User malik younsi - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T14:29:01Z http://mathoverflow.net/feeds/user/1162 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26162/what-can-be-said-about-pairs-of-matrices-p-q-that-satisfies-p-1t-circ-p What can be said about pairs of matrices P,Q that satisfies $(P^{-1})^T \circ P = (Q^{-1})^T \circ Q$ ? Malik Younsi 2010-05-27T16:07:49Z 2013-05-22T18:24:22Z <p>Let $P,Q$ be $n$ by $n$ invertible matrices. Suppose further that $P$ and $Q$ satisfies the following equation :</p> <p>$$(P^{-1})^T \circ P = (Q^{-1})^T \circ Q$$</p> <p>where $\circ$ denotes the Hadamard matrix product, which is simply the entrywise product. </p> <p>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</p> <p>$$tr(P^{-1}DPE) = tr(Q^{-1}DQE)$$ for all diagonal matrices $D$ and $E$.</p> <p>References in the litterature about matrices of the form $(P^{-1})^T \circ P$ would help too. Thank you, Malik</p> http://mathoverflow.net/questions/126420/functions-of-one-complex-variable-geometric-theory/126437#126437 Answer by Malik Younsi for functions of one complex variable: geometric theory Malik Younsi 2013-04-03T19:26:52Z 2013-04-03T19:26:52Z <p>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).</p> http://mathoverflow.net/questions/124146/on-the-set-of-zero-radial-limits-of-bounded-analytic-functions On the set of zero radial limits of bounded analytic functions Malik Younsi 2013-03-10T12:16:30Z 2013-03-10T12:58:29Z <p>Hi,</p> <p>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 :</p> <p><code>$$Z_f:= \{e^{i\theta} \in \mathbb{T} : \lim_{r \rightarrow 1}f(re^{i\theta})=0\}.$$</code></p> <p>Then it is also well-known that $Z_f$ has measure zero.</p> <p>My question is the following :</p> <p>For which sets $E \subseteq \mathbb{T}$ of measure zero does there exist a bounded analytic function $f$ in $\mathbb{D}$ such that $E=Z_f$?</p> <p>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 $f$.</p> http://mathoverflow.net/questions/120118/when-does-continuity-imply-holomorphy/120122#120122 Answer by Malik Younsi for When does continuity imply holomorphy? Malik Younsi 2013-01-28T16:10:28Z 2013-01-28T16:29:00Z <p>The following is too long for a comment. I will suppose here that your set of measure zero is compact.</p> <p>In this case, Your question is closely related to so-called <em>continuous analytic capacity</em>. Let $K$ be a compact set in the plane, and let $\Omega$ be the complement of $K$ with respect to $\mathbb{C}_\infty$. The <em>continuous analytic capacity</em> $\alpha(K)$ is defined as</p> <p>$$\alpha(K):= \sup { |f'(\infty)| : f \in A(\Omega), \|f\|_{\infty} \leq 1 },$$</p> <p>where $A(\Omega)$ is the set of all functions holomorphic in $\Omega$ that extend continuously to $\mathbb{C}_\infty$.</p> <p>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 $U$. </p> <p>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.</p> <p>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. </p> <p>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."</p> http://mathoverflow.net/questions/119499/on-the-universality-of-the-riemann-zeta-function On the Universality of the Riemann zeta-function Malik Younsi 2013-01-21T18:14:45Z 2013-01-28T10:46:09Z <p>Hi,</p> <p>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.</p> <p>First, recall Voronin's remarkable theorem on the Universality of the Riemann zeta-function :</p> <p>Let $K$ be a compact subset with connected complement lying in the strip ${1/2 &lt; \operatorname{Re}(z)&lt;1}$, 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 $$\max_{z \in K} |\zeta(z+it)-f(z)|&lt;\epsilon.$$ Even more : the lower density of the set of such $t$'s is positive..!</p> <p>Note that of course, the hypothesis that the complement of $K$ is connected is essential in the above theorem.</p> <p>My question is the following :</p> <p><strong>Is there some sort of (modified) zeta-function universality-like result for compact sets $K$ with <em>disconnected</em> complements? For example, if $\mathbb{C}_\infty \setminus K$ has a finite number of components?</strong></p> <p><strong>EDIT</strong></p> <p>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 <strong>some</strong> 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 $K$. </p> <p><strong>2nd EDIT</strong></p> <p>Let me explain what I was looking for here. Basically, I'd like to know if there exists a result of the following form :</p> <p>Let $K$ be a compact subset <strong>whose complement has finitely many components</strong> lying in the strip ${1/2 &lt; \operatorname{Re}(z)&lt;1}$, 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...</p> <p><em>Here insert some uniform approximation of $f$ on $K$ by a function involving the Riemann zeta-function</em></p> <p>Furthermore, in the case when $K$ has <strong>connected</strong> complement, I would like the above result to reduce to Voronin's Theorem.</p> <p>In summary, I want to know if there exists a generalization of Voronin's Theorem to compact sets whose complement have finitely many components.</p> <p>Thank you, Malik</p> http://mathoverflow.net/questions/117633/a-question-about-the-limit-of-a-sequence-of-pointwise-convergent-analytic-funtion/117643#117643 Answer by Malik Younsi for A question about the limit of a sequence of pointwise convergent analytic funtions Malik Younsi 2012-12-30T15:38:07Z 2012-12-30T20:48:55Z <p>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. </p> <p>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. <a href="http://math.stackexchange.com/questions/113240/pointwise-convergence-of-sequences-of-holomorphic-functions-to-holomorphic-funct" rel="nofollow">this question</a></p> <p><strong>EDIT</strong></p> <p>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 $p_n(0) \rightarrow 1$. </p> http://mathoverflow.net/questions/108550/inequivalent-complete-norms-and-the-axiom-of-choice Inequivalent complete norms and the axiom of choice Malik Younsi 2012-10-01T15:28:17Z 2012-10-01T15:54:02Z <p>Hi,</p> <p>I've been wondering about the following :</p> <p>Is it possible, without the axiom of choice, to have two inequivalent complete norms on a vector space?</p> <p>All the examples of inequivalent complete norms I've seen rely on the existence of Hamel bases...</p> <p>This is most likely well-known, but I'd be glad if someone could provide a good reference.</p> <p>Thank you, Malik</p> http://mathoverflow.net/questions/107508/is-there-an-explicit-formula-for-the-modulus-of-an-annulus-given-a-parameterizati/107560#107560 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? Malik Younsi 2012-09-19T12:52:02Z 2012-09-19T12:52:02Z <p>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. </p> <p>For an introduction to these methods, I suggest you take a look at <a href="http://194.42.1.1/~nickp/lectncm.pdf" rel="nofollow">this document</a>, sections 1.3 and 2.5.</p> http://mathoverflow.net/questions/79265/bounded-spherical-derivative-implies-finite-order Bounded spherical derivative implies finite order Malik Younsi 2011-10-27T13:48:56Z 2012-08-03T19:00:16Z <p>Hi,</p> <p>Let $f$ be an entire function. The <em>spherical derivative</em> $\rho(f)$ is defined by $$\rho(f)(z):= \frac{|f'(z)|}{1+|f(z)|^2}.$$</p> <p>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.</p> <p>My question is the following :</p> <p>Is there an elementary proof that if $\rho(f)$ is bounded, then $f$ is of <strong>finite order</strong>?</p> <p>(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 $z$.</p> <p><strong>Motivation :</strong> 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 <em>bounded</em> 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.</p> <p>Any reference is welcome. Thank you, Malik.</p> <p>EDIT I asked the <a href="http://math.stackexchange.com/questions/77886/bounded-spherical-derivative-implies-finite-order/78182#78182" rel="nofollow">question </a> on math.stackexchange.</p> http://mathoverflow.net/questions/102685/is-the-following-function-decreasing-on-0-1 Is the following function decreasing on $(0,1)$? Malik Younsi 2012-07-19T17:13:48Z 2012-07-24T15:17:44Z <p>Hi,</p> <p>I asked some time ago the following <a href="http://math.stackexchange.com/questions/167498/is-this-function-decreasing-on-0-1" rel="nofollow">question</a> on math.stackexchange, but I ask it here too since it remains unanswered.</p> <p>The question concerns a function I encountered during research :</p> <p>$$f(k):= k K(k) \sinh \left(\frac{\pi}{2} \frac{K(\sqrt{1-k^2})}{K(k)}\right)$$ for $k \in (0,1)$. </p> <p>Here $K$ is the <a href="http://en.wikipedia.org/wiki/Elliptic_integral#Complete_elliptic_integral_of_the_first_kind" rel="nofollow">Complete elliptic integral of the first kind</a>, defined by $$K(k):= \int_{0}^{1} \frac{dt}{\sqrt{1-t^2} \sqrt{1-k^2t^2}}.$$</p> <p>More specifically, my question is the following :</p> <p><strong>Is $f$ decreasing on $(0,1)$?</strong></p> <p>This seems to be true, as the graph below suggests (obtained with Maple) :</p> <p><img src="http://i.stack.imgur.com/cOtoK.jpg" alt="graph of $f$"></p> <p>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 $16^j$...</p> <p>Any comment or relevant reference is welcome.</p> <p>Thank you, Malik</p> <p><strong>EDIT (20-07-2012)</strong> 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 <a href="http://mathworld.wolfram.com/Nome.html" rel="nofollow">Elliptic nome</a>. Maybe this is useful...</p> http://mathoverflow.net/questions/102685/is-the-following-function-decreasing-on-0-1/103010#103010 Answer by Malik Younsi for Is the following function decreasing on $(0,1)$? Malik Younsi 2012-07-24T15:17:44Z 2012-07-24T15:17:44Z <p>This is more of a long comment to the remark in GH's answer. I did some calculations regarding the expression $$-\frac{q}{1-q} - \sum_{n=1}^{\infty} \frac{8nq^{4n}}{1-q^{4n}} + \sum_{n=1}^{\infty}\frac{4nq^{2n}}{1+q^{2n}}.$$</p> <p>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..) :</p> <p>$$-\frac{q}{1-q} - \sum_{n=1}^{k-1} \frac{8nq^{4n}}{1-q^{4n}} + \sum_{n=1}^{k-1}\frac{4nq^{2n}}{1+q^{2n}} + \frac{8(k-1)q^{4(k+1)}-8kq^{4k}}{(1-q^4)^2} + \frac{4kq^{2k}-4(k-1)q^{2(k+1)}}{(1-q^2)^2}.$$</p> <p>Given small $\epsilon$, we want to find $k$ such that the above is negative on $(0,1-\epsilon)$.</p> <p>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 $\alpha>0.9$.</p> http://mathoverflow.net/questions/100265/not-especially-famous-long-open-problems-which-anyone-can-understand/100312#100312 Answer by Malik Younsi for Not especially famous, long-open problems which anyone can understand Malik Younsi 2012-06-21T23:46:38Z 2012-06-21T23:46:38Z <p>I always enjoyed telling people about the <a href="http://en.wikipedia.org/wiki/Inscribed_square_problem" rel="nofollow">Inscribed square problem</a> :</p> <p>Does every (Jordan) curve in the plane contain all four vertices of some square?</p> http://mathoverflow.net/questions/99454/on-holomorphic-branched-coverings-of-a-domain-in-the-plane-to-the-unit-disk On holomorphic branched coverings of a domain in the plane to the unit disk Malik Younsi 2012-06-13T14:39:28Z 2012-06-18T16:40:46Z <p>Hi,</p> <p>This question is partly motivated by my answer to <a href="http://math.stackexchange.com/questions/95570/does-constant-modulus-on-boundary-of-annulus-imply-constant-function/95581#95581" rel="nofollow">this question</a> on math.stackexchange.</p> <p>Let $\Omega$ be a bounded $n$-connected domain in the plane, bounded by $n$ pairwise disjoint Jordan curves. </p> <p>It was proved by Ahlfors that by solving an extremal problem (related to so-called <em>analytic capacity</em>), one can obtain a function $f$ holomorphic in $\Omega$ with the following properties :</p> <ol> <li><p>$f$ is an $n$-to-$1$ branched covering of $\Omega$ onto the unit disk $\mathbb{D}$,</p></li> <li><p>$f$ extends continuously to the boundary of $\Omega$, and maps each boundary curve homeomorphically onto the unit circle.</p></li> </ol> <p>See e.g. Krantz's <em>Geometric function theory: explorations in complex analysis</em>, theorem 4.5.9.</p> <p>My question is :</p> <p><strong>Is there another (perhaps more intuitive) way to see that such a function necessarily exists?</strong></p> <p>Note :</p> <p>The case $n=1$ is of the above result is essentially the Riemann mapping theorem (for Jordan domains) + <a href="http://en.wikipedia.org/wiki/Carath%25C3%25A9odory%2527s_theorem_%2528conformal_mapping%2529" rel="nofollow">Carathéodory's theorem</a> on the boundary behaviour of conformal mappings.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99544#99544 Answer by Malik Younsi for Blackbox Theorems Malik Younsi 2012-06-14T01:36:13Z 2012-06-14T01:36:13Z <p>I think the <a href="http://en.wikipedia.org/wiki/Uniformization_theorem" rel="nofollow"><strong>Uniformization theorem</strong></a> is an example of <em>blackbox theorem</em> : any simply connected Riemann surface is conformally equivalent to either the open unit disk, the complex plane or the Riemann sphere.</p> http://mathoverflow.net/questions/97654/on-the-set-of-divergence-to-infinity-for-sequences-of-positive-continuous-functio On the set of divergence to infinity for sequences of positive continuous functions Malik Younsi 2012-05-22T12:39:10Z 2012-05-24T14:06:16Z <p>Hi,</p> <p>I have asked this question on <a href="http://math.stackexchange.com/questions/146280/on-the-set-of-divergence-to-infinity-for-sequences-of-positive-continuous-functi" rel="nofollow">math.stackexchange</a> but it has not received much attention, so I ask it here.</p> <p>This question is partly motivated by this <a href="http://mathoverflow.net/questions/34371/on-the-existence-of-a-sequence-of-positive-continuous-functions" rel="nofollow">one</a>, 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 $x \in \mathbb{Q}$.</p> <p>My question is the following :</p> <p>For a given sequence of positive continuous functions $(f_n)$ on $\mathbb{R}$, denote by $S((f_n))$ the set of divergence to $\infty$ :</p> <p>$$S((f_n)):={ x \in \mathbb{R}: f_n(x) \rightarrow \infty }$$</p> <p>Is there a necessary and sufficient condition for a given set $S$ to be $S((f_n))$ for some sequence $(f_n)$?</p> <p>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?</p> <p>Thank you, Malik</p> http://mathoverflow.net/questions/93891/what-is-the-status-of-the-subadditivity-problem-for-analytic-capacity What is the status of the subadditivity problem for analytic capacity? Malik Younsi 2012-04-12T18:14:07Z 2012-04-12T18:14:07Z <p>Hi,</p> <p>Here is another question that concerns analytic capacity. For a compact set $K$ in the plane, define the <em>analytic capacity</em> 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</p> <p>$$f'(∞)=\lim_{z \rightarrow \infty}z(f(z)−f(\infty)).$$</p> <p>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 $$\gamma(E \cup F) \leq C(\gamma(E) + \gamma(F)).$$</p> <p>This was proved by Tolsa in 2003. From what I know, wether or not analytic capacity is <em>subadditive</em>, 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 :</p> <p><strong>What is the status on this problem? Has it been proved that analytic capacity is indeed subadditive in some particular cases?</strong></p> <p>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.</p> <p>Any reference about this is welcome.</p> <p>Thank you, Malik</p> http://mathoverflow.net/questions/86220/is-analytic-capacity-continuous-from-below Is analytic capacity continuous from below? Malik Younsi 2012-01-20T15:54:03Z 2012-01-31T21:36:29Z <p>I've been wondering about the following, I don't know if anyone knows the answer :</p> <p>For a compact set $K$ in the complex plane, define the <em>analytic capacity</em> 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</p> <p>$$f'(\infty) = \lim_{z \rightarrow \infty} z(f(z)-f(\infty)).$$</p> <p>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 <em>Ahlfors function</em> of $K$, such that $F \in H^{\infty}(\mathbb{C}_ {\infty} \setminus K)$, $\|F\|_{\infty} \leq 1$, and $F'(\infty)=\gamma(K)$.</p> <p>It's not hard to show that $\gamma$ is <em>continuous from above</em> : 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 $E \subseteq F$.</p> <p>My question is the following :</p> <p>Is analytic capacity <em>continuous from below</em>? 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 $\gamma(K) = \lim_{n \rightarrow \infty} \gamma(K_n)?$</p> <p>I could not find anything in the litterature.</p> <p>Thank you, Malik</p> <p><strong>EDIT ( 12-01-2012)</strong> I edit the question to add what I know so far about this question :</p> <ol> <li><p>As pointed out by Fedja in the comments, analytic capacity is <em>comparable</em> to a quantity which is continuous from below, see the article "Painleve's problem and the semiadditivity of analytic capacity" by Xavier Tolsa.</p></li> <li><p>The answer is yes if the compact sets $K_n$ and $K$ are connected. Indeed, for connected compact sets, analytic capacity is equal to <em>logarithmic capacity</em>, and logarithmic capacity is continuous from below.</p></li> <li><p>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.</p></li> </ol> <p><strong>EDIT ( 31-01-2012)</strong> 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. </p> <p>I'll leave the question open though, because I'd be very interested to hear about sufficient conditions or similar results.</p> http://mathoverflow.net/questions/54647/reference-for-complex-analysis-jargon/82422#82422 Answer by Malik Younsi for Reference for complex analysis jargon Malik Younsi 2011-12-02T01:56:50Z 2011-12-02T01:56:50Z <p>I really recommend the book <a href="http://books.google.ca/books?id=bukn-Rs-t3sC&amp;pg=PR9&amp;lpg=PR9&amp;dq=potential+theory+in+the+complex+plane&amp;source=bl&amp;ots=_kLyeM3BMt&amp;sig=7DuH0JZqkx9_qB_1NfesPslI0dY&amp;hl=en&amp;ei=wi_YTtK2IIHv0gGWyJDLDQ&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=5&amp;ved=0CEMQ6AEwBA#v=onepage&amp;q&amp;f=false" rel="nofollow">Potential Theory in the complex plane" </a> by Thomas Ransford.</p> <p>It's a very nice book with exercises and it covers each of the 5 points you mentioned.</p> http://mathoverflow.net/questions/78560/holomorphic-function-with-special-decreasing-property/78589#78589 Answer by Malik Younsi for holomorphic function with special decreasing property Malik Younsi 2011-10-19T15:56:14Z 2011-10-19T15:56:14Z <p>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 :</p> <p>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</p> <p>From the abstract :</p> <p>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 </p> <p>$$\limsup_{z \rightarrow \infty} |z|\rho(f(z)) &lt; k+\epsilon.$$</p> http://mathoverflow.net/questions/61340/removable-sets-for-harmonic-functions-and-hardy-spaces-of-general-domains Removable sets for harmonic functions and hardy spaces of general domains Malik Younsi 2011-04-11T21:37:42Z 2011-04-22T21:49:09Z <p>Hi,</p> <p>Let $\Omega$ be a domain of the complex plane. The Hardy space $H^p(\Omega)$ is defined, for $1 \leq p&lt;\infty$, 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 $z \in \Omega$.</p> <p>For $p=\infty$, $H^\infty(\Omega)$ is the class of bounded holomorphic functions on $\Omega$.</p> <p>I came upon the following question :</p> <p>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?</p> <p>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.</p> <p>What about the case $p &lt; \infty$? Any reference would be quite useful.</p> <p>Thank you, Malik</p> http://mathoverflow.net/questions/62218/german-mathematical-terms-like-nullstellensatz/62350#62350 Answer by Malik Younsi for German mathematical terms like "Nullstellensatz" Malik Younsi 2011-04-19T23:44:26Z 2011-04-19T23:44:26Z <p>There is <strong>Ahlfor's scheibensatz</strong> in complex function theory, which is a generalization of <a href="http://mathworld.wolfram.com/AhlforsFiveIslandTheorem.html" rel="nofollow">Ahlfors five islands theorem</a></p> http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/61730#61730 Answer by Malik Younsi for Examples of common false beliefs in mathematics. Malik Younsi 2011-04-14T18:19:31Z 2011-04-14T18:19:31Z <p>I have heard the following a few times :</p> <p>"If $f$ is holomorphic on a region $\Omega$ and not one-to-one, then $f'$ must vanish somewhere in $\Omega$."</p> <p>$f(z)=e^z$ of course is a counterexample.</p> http://mathoverflow.net/questions/34390/on-proving-that-a-certain-set-is-not-empty-by-proving-that-it-is-actually-large On proving that a certain set is not empty by proving that it is actually large Malik Younsi 2010-08-03T14:32:24Z 2010-08-04T11:42:31Z <p>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.</p> <p>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.</p> <p>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.</p> <p>What are some good examples of such proofs?</p> http://mathoverflow.net/questions/29323/math-puzzles-for-dinner/29372#29372 Answer by Malik Younsi for Math puzzles for dinner Malik Younsi 2010-06-24T13:00:27Z 2010-06-24T13:00:27Z <p>I really like the following puzzle, called the blue-eyed islanders problem, taken from Professor Tao's blog :</p> <p>"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).</p> <p>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).</p> <p>One day, a blue-eyed foreigner visits to the island and wins the complete trust of the tribe.</p> <p>One evening, he addresses the entire tribe to thank them for their hospitality.</p> <p>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”.</p> <p>What effect, if anything, does this faux pas have on the tribe?"</p> <p>For those of you interested, there is a huge discussion of the problem at <a href="http://terrytao.wordpress.com/2008/02/05/the-blue-eyed-islanders-puzzle/" rel="nofollow">http://terrytao.wordpress.com/2008/02/05/the-blue-eyed-islanders-puzzle/</a></p> <p>Malik</p> http://mathoverflow.net/questions/28530/on-matrices-that-almost-have-the-same-eigenvalues On matrices that almost have the same eigenvalues Malik Younsi 2010-06-17T16:16:32Z 2010-06-17T21:16:13Z <p>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.</p> <p>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 $tr(A^3) \neq tr(B^3)$.</p> <p>Does anyone know how to generate such matrices? Have they ever been studied? A reference would be nice.</p> <p>Thank you, Malik</p> http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/26769#26769 Answer by Malik Younsi for Examples of common false beliefs in mathematics. Malik Younsi 2010-06-01T23:52:30Z 2010-06-01T23:52:30Z <p>As a teaching assistant in an elementary number theory course, I've seen the following quite often :</p> <p>If $a$ divides $bc$ and $a$ does not divide $b$, then $a$ divides $c$.</p> <p>That's of course true if $a$ is prime, but people seem to forget that hypothesis.</p> http://mathoverflow.net/questions/2369/in-a-banach-algebra-do-ab-and-ba-have-almost-the-same-exponential-spectrum In a Banach algebra, do ab and ba have almost the same exponential spectrum? Malik Younsi 2009-10-24T21:39:14Z 2010-04-06T02:04:19Z <p>Let $A$ be a complex Banach algebra with identity 1. Define the exponential spectrum $e(x)$ of an element $x\in A$ by <code>$$e(x)= \{\lambda\in\mathbb{C}: x-\lambda1 \notin G_1(A)\},$$</code> where $G_1(A)$ is the connected component of the group of invertibles $G(A)$ that contains the identity.</p> <blockquote> <p>Is it true that <code>$e(ab)\cup\{0\} = e(ba)\cup\{0\}$</code> for all $a,b \in A$?</p> </blockquote> <p>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 $a,b \in A$?</p> <p>Note: The usual spectrum has this property.</p> <p>Just an additional note:</p> <p>We have <code>$e(ab)\cup\{0\} = e(ba)\cup\{0\}$</code> for all $a,b \in A$ if</p> <p>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.</p> <p>2) The set <code>$Z(A)G(A) = \{ab: a \in Z(A), b\in G(A)\}$</code> is dense in $A$, where $Z(A)$ is the center of $A$. (One can prove this). In particular, we have <code>$e(ab)\cup\{0\} = e(ba)\cup\{0\}$</code> for all $a,b \in A$ if the invertibles are dense in $A$.</p> <p>3) $A$ is commutative, clearly.</p> <p>But what about other Banach algebras? Can someone provide a counterexample?</p> http://mathoverflow.net/questions/7155/famous-mathematical-quotes/7194#7194 Answer by Malik Younsi for Famous mathematical quotes Malik Younsi 2009-11-29T21:53:52Z 2009-11-29T21:53:52Z <p>"The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?" — Jerry Bona</p> http://mathoverflow.net/questions/3079/most-helpful-heuristic/3087#3087 Answer by Malik Younsi for Most helpful heuristic? Malik Younsi 2009-10-28T16:52:11Z 2009-10-28T16:52:11Z <p>A very interesting heuristic is a principle in complex variables called "Bloch's heuristic principle". </p> <p>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.</p> <p>Bloch's principle goes as follows :</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>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".</p> <p>I wouldn't qualify Bloch's principle as "most helpful", but it is certainly interesting.</p> <p>Malik</p> http://mathoverflow.net/questions/2944/which-sequences-can-be-extended-to-analytic-functions-e-g-ackermanns-functi/3052#3052 Answer by Malik Younsi for Which sequences can be extended to analytic functions? (e. g., Ackermann's function) Malik Younsi 2009-10-28T13:50:42Z 2009-10-28T13:50:42Z <p>Actually, there is an even stronger result, often called the interpolation theorem, which follows from a well known theorem of Mittag-Leffler :</p> <p>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 &lt;= k &lt;= l(n) ) complex numbers. Then there exists an entire function g(z) such that</p> <p>g^(k)(z_n) = a_nk</p> <p>for every n>=1 and every 0 &lt;= k &lt;= l(n)</p> <p>That is, you can fix values for the derivative at the z_j's.</p> http://mathoverflow.net/questions/124146/on-the-set-of-zero-radial-limits-of-bounded-analytic-functions Comment by Malik Younsi Malik Younsi 2013-03-12T20:20:15Z 2013-03-12T20:20:15Z @Giuseppe : Thank you! http://mathoverflow.net/questions/124146/on-the-set-of-zero-radial-limits-of-bounded-analytic-functions/124150#124150 Comment by Malik Younsi Malik Younsi 2013-03-12T20:19:33Z 2013-03-12T20:19:33Z Thank you very much for the interesting references, I will take a look at them. http://mathoverflow.net/questions/124146/on-the-set-of-zero-radial-limits-of-bounded-analytic-functions Comment by Malik Younsi Malik Younsi 2013-03-10T12:17:25Z 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? http://mathoverflow.net/questions/122406/undecidability-and-holomorphic-functions-reference-request/122428#122428 Comment by Malik Younsi Malik Younsi 2013-02-20T20:24:48Z 2013-02-20T20:24:48Z The proof is very elegant, and it appears in &quot;Proof from the Book&quot; by Aigner and Ziegler. http://mathoverflow.net/questions/120118/when-does-continuity-imply-holomorphy/120122#120122 Comment by Malik Younsi Malik Younsi 2013-01-28T17:44:13Z 2013-01-28T17:44:13Z @EmilJeř&#225;bek : I didn't know about the removability criterion for the H&#246;lder functions. This is very interesting. http://mathoverflow.net/questions/119499/on-the-universality-of-the-riemann-zeta-function/119759#119759 Comment by Malik Younsi Malik Younsi 2013-01-26T20:15:40Z 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 ! http://mathoverflow.net/questions/119499/on-the-universality-of-the-riemann-zeta-function Comment by Malik Younsi Malik Younsi 2013-01-25T20:05:03Z 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). http://mathoverflow.net/questions/119499/on-the-universality-of-the-riemann-zeta-function/119759#119759 Comment by Malik Younsi Malik Younsi 2013-01-25T20:03:56Z 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 $\log$..? http://mathoverflow.net/questions/119499/on-the-universality-of-the-riemann-zeta-function/119642#119642 Comment by Malik Younsi Malik Younsi 2013-01-24T15:57:25Z 2013-01-24T15:57:25Z @MarcPalm : yes indeed, but these generalizations of the universality theorem still require $K$ to have connected complement. http://mathoverflow.net/questions/119499/on-the-universality-of-the-riemann-zeta-function/119759#119759 Comment by Malik Younsi Malik Younsi 2013-01-24T15:50:50Z 2013-01-24T15:50:50Z See my new edit. I hope it clarifies what I'm looking for here. http://mathoverflow.net/questions/119499/on-the-universality-of-the-riemann-zeta-function/119759#119759 Comment by Malik Younsi Malik Younsi 2013-01-24T15:41:16Z 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 <i>disconnected</i> complements. Also, just to clarify : I don't want to remove the non-vanishing assumption. I'm fine with it! http://mathoverflow.net/questions/119499/on-the-universality-of-the-riemann-zeta-function/119642#119642 Comment by Malik Younsi Malik Younsi 2013-01-23T16:30:01Z 2013-01-23T16:30:01Z Yes, after searing through the literature, it appears that it is still open wether or not one can replace &quot;non-vanishing on $K$&quot; by &quot;non-vanishing in the <i>interior</i> of $K$ &quot; in the statement of the Universality Theorem. This is quite interesting! http://mathoverflow.net/questions/119499/on-the-universality-of-the-riemann-zeta-function/119642#119642 Comment by Malik Younsi Malik Younsi 2013-01-23T15:52:48Z 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 <i>interior</i> of $K$. This is because $f$ is holomorphic there. But a priori $f$ <i>could</i> vanish somewhere on the boundary of $K$, without contradicting RH! http://mathoverflow.net/questions/119499/on-the-universality-of-the-riemann-zeta-function/119642#119642 Comment by Malik Younsi Malik Younsi 2013-01-23T13:55:51Z 2013-01-23T13:55:51Z Ok, it seems that the answer to my question 2 is unknown! See Conjecture 1 in the article <a href="http://arxiv.org/abs/1010.0850" rel="nofollow">arxiv.org/abs/1010.0850</a> http://mathoverflow.net/questions/119499/on-the-universality-of-the-riemann-zeta-function/119642#119642 Comment by Malik Younsi Malik Younsi 2013-01-23T13:53:21Z 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?