Timeline for Proof of existence and uniqueness of solution to f(c)=0
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Jun 4, 2018 at 13:24 | comment | added | Pietro Majer | We need to know what is the behaviour of f on the three edges of $\Delta_\epsilon$: does f send the edge $[V_1,V_2]$ of the triangle to a curve close to the angle bounded by the first and the second axes? I'm not sure, maybe yes, thanks to the assumptions on the partial derivatives. Here if "close" is "close enough" then a linking argument (or topological degree, or even Brouwer fixed point theorem) should be sufficient for the existence. | |
| Jun 4, 2018 at 13:24 | comment | added | Pietro Majer | let's focus on the case n=3. By 4 it is sufficient to consider $f$ on the open equilateral triangle $\Delta:=\{x\in\mathbb{R}^3: \forall i\, x_i>0, x_1+x_2+x_3=1$. For small $\epsilon$, $\Delta_\epsilon\subset \Delta$ is a closed equilateral triangle a bit shrunk. By 2 and 3 we know that f sends each vertex of $\Delta_\epsilon$ close to the corresponding axis, e.g $f(\epsilon,\epsilon,1-2\epsilon)=f(1,1,1/\epsilon-2)$ is close to the third semi-axis, $(0)\times(0)\times\mathbb{R}$ (that is, the first two coordinates are bounded, the third is large). | |
| Jun 4, 2018 at 12:44 | comment | added | Jürg W. Spaak | On the boundary of $\Delta_\epsilon$ $f(x)$ will have at least one coordinate that is close to $A_i$ | |
| Jun 4, 2018 at 12:38 | comment | added | Pietro Majer | I put it in another way: it would be useful knowing something on $f(x)$ for x in the boundary of $\Delta_\epsilon$, that is, at least one coordinate of x is $\epsilon$. The assumptions tell us something when x is a vertex of $\Delta_\epsilon$, and the behaviour is in the right direction to get the result, as you wrote in the post, but I'm not sure if it is enough. | |
| Jun 4, 2018 at 12:27 | comment | added | Pietro Majer | Uniform close = close in the uniform distance en.wikipedia.org/wiki/Uniform_norm | |
| Jun 4, 2018 at 10:25 | comment | added | Jürg W. Spaak | @PietroMajer My functions are essentially growthrates of species in biology, so you can assume that they do not behave very weird (however I never put mathematical terms to "very weird"). They are potentially $C^\infty$. I do however not really understand what you mean by uniformly close. | |
| Jun 4, 2018 at 9:53 | comment | added | Pietro Majer | @JürgMerlinSpaak are these assumptions 1,2,3,4 really the only features of your maps, or are there possibly other ones? To conclude some uniformity is needed, for instance on the behavior of $f$ on the boundary of $$\Delta_\epsilon:=\{x\in\mathbb{R}^n: \forall i \ x_i\ge\epsilon , \sum_{i =1}^nx_i=1\}.$$ Can you e.g. assume that $f_{|\Delta\epsilon}$ is uniformly close (or "not too far") to a multiple of the identity map etc? | |
| Jun 2, 2018 at 18:34 | comment | added | Jürg W. Spaak | @PietroMajer They are assumed to be fixed. I guess uniformity of these limits would be inconsystent with 4. ( just intuition, without any deep thoughts). | |
| Jun 2, 2018 at 17:27 | comment | added | Pietro Majer | In the limits in 2 (resp. 3) are the coordinates $c_l$ of $c$ just assumed fixed for $l\neq i$ (resp. $l\neq j$) as $c_i\to\infty$ (resp. as $c_j\to\infty$) ? Or are those limits uniform in $c$? | |
| Jun 1, 2018 at 18:16 | comment | added | Jürg W. Spaak | Not so sure about algebraic topology. I had a class of algebraic topology about 3 years ago or so. I thought that algebraic topology includes the extension of simply connectedness. My functions are not algebraic. In my commet to Meisman there is an example of $f$. 3. That might be a problem. Is it true, that the image of a injective function will be simply connected, when the preimage is simply connected? Because my $f$ can be made injectiv, by assuming $c_1$ = 1. | |
| Jun 1, 2018 at 18:13 | comment | added | Jürg W. Spaak | @MeisamSoleimaniMalekan The following functions proof, that 2 and 4 are not incompatible: $f_i (c) = \frac{c_i^n}{\Pi_j c_j}$ | |
| Jun 1, 2018 at 16:14 | answer | added | Neil Strickland | timeline score: 1 | |
| Jun 1, 2018 at 14:11 | comment | added | Iosif Pinelis | 1. Why algebraic topology? Is your $f$ algebraic? 2. Do you have an example of such an $f$? 3. Quoting from en.wikipedia.org/wiki/Simply_connected_space : "The image of a simply connected set under a continuous [or even entire analytic -- I.P.] function need not be simply connected. Take for example the complex plane under the exponential map: the image is C - {0}, which is not simply connected". | |
| Jun 1, 2018 at 13:34 | comment | added | Jürg W. Spaak | @MeisamSoleimaniMalekan The standard vectors are not allowed, as they contain non positive entries ($c\in\mathbb{R}^n_+$). Basically assumption 4 is why only positive vecotrs are allowed. | |
| Jun 1, 2018 at 13:32 | history | edited | Jürg W. Spaak | CC BY-SA 4.0 |
Change $e_i$
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| Jun 1, 2018 at 13:29 | comment | added | MSMalekan | Conditions 2 and 4 are incompatible; let $\{\mathbf e_k\}$ be the standard basis of $\mathbb R^n$. By 2 and 4, $$f_1(\mathbf e_1)=f_1(2^n\mathbf e_1)=\lim_nf_1(2^n\mathbf e_1)=\infty$$ | |
| Jun 1, 2018 at 12:58 | review | First posts | |||
| Jun 1, 2018 at 13:06 | |||||
| Jun 1, 2018 at 12:55 | history | asked | Jürg W. Spaak | CC BY-SA 4.0 |