Operator theory – low convergence via relatively compact systems

I have problems with a proof. Let $ mathcal {A} = L _ { infty} ( mu) $ either von Neumann's algebra of essentially related measurable functions. ($ mu $ finished) I have a sequence of continuous maps $$ V_m:[0,T] rightarrow mathcal {A}. $$ I want to show that this sequence converges weakly to a $ V:[0,T] rightarrow mathcal {A}. $ To do this, I show that the images $ bigcup_m V_m ([0,T]$ are weakly relatively compact.
I know that I have foretold $ mathcal {A} _ {*} = L_1 ( mu) $ of $ mathcal {A} $ this $ mathcal {A} subset mathcal {A} _ {*}. $ So I can consider my $ V_m $ as mappings in the prediction $ mathcal {A} _ {*} $. That's what I was advised to do.

1) What I do not understand, that's why should I make the change to work in the predelict $ mathcal {A} _ {*} $ instead of $ mathcal {A} $? I need uniform integrability, but why can not I consider the images $ V_m ([0,T]$ as subsets of $ L_1 ([0,T], mathcal {A}) $ instead of $ L_1 ([0,T], mathcal {A} _ {*}) $?

2) In addition, I do not understand why the low limit of $ V_m $ would need to be continuous, which is apparently, as I was advised to show …

Thank you so much 🙂

Operator Algebras – Trace Extension on von Neumann Subalgebra

Not if $ Gamma $ is a group without torsion and with infinite conjugacy class property then algebra $ L Gamma $ is a type factor $ II $ and has only one normal tracial condition.

Now let $ x $ in $ Gamma $ different from the identity. The von Neumann algebra $ S $ generated by $ x $ is isomorphic to $ L ^ infty (S ^ 1) $, the algebra of measurable functions on the circle. It has a lot of normal tracial states.

c ++ – Calling a recursive function with the decrement operator

I wrote a code to solve a problem of factorial sum. When I created the recursive function for the factorial calculation, I used the decrement operator "-" followed by the number in the function call, which generated an error in the answer. Can any one explain the reason for using the operator to modify the answer? Follow the code below.

#include 
#include 
using namespace std;

long long factorial inte (long long int num) {
if (num == 1 || num == 0) returns 1;
return num * factorial (- num);
}

int main () {
long long int M, N;
while (scanf ("% lli% lli", & M, & N)! = EOF)
factorial (M) + factorial (N) << endl;
returns 0;
}

Linear algebra – How to calculate the gradient of the cost function with the trace operator?

The cost function is presented as follows:
begin {equation}
min _ { mathbf {w}} left {J_ {1} ( mathbf {w}) = frac {1} {2} operatorname {tr} left { mathbf {R} _ { x} ^ {- 1} left[left(mathbf{w}^{T} mathbf{R}_{x} mathbf{w}right)^{2}-mathbf{w}^{T} mathbf{R}_{y} mathbf{w}right] right } right }
end {equation}

My basic idea for the gradient solution is presented as follows:

(1) inside $ J_ {1} ( mathbf {w}) $, the term $ ( left ( mathbf {w} ^ {T} mathbf {R} _ {x} mathbf {w} right) ^ {2} – mathbf {w} ^ {T} mathbf {R} _ {y} mathbf {w}) $ is a scalar,
and $ mathbf {R} _ {x} ^ {- 1} $ is the matrix.
With the law $ Delta (Tr (X)) = Tr ( Delta (X)) $, so the derivative of $ J_ {1} ( mathbf {w}) $ more than $ mathbf {w} $ is:
begin {align}
Delta_w J_ {1} ( mathbf {w}) & = frac {1} {2} operatorname {tr} left { mathbf {R} _ {x} ^ {- 1} Delta _ { mathbf {w}} left[left(mathbf{w}^{T} mathbf{R}_{x} mathbf{w}right)^{2}-mathbf{w}^{T} mathbf{R}_{y} mathbf{w}right] right } \
& = frac {1} {2} operatorname {tr} left { mathbf {R} _ {x} ^ {- 1} left[
2(mathbf{w}^{T} mathbf{R}_{x} mathbf{w}) mathbf{R}_{x} mathbf{w}- 2mathbf{R}_{y} mathbf{w}
right] right}\
& = operatorname {tr} left {( mathbf {w} ^ {T} mathbf {R} _ {x} mathbf {w}) mathbf {w} –
mathbf {R} _ {x} ^ {- 1} mathbf {R} _ {y} mathbf {w} right }
end {align}

(2) the result obtained is the trace on a column vector $ ( mathbf {w} ^ {T} mathbf {R} _ {x} mathbf {w}) mathbf {w} –
mathbf {R} _ {x} ^ {- 1} mathbf {R} _ {y} mathbf {w} $
So what is the trace on the column vector? or basically, my idea is not correct.

(3) the correct answer as proposed by the paper $ Delta J_ {1} ( mathbf {w}) = ( mathbf {w} ^ {T} mathbf {R} _ {x} mathbf {w}) mathbf {w} –
mathbf {R} _ {x} ^ {- 1} mathbf {R} _ {y} mathbf {w} $

Can you tell me the right way to get the right answer, thanks!

Explicitly find the spectral projection of a positive operator

Let $ a $ to be an element in $ mathcal B ( mathcal H) $. How can we show that $ chi _ {(0, infty)} (yy ^ *) =[amathcal H]$? Right here $[amathcal H]$ is the projection corresponding to the closed subspace $ overline {a mathcal H} $.

views – How to apply the "superior to equals" operator to a field (whose type is text) in the drupal 8 exposed filter

I have a field "Date" in my type story, which is of type "text". I know that it should be of type date, so for that I converted this type of field to "date" by modifying it using the code below: –

function MODULE_NAME_form_views_exposed_form_alter (& $ form,
 Drupal  Core  Form  FormStateInterface $ form_state, $ form_id)
{

if ($ form['#id'] == & # 39; VIEW_ID & # 39;) {
$ form['FIELD_NAME']['#type']            = & # 39; date & # 39 ;;
}
}

Now I've used this field in the filter exposed inside a view. The problem is that I want a range for this field, for which I want to use the "Greater Than Equal To" operator, but the options available to me do not include "greater than equals". Refer to the link below.

Click here to see the picture.

reference request – eigenvalues ​​and domain of the operator Laplace-Beltrami

There are different approaches. First: consider $ Delta $ as an unlimited operator on $ L ^ 2 (M) $ with domain $ W ^ {2,2} (M) $. It is closed, densely defined and $ – Delta $ is self-adjoint, positive. There is a well-defined spectral theory for this class, which you should find somewhere in Reed & Simon. That spectrum is unobtrusive and accumulates to infinity stems from the fact that $ – Delta + 1 $ has a compact reverse.

Second: the eigenvalues ​​of $ – Delta $ are the critical points of the functional (Rayleigh ratio)
$$ I[u]= frac { int_M | u | ^ 2 , { rm Vol} _g} { int_Mg ( nabla u, nabla u) , { rm Vol} _g}, $$
which is well defined on $ W ^ {1,2} (M) setminus {0 } $. Again, you have to use the compactness of the integration $ W ^ {1,2} (M) subset L ^ 2 (M) $.

Overall, the way you get the eigenvalues ​​does not really matter because once you have $ – Delta u = lambda u $, the elliptical regularity plus a bootstrap argument tell you that $ u $ is $ C ^ infty $, so is a clean function in every way you could imagine.

php – Chatbot how to pass the conversation to a switchboard operator?

I develop a chatbot using bootman frame together with php and mysqlrather than putting predefined questions and answers in a file php I put this information in my mysqlSo, everything that works when I ask my cat automatically responds to the user when the query of it has more than one answer, my application informs the user that, for the Questioning of the same, there is more than one answer so far, everything is fine, but need to pass the care to a real person Has anyone ever implemented this with botman?

The idea was that when I was told to talk to a real clerk, my clerk was automatically informed and started to attend.

Here is an example of what I did:

<? php

require_once & # 39; vendor / autoload.php & # 39 ;;

use BotMan BotMan BotMan;
use BotMan BotMan BotManFactory;
use BotMan BotMan Drivers DriverManager;

$ config = [
// Your driver-specific configuration
// “telegram” => [
// “token” => “TOKEN”
// ]
];

DriverManager :: loadDriver ( BotMan Drivers Web WebDriver :: class);

$ botman = BotManFactory :: create ($ config);

$ botman-> hears (& # 39 ;. *; function (BotMan $ bot) {
$ s = "";

                $ s = $ bot-> getMessage () -> getText ();
$ s = $ bot-> delete accounts ($ s);

$ con = $ bot-> connection ();
$ result = $ bot-> executetaquery ($ s, $ con);

$ count = mysqli_num_rows ($ result);
switch ($ account) {
case ($ count> 1):
$ bot-> reply ("Several answers to your question will be listed");
while ($ data = mysqli_fetch_assoc ($ result))
{
// $ bot-> typesAndWaits (2);
$ bot-> reply ($ data['resposta'])
}
pause
case ($ count = 1):
while ($ data = mysqli_fetch_assoc ($ result))
{
// $ bot-> typesAndWaits (2);
$ bot-> reply ($ data['resposta'])
}
pause
default:
$ bot-> reply ('Sorry, I did not understand your question.');
}



mysqli_free_result ($ result);           

});

// start listening
$ botman-> listen ();
?>

Query with the AND operator on Elasticsearch

Could any one help me understand how the query under elasticsearch would use the AND operator to find both words (car AND speed), since it should contain both words in the result, that the two words appear in the title or that the two words are in the description or if a word is in the title and the other in the description, this is not a word. importance, but you must have both words in the result. Because the query below works as if (OR) brought results containing both words and results containing only one of the words.

{
"from: 0," size ": 10,

"request": {

"bool": {
"must": [
           {
               "match": {
                   "titulo": {
                   "query": "carro veloz",
                   "fuzziness": 1

                   }
               }
           },
           {
               "match": {
                   "descricao": {
                       "query": "carro veloz",
                   "fuzziness": 1
                   }
               }
           }
   ]
}

}
}

Operator theory – A question related to spectral decomposition

Assume that $ ( mathcal {A}, Phi) $ is a semi-finite von Neumann algebra with trace $ Phi $, $ B $ is a positive self-adjoint operator affiliated with $ mathcal {A} $, $ X $ is a positive operator with $ Phi (X) < infty $ who commutes with $ B $, and $ E_n $"S are an increasing sequence of spectral projections of $ X $ with $ Phi (E_n) < infty $ and $ lim E_n = s (X) $, or $ s (X) $ denotes the support of $ X $. Let $$ BX = int_0 ^ infty lambda dP ( lambda) $$ to be the spectral resolution of $ BX $. then $$ BXE_n = int_0 ^ infty lambda dP_n ( lambda) $$ or $ I-P_n ( lambda) = (I-P ( lambda)) E_n $. Why is it true that $$ Phi (BXE_n) = int_0 ^ infty Phi (I-P_n ( lambda)) d lambda? $$