there are additional relations between $P$ and $Q$. For example, one can show that the condition $(P^{-1})^T \circ P = (Q^{-1})^T \circ Q$ implies

$$tr(P^{-1}DPE) = tr(Q^{-1}DQE)$$ for all diagonal matrices $D$ and $E$.

References in the litterature about matrices of the form $(P^{-1})^T \circ P$ would help too. Thank you, Malik

Answer by Malik Younsi for functions of one complex variable: geometric theory

2013-04-03T19:26:52Z

I suggest "Functions of One Complex Variable" I and II, by John B. Conway. It is especially well-written and cover a lot of topics, from elementary ones (volume I) to more advanced (volume II).

On the set of zero radial limits of bounded analytic functions

2013-03-10T12:16:30Z

Hi,

Let $f$ be a non-identically zero bounded analytic function in the open unit disk $\mathbb{D}$. It is well-known that $f$ has radial limits almost everywhere on the unit circle $\mathbb{T}$. Let $Z_f$ be the set of points in $\mathbb{T}$ where $f$ has zero radial limit :

$$Z_f:= \{e^{i\theta} \in \mathbb{T} : \lim_{r \rightarrow 1}f(re^{i\theta})=0\}.$$

Then it is also well-known that $Z_f$ has measure zero.

My question is the following :

For which sets $E \subseteq \mathbb{T}$ of measure zero does there exist a bounded analytic function $f$ in $\mathbb{D}$ such that $E=Z_f$?

Remark : It is easy to see that every $Z_f$ is a $F_{\sigma \delta}$, so the question could be wether or not every $F_{\sigma \delta}$ of measure zero is $Z_f$ for some $f$.

Answer by Malik Younsi for When does continuity imply holomorphy?

2013-01-28T16:10:28Z

The following is too long for a comment. I will suppose here that your set of measure zero is compact.

In this case, Your question is closely related to so-called continuous analytic capacity. Let $K$ be a compact set in the plane, and let $\Omega$ be the complement of $K$ with respect to $\mathbb{C}_\infty$. The continuous analytic capacity $\alpha(K)$ is defined as

$$\alpha(K):= \sup { |f'(\infty)| : f \in A(\Omega), \|f\|_{\infty} \leq 1 },$$

where $A(\Omega)$ is the set of all functions holomorphic in $\Omega$ that extend continuously to $\mathbb{C}_\infty$.

It is not difficult to prove that $$\alpha(K)=0$$ if and only if for every open set $U$ with $K \subseteq U$, every $f$ holomorphic on $U \setminus K$, continuous on $U$, extend analytically to the whole of $U$.

So what you're looking for is a characterization of the compact sets $K$ with $\alpha(K)=0$. This is certainly very difficult, since it took more than a hundred years to solve a similar problem (but with analytic capacity $\gamma$ instead of $\alpha$), the so-called Painlevé's problem.

It is easy to prove however that if $\alpha(K)=0$, then the area of $K$ must be zero. You're asking if this condition is sufficient. Most likely it is not, but I don't have a counterexample right now.. Usually, for problems related to removable sets for holomorphic functions, it is more useful to consider hausdorff measure instead.

Anyway, I suggest you look in Garnett's book "Analytic Capacity and measure". See also the recent book by Dudziak, "Vitushkin's conjecture for removable sets."

On the Universality of the Riemann zeta-function

2013-01-21T18:14:45Z

Hi,

I have a question regarding the universality property of the Riemann zeta-function. I am no expert on this, so I'd be glad for any relevant reference.

First, recall Voronin's remarkable theorem on the Universality of the Riemann zeta-function :

Let $K$ be a compact subset with connected complement lying in the strip ${1/2 < \operatorname{Re}(z)<1}$, and let $f : K \rightarrow \mathbb{C}$ be continuous, holomorphic on the interior of $K$, and zero-free on $K$. Then for each $\epsilon>0$, there exists $t>0$ such that $$\max_{z \in K} |\zeta(z+it)-f(z)|<\epsilon.$$ Even more : the lower density of the set of such $t$'s is positive..!

Note that of course, the hypothesis that the complement of $K$ is connected is essential in the above theorem.

My question is the following :

Is there some sort of (modified) zeta-function universality-like result for compact sets $K$ with disconnected complements? For example, if $\mathbb{C}_\infty \setminus K$ has a finite number of components?

EDIT

Of course I know that a sequence of the form $f_n(z):=\zeta(z+it_n)$ won't work in the case when the complement of $K$ is disconnected (such a sequence cannot approximate uniformly say $1/z$ on an annulus centered at $0$). I'm asking wether there is some sequence of functions, involving the Riemann zeta-function, that could work in this case, and generalize Voronin's Theorem. Note that such functions will necessarily have poles in each component of the complement of $K$.

2nd EDIT

Let me explain what I was looking for here. Basically, I'd like to know if there exists a result of the following form :

Let $K$ be a compact subset whose complement has finitely many components lying in the strip ${1/2 < \operatorname{Re}(z)<1}$, and let $f : K \rightarrow \mathbb{C}$ be continuous, holomorphic on the interior of $K$, and zero-free on $K$. Then for each $\epsilon>0$, there exists...

Here insert some uniform approximation of $f$ on $K$ by a function involving the Riemann zeta-function

Furthermore, in the case when $K$ has connected complement, I would like the above result to reduce to Voronin's Theorem.

In summary, I want to know if there exists a generalization of Voronin's Theorem to compact sets whose complement have finitely many components.

Thank you, Malik

Answer by Malik Younsi for A question about the limit of a sequence of pointwise convergent analytic funtions

2012-12-30T15:38:07Z

Of course (1) does not imply (2) : the functions $f_n:=z^n$ converge pointwisely to $0$ on the unit disk, but the convergence is not uniform.

In fact, assuming (1), the convergence need not be locally uniform : using Runge's Theorem, it is possible to find a sequence of polynomials $p_n$ such that $p_n \rightarrow 0$ pointwisely on the unit disk, but the convergence is not uniform in any neighborhood of $0$ : see e.g. this question

EDIT

A similar argument (using Runge's Theorem) answers your question (3) negatively : It is possible to construct a sequence of polynomials $p_n$'s with $p_n(z) \rightarrow 0$ for every $z \neq 0$ but $p_n(0) \rightarrow 1$.

Inequivalent complete norms and the axiom of choice

2012-10-01T15:28:17Z

Hi,

I've been wondering about the following :

Is it possible, without the axiom of choice, to have two inequivalent complete norms on a vector space?

All the examples of inequivalent complete norms I've seen rely on the existence of Hamel bases...

This is most likely well-known, but I'd be glad if someone could provide a good reference.

Thank you, Malik

Answer by Malik Younsi for Is there an explicit formula for the modulus of an annulus given a parameterization of the inner and outer boundries?

2012-09-19T12:52:02Z

I also think there is no "explicit formula" for the conformal modulus, even in simple cases. However, as mentioned in Igor Rivin's answer, there are methods for approximating the conformal map and the conformal modulus. These methods are generelizations to doubly-connected domains of the well-known Bergman Kernel Method for simply connected domains.

For an introduction to these methods, I suggest you take a look at this document, sections 1.3 and 2.5.

Bounded spherical derivative implies finite order

2011-10-27T13:48:56Z

Hi,

Let $f$ be an entire function. The spherical derivative $\rho(f)$ is defined by $$\rho(f)(z):= \frac{|f'(z)|}{1+|f(z)|^2}.$$

A result from Clunie and Hayman states that if $\rho(f)$ is bounded, then $f$ is of exponential type. The proof uses the machinery of Nevanlinna's theory of value distribution.

My question is the following :

Is there an elementary proof that if $\rho(f)$ is bounded, then $f$ is of finite order?

(Note that this is a weaker result, since I'm only asking for finite order here). Finite order means that there exists constants $K$ and $\alpha$ such that $$|f(z)| \leq Ke^{|z|^\alpha}$$ for all $z$.

Motivation : I'm interested in this because it would lead to a quick proof of Picard's little theorem. Indeed, if there exists a non-constant entire function which omits $0$ and $1$, then it is possible to obtain (using normal families techniques) a non-constant entire function $f$ which omits $0$ and $1$ and that has bounded spherical derivative. Write $f = e^g$ for some entire function $g$. Since $f$ is of finite order, $g$ is a polynomial. But $f$ does not take the value $1$, so $g$ must be constant, a contradiction.

Any reference is welcome. Thank you, Malik.

EDIT I asked the question on math.stackexchange.

Is the following function decreasing on $(0,1)$?

2012-07-19T17:13:48Z

Hi,

I asked some time ago the following question on math.stackexchange, but I ask it here too since it remains unanswered.

The question concerns a function I encountered during research :

$$f(k):= k K(k) \sinh \left(\frac{\pi}{2} \frac{K(\sqrt{1-k^2})}{K(k)}\right)$$ for $k \in (0,1)$.

Here $K$ is the Complete elliptic integral of the first kind, defined by $$K(k):= \int_{0}^{1} \frac{dt}{\sqrt{1-t^2} \sqrt{1-k^2t^2}}.$$

More specifically, my question is the following :

Is $f$ decreasing on $(0,1)$?

This seems to be true, as the graph below suggests (obtained with Maple) :

In fact, as remarked by Henry Cohn, much more seems to be true : all the derivatives of $f$ seem to be negative. This can be seen by looking at the Taylor series expansion of $f$ (see the link to math.stackexchange). The Taylor series expansion seems to have all negative coefficients (except the constant term), and the coefficient of $k^{2j}$ seems to be $\pi$ times a rational number with denominator dividing $16^j$...

Any comment or relevant reference is welcome.

Thank you, Malik

EDIT (20-07-2012) It was remarked by J.M. on M.SE that $f$ can be written as $$f(k)=kK(k)\frac{1-q(k)}{2\sqrt{q(k)}},$$ where $q(k)$ is the Elliptic nome. Maybe this is useful...

Answer by Malik Younsi for Is the following function decreasing on $(0,1)$?

2012-07-24T15:17:44Z

This is more of a long comment to the remark in GH's answer. I did some calculations regarding the expression $$-\frac{q}{1-q} - \sum_{n=1}^{\infty} \frac{8nq^{4n}}{1-q^{4n}} + \sum_{n=1}^{\infty}\frac{4nq^{2n}}{1+q^{2n}}.$$

Keeping the first terms in the two series and estimating the remaining tail as in GH's answer, we obtain the following expression (if I didn't make any mistake..) :

$$-\frac{q}{1-q} - \sum_{n=1}^{k-1} \frac{8nq^{4n}}{1-q^{4n}} + \sum_{n=1}^{k-1}\frac{4nq^{2n}}{1+q^{2n}} + \frac{8(k-1)q^{4(k+1)}-8kq^{4k}}{(1-q^4)^2} + \frac{4kq^{2k}-4(k-1)q^{2(k+1)}}{(1-q^2)^2}.$$

Given small $\epsilon$, we want to find $k$ such that the above is negative on $(0,1-\epsilon)$.

With $k=1$, we get GH's result that the function is negative for $q \in (0,0.37795)$. With $k=8$, Maple gives that the above is negative for $q \in (0, 0.78177)$. Large values of $k$ take longer to solve explicitely, but plotting the function with $k=30$ gives a negative function for $q \in (0,\alpha)$ where $\alpha>0.9$.

Answer by Malik Younsi for Not especially famous, long-open problems which anyone can understand

2012-06-21T23:46:38Z

I always enjoyed telling people about the Inscribed square problem :

Does every (Jordan) curve in the plane contain all four vertices of some square?

On holomorphic branched coverings of a domain in the plane to the unit disk

2012-06-13T14:39:28Z

Hi,

This question is partly motivated by my answer to this question on math.stackexchange.

Let $\Omega$ be a bounded $n$-connected domain in the plane, bounded by $n$ pairwise disjoint Jordan curves.

It was proved by Ahlfors that by solving an extremal problem (related to so-called analytic capacity), one can obtain a function $f$ holomorphic in $\Omega$ with the following properties :

$f$ is an $n$-to-$1$ branched covering of $\Omega$ onto the unit disk $\mathbb{D}$,
$f$ extends continuously to the boundary of $\Omega$, and maps each boundary curve homeomorphically onto the unit circle.

See e.g. Krantz's Geometric function theory: explorations in complex analysis, theorem 4.5.9.

My question is :

Is there another (perhaps more intuitive) way to see that such a function necessarily exists?

Note :

The case $n=1$ is of the above result is essentially the Riemann mapping theorem (for Jordan domains) + Carathéodory's theorem on the boundary behaviour of conformal mappings.

Answer by Malik Younsi for Blackbox Theorems

2012-06-14T01:36:13Z

I think the Uniformization theorem is an example of blackbox theorem : any simply connected Riemann surface is conformally equivalent to either the open unit disk, the complex plane or the Riemann sphere.

On the set of divergence to infinity for sequences of positive continuous functions

2012-05-22T12:39:10Z

Hi,

I have asked this question on math.stackexchange but it has not received much attention, so I ask it here.

This question is partly motivated by this one, which contains an example of a sequence $(f_n)$ of positive continuous functions on $\mathbb{R}$ such that $$f_n(x) \rightarrow \infty$$ if and only if $x \in \mathbb{Q}$.

My question is the following :

For a given sequence of positive continuous functions $(f_n)$ on $\mathbb{R}$, denote by $S((f_n))$ the set of divergence to $\infty$ :

$$S((f_n)):={ x \in \mathbb{R}: f_n(x) \rightarrow \infty }$$

Is there a necessary and sufficient condition for a given set $S$ to be $S((f_n))$ for some sequence $(f_n)$?

As noted in the comments at math.stackexchange, a necessary condition is that $S$ must be a countable interesection of $F_{\sigma}$'s... But is this sufficient?

Thank you, Malik

What is the status of the subadditivity problem for analytic capacity?

2012-04-12T18:14:07Z

Hi,

Here is another question that concerns analytic capacity. For a compact set $K$ in the plane, define the analytic capacity of $K$ by $$\gamma(K):=\sup|f'(\infty)|,$$ where the supremum is taken over all functions $f$ holomorphic and bounded by $1$ in the complement of $K$. Here

$$f'(∞)=\lim_{z \rightarrow \infty}z(f(z)−f(\infty)).$$

It was conjectured by Vitushkin, in view to applications in approximation theory, that analytic capacity is semi-additive, i.e. there exists a universal constant $C$ such that $$\gamma(E \cup F) \leq C(\gamma(E) + \gamma(F)).$$

This was proved by Tolsa in 2003. From what I know, wether or not analytic capacity is subadditive, i.e. if we can take $C=1$, is still open. However, I can't find much about this problem in the literature, hence the question :

What is the status on this problem? Has it been proved that analytic capacity is indeed subadditive in some particular cases?

The only thing I found about this is an article by Suita, "On subadditivity of analytic capacity for two continua.", in which it is proved that analytic capacity is subadditive in the case where $E,F$ are disjoint connected compact sets. The proof relies deeply on the fact that for compact connected sets, analytic capacity equals logarithmic capacity, which does not hold in the disconnected case.

Any reference about this is welcome.

Thank you, Malik

Is analytic capacity continuous from below?

2012-01-20T15:54:03Z

I've been wondering about the following, I don't know if anyone knows the answer :

For a compact set $K$ in the complex plane, define the analytic capacity of $K$ by $$\gamma(K) := \sup |f'(\infty)|$$ where the supremum is taken over all functions $f$ holomorphic and bounded by $1$ in the complement of $K$ : $f \in H^{\infty}(\mathbb{C}_ {\infty} \setminus K)$, $\|f\|_{\infty} \leq 1$. Here

$$f'(\infty) = \lim_{z \rightarrow \infty} z(f(z)-f(\infty)).$$

A theorem due to Ahlfors states that for each compact $K$, there always exists a unique (in the unbounded component of the complement of $K$) function $F$, called the Ahlfors function of $K$, such that $F \in H^{\infty}(\mathbb{C}_ {\infty} \setminus K)$, $\|F\|_{\infty} \leq 1$, and $F'(\infty)=\gamma(K)$.

It's not hard to show that $\gamma$ is continuous from above : if $(K_n)$ is a decreasing sequence of compact sets, then $$\gamma(\cap_n K_n) = \lim_{n\rightarrow \infty} \gamma(K_n).$$ This essentially follows from Montel's theorem and the fact that $\gamma(E) \subseteq \gamma(F)$ whenever $E \subseteq F$.

My question is the following :

Is analytic capacity continuous from below? More precisely, if $(K_n)$ is a sequence of compact sets such that $$K_1 \subseteq K_2 \subseteq K_3 \subseteq \dots$$ and such that $K:=\cup_n K_n$ is compact, then is it true that $\gamma(K) = \lim_{n \rightarrow \infty} \gamma(K_n)?$

I could not find anything in the litterature.

Thank you, Malik

EDIT ( 12-01-2012) I edit the question to add what I know so far about this question :

As pointed out by Fedja in the comments, analytic capacity is comparable to a quantity which is continuous from below, see the article "Painleve's problem and the semiadditivity of analytic capacity" by Xavier Tolsa.
The answer is yes if the compact sets $K_n$ and $K$ are connected. Indeed, for connected compact sets, analytic capacity is equal to logarithmic capacity, and logarithmic capacity is continuous from below.
The answer is yes if $K$ is a compact set whose boundary consists of a finite number of analytic and pairwise disjoint Jordan curves, provided we replace the condition $K:=\cup_n K_n$ by the condition that each compact subset of the interior of $K$ is eventually contained in some $K_n$. This mainly follows from the fact that in this case, the Ahlfors function of $K$ extends analytically across the boundary of $K$. See for example the book "analytic capacity and measure" by Garnett, p. 18.

EDIT ( 31-01-2012) I contacted Xavier Tolsa, and according to him, it's an open problem, related to the so called capacitability problem. It's not known if the Borel sets are capacitable.

I'll leave the question open though, because I'd be very interested to hear about sufficient conditions or similar results.

Answer by Malik Younsi for Reference for complex analysis jargon

2011-12-02T01:56:50Z

I really recommend the book Potential Theory in the complex plane" by Thomas Ransford.

It's a very nice book with exercises and it covers each of the 5 points you mentioned.

Answer by Malik Younsi for holomorphic function with special decreasing property

2011-10-19T15:56:14Z

The answer seems to be no.The quantity $$\rho(f(z)):= \frac{|f'(z)|}{1+|f(z)|^2}$$ is called the spherical derivative of $f$. Since you're interested in the behaviour of $z\rho(f(z))$ near $\infty$, then you should really take a look at Lehto and Virtanen's article :

MR0087747 (19,404a) Lehto, Olli; Virtanen, K. I. On the behaviour of meromorphic functions in the neighbourhood of an isolated singularity. Ann. Acad. Sci. Fenn. Ser. A. I. 1957 (1957), no. 240, 9 pp. (Reviewer: A. J. Lohwater), 30.0X

From the abstract :

It is proved first that if $f(z)$ is single-valued and meromorphic in a neighborhood of the isolated essential singularity $z=∞$, then there exists a universal constant $k>0$ such that $$\limsup_{z→∞}|z|\rho(f(z))≥k$$ for all such $f(z)$, while, for arbitrary $\epsilon>0$, there exist functions for which

$$\limsup_{z \rightarrow \infty} |z|\rho(f(z)) < k+\epsilon.$$

Removable sets for harmonic functions and hardy spaces of general domains

2011-04-11T21:37:42Z

Hi,

Let $\Omega$ be a domain of the complex plane. The Hardy space $H^p(\Omega)$ is defined, for $1 \leq p<\infty$, as the class of functions $f$ that are holomorphic on $\Omega$ such that $|f|^p$ has a harmonic majorant on $\Omega$, i.e. there is a function $u$ harmonic on $\Omega$ such that $$|f(z)|^p \leq u(z) $$ for all $z \in \Omega$.

For $p=\infty$, $H^\infty(\Omega)$ is the class of bounded holomorphic functions on $\Omega$.

I came upon the following question :

Let $E$ be a compact subset of the real line, and suppose that $E$ has zero length. Let $\Omega$ be the complement of $E$. Does $H^p(\Omega)$ consist only of the constant functions?

For $p=\infty$, the answer is yes : one can use Cauchy's formula to extend any bounded holomorphic function on $\Omega$ to a bounded holomorphic function on $\mathbb{C}$, and that function is now constant by Liouville's theorem.

What about the case $p < \infty$? Any reference would be quite useful.

Thank you, Malik

Answer by Malik Younsi for German mathematical terms like "Nullstellensatz"

2011-04-19T23:44:26Z

There is Ahlfor's scheibensatz in complex function theory, which is a generalization of Ahlfors five islands theorem

Answer by Malik Younsi for Examples of common false beliefs in mathematics.

2011-04-14T18:19:31Z

I have heard the following a few times :

"If $f$ is holomorphic on a region $\Omega$ and not one-to-one, then $f'$ must vanish somewhere in $\Omega$."

$f(z)=e^z$ of course is a counterexample.

On proving that a certain set is not empty by proving that it is actually large

2010-08-03T14:32:24Z

It happens occasionally that one can prove that a given set is not empty by proving that it is actually large. The word "large" here may refer to different properties.

For example, one can prove that a certain set is not empty by proving that its cardinality is big, as in the proof that there exist transcendental numbers : The set of algebraic numbers is countable, but the set of real numbers is uncountable, so there is uncountably many transcendental numbers.

One could also prove that a certain set is not empty by proving, for example, that it has positive measure, that it is dense, etc.

What are some good examples of such proofs?

Answer by Malik Younsi for Math puzzles for dinner

2010-06-24T13:00:27Z

I really like the following puzzle, called the blue-eyed islanders problem, taken from Professor Tao's blog :

"There is an island upon which a tribe resides. The tribe consists of 1000 people, with various eye colours. Yet, their religion forbids them to know their own eye color, or even to discuss the topic; thus, each resident can (and does) see the eye colors of all other residents, but has no way of discovering his or her own (there are no reflective surfaces). If a tribesperson does discover his or her own eye color, then their religion compels them to commit ritual suicide at noon the following day in the village square for all to witness. All the tribespeople are highly logical and devout, and they all know that each other is also highly logical and devout (and they all know that they all know that each other is highly logical and devout, and so forth).

Of the 1000 islanders, it turns out that 100 of them have blue eyes and 900 of them have brown eyes, although the islanders are not initially aware of these statistics (each of them can of course only see 999 of the 1000 tribespeople).

One day, a blue-eyed foreigner visits to the island and wins the complete trust of the tribe.

One evening, he addresses the entire tribe to thank them for their hospitality.

However, not knowing the customs, the foreigner makes the mistake of mentioning eye color in his address, remarking “how unusual it is to see another blue-eyed person like myself in this region of the world”.

What effect, if anything, does this faux pas have on the tribe?"

For those of you interested, there is a huge discussion of the problem at http://terrytao.wordpress.com/2008/02/05/the-blue-eyed-islanders-puzzle/

Malik

On matrices that almost have the same eigenvalues

2010-06-17T16:16:32Z

Let $A$ and $B$ be two $4$ by $4$ matrices. Using Newton's identites, one can prove that if $$det(A) = det(B)$$ and $$tr(A^i) = tr(B^i)$$ for $i=1,2,3$, then $A$ and $B$ have the same characteristic polynomial, thus the same eigenvalues.

I'm interested in pairs of matrices $A$ and $B$ that satisfy all those equations except the last one, i.e. $$det(A)=det(B)$$ $$tr(A)=tr(B)$$ $$tr(A^2)=tr(B^2)$$ but $tr(A^3) \neq tr(B^3)$.

Does anyone know how to generate such matrices? Have they ever been studied? A reference would be nice.

Thank you, Malik

Answer by Malik Younsi for Examples of common false beliefs in mathematics.

2010-06-01T23:52:30Z

As a teaching assistant in an elementary number theory course, I've seen the following quite often :

If $a$ divides $bc$ and $a$ does not divide $b$, then $a$ divides $c$.

That's of course true if $a$ is prime, but people seem to forget that hypothesis.

In a Banach algebra, do ab and ba have almost the same exponential spectrum?

2009-10-24T21:39:14Z

Let $A$ be a complex Banach algebra with identity 1. Define the exponential spectrum $e(x)$ of an element $x\in A$ by $$e(x)= \{\lambda\in\mathbb{C}: x-\lambda1 \notin G_1(A)\},$$ where $G_1(A)$ is the connected component of the group of invertibles $G(A)$ that contains the identity.

Is it true that $e(ab)\cup\{0\} = e(ba)\cup\{0\}$ for all $a,b \in A$?

Equivalently, is it true that $1-ab$ is in $G_1(A)$ if and only if $1-ba$ is in $G_1(A)$, for all $a,b \in A$?

Note: The usual spectrum has this property.

Just an additional note:

We have $e(ab)\cup\{0\} = e(ba)\cup\{0\}$ for all $a,b \in A$ if

1) The group of invertibles of $A$ is connected, because then the exponential spectrum of any element is just the usual spectrum of that element.

2) The set $Z(A)G(A) = \{ab: a \in Z(A), b\in G(A)\}$ is dense in $A$, where $Z(A)$ is the center of $A$. (One can prove this). In particular, we have $e(ab)\cup\{0\} = e(ba)\cup\{0\}$ for all $a,b \in A$ if the invertibles are dense in $A$.

3) $A$ is commutative, clearly.

But what about other Banach algebras? Can someone provide a counterexample?

Answer by Malik Younsi for Famous mathematical quotes

2009-11-29T21:53:52Z

"The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?" — Jerry Bona

Answer by Malik Younsi for Most helpful heuristic?

2009-10-28T16:52:11Z

A very interesting heuristic is a principle in complex variables called "Bloch's heuristic principle".

Bloch's principle is about families of analytic functions called "normal". A family F of analytic functions on a domain D is called normal on D if every sequence of functions of F has a subsequence that converges uniformly on compact subsets (either to an analytic function or to infinity). Normal families are very studied for their applications in complex dynamics.

Bloch's principle goes as follows :

A family of analytic functions on a domain D having a property in common is most likely to be normal if there is no non-constant entire function having this property on the whole complex plane.

There are many examples of Bloch's principle. For example, take the property of being bounded : a well known theorem of Montel says that a family of analytic functions on a domain D which is uniformly bounded is necessarily normal on D, and Liouville's theorem says that there is no non-constant entire bounded function.

Or, take the property of omitting two distinct complex values. Again, a theorem of Montel says that a family of analytic functions on a domain D such that each function omits a,b in C, a different than b, is normal on D. The version for the whole complex plane is a well known theorem of Picard, that says that there is no non-constant entire function that omits two distinct complex values.

However, there are many counter-examples to Bloch's principle as it is stated, but it can be transformed into a rigourous theorem that goes like "If a property satisfies these conditions, then bloch's principle is respected".

I wouldn't qualify Bloch's principle as "most helpful", but it is certainly interesting.

Malik

Answer by Malik Younsi for Which sequences can be extended to analytic functions? (e. g., Ackermann's function)

2009-10-28T13:50:42Z

Actually, there is an even stronger result, often called the interpolation theorem, which follows from a well known theorem of Mittag-Leffler :

Let (z_n) be sequence of complex numbers with no accumulation point. For each n, let l(n) be any integer greater or equal to 1 and (a_nk) (0 <= k <= l(n) ) complex numbers. Then there exists an entire function g(z) such that

g^(k)(z_n) = a_nk

for every n>=1 and every 0 <= k <= l(n)

That is, you can fix values for the derivative at the z_j's.

Comment by Malik Younsi

2013-03-12T20:20:15Z

@Giuseppe : Thank you!

Comment by Malik Younsi

2013-03-12T20:19:33Z

Thank you very much for the interesting references, I will take a look at them.

Comment by Malik Younsi

2013-03-10T12:17:25Z

I'm having some problem with LaTex in the definition of $Z_f$... Does someone know how to fix that?

Comment by Malik Younsi

2013-02-20T20:24:48Z

The proof is very elegant, and it appears in "Proof from the Book" by Aigner and Ziegler.

Comment by Malik Younsi

2013-01-28T17:44:13Z

@EmilJeřábek : I didn't know about the removability criterion for the Hölder functions. This is very interesting.

Comment by Malik Younsi

2013-01-26T20:15:40Z

@MarcPalm : Yes, indeed, I see now why we need $\log$. But still, it is a good idea to work with joint universality. Thanks, +1 !

Comment by Malik Younsi

2013-01-25T20:05:03Z

I mean, the result I'm looking for should contain Voronin's Universality Theorem as a particular case (the case when the complement of $K$ is connected).

Comment by Malik Younsi

2013-01-25T20:03:56Z

Oh I see, indeed I misread $K_0$ for $K$, sorry about that! But I don't understand why you work with $\log$..?

Comment by Malik Younsi

2013-01-24T15:57:25Z

@MarcPalm : yes indeed, but these generalizations of the universality theorem still require $K$ to have connected complement.

Comment by Malik Younsi

2013-01-24T15:50:50Z

See my new edit. I hope it clarifies what I'm looking for here.

Comment by Malik Younsi

2013-01-24T15:41:16Z

This joint universality result is interesting, but what I'm looking for is really a result for compact sets with disconnected complements. Also, just to clarify : I don't want to remove the non-vanishing assumption. I'm fine with it!

Comment by Malik Younsi

2013-01-23T16:30:01Z

Yes, after searing through the literature, it appears that it is still open wether or not one can replace "non-vanishing on $K$" by "non-vanishing in the interior of $K$ " in the statement of the Universality Theorem. This is quite interesting!

Comment by Malik Younsi

2013-01-23T15:52:48Z

Thank you for your answers. Concerning Question 2, yes I know that RH implies that if you approximate uniformly $f$ on $K$, then $f$ cannot vanish in the interior of $K$. This is because $f$ is holomorphic there. But a priori $f$ could vanish somewhere on the boundary of $K$, without contradicting RH!

Comment by Malik Younsi

2013-01-23T13:55:51Z

Ok, it seems that the answer to my question 2 is unknown! See Conjecture 1 in the article arxiv.org/abs/1010.0850

Comment by Malik Younsi

2013-01-23T13:53:21Z

In fact, this is an open problem I've seen in a few articles. 3. Do you think that your proof can be somehow yield information on the lower density of such $t$'s that work?