If $g:X to [0,infty)$ is defined by $g(x)=d(x,f(x))$, prove that $g$ is uniformly continuous on $X$.












0












$begingroup$


Let $f:X to X$ be a function on a compact metric space $(X,d)$ such that



$d(f(x),f(y)) < d(x,y)$ for all $xneq y$



a) If $g:X to [0,infty)$ is defined by $g(x)=d(x,f(x))$, prove that $g$ is uniformly continuous on $X$.



b) Prove that $f$ has a unique fixed point in X.



My attempt:



a) We must show that $g$ is uniformly continuous on X.
Let $g:X to [0,infty)$ be defined by $g(x)=d(x,f(x))$



Choose $epsilon>0$. Set $delta=epsilon/2$.



Then, let $delta>0$ such that $d(x,y)<delta$,
we have



|$g(x)-g(y)$| = $d(x,f(x))-d(y,f(y))$



=$||x-f(x)|-|y-f(y)||$



$=||x-f(x)+f(x)|-|y-f(y)+f(y)||$



$leq|(|x-f(x)|+f(x))-(|y-f(y)|+f(y))|$



However, I'm stuck in this step,



what I need to do is to somehow make the following inequality



|$g(x)-g(y)| < d(x,y) + d(f(x),f(y))$



which would lead to



$< 2d(x,y) < 2delta leq 2(epsilon/2) leq epsilon$



However, I can't seem to get pass the above step. Any help or suggestions would be greatly appreciated.



For (b) We must show that there exists a unique fixed point $x in X$ s.t. $f(x) = x$, i.e. such that $g(x) = 0$



I know that by compactness there exists a $z in X$ such that $inf_{x in X} g(x) = g(z)$ by Theorem. So existence of this x is given by compactness and continuity, however, I'm stuck following this. I know I still need to prove uniqueness as well, but does the previous statement serve as a proof for existence?



Any help would be greatly appreciated.






real-analysis uniform-continuity fixed-point-theorems fixedpoints







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $f$ is continuous and so does $d$. So $g$ is continuous on compact space $X$, which means that $g$ is uniformly continuous.
    $endgroup$
    – yoyo
    Dec 8 '18 at 11:37






  • 1




    $begingroup$
    If $g(z)=d(z,f(z))>0$, then $g(f(z))=d(f(z),f^2(z))<d(z,f(z))=g(z)$. Then $g(z)$ is not the minimum value, since $g(f(z))<g(z)$.
    $endgroup$
    – yoyo
    Dec 8 '18 at 11:41


















0












$begingroup$


Let $f:X to X$ be a function on a compact metric space $(X,d)$ such that



$d(f(x),f(y)) < d(x,y)$ for all $xneq y$



a) If $g:X to [0,infty)$ is defined by $g(x)=d(x,f(x))$, prove that $g$ is uniformly continuous on $X$.



b) Prove that $f$ has a unique fixed point in X.



My attempt:



a) We must show that $g$ is uniformly continuous on X.
Let $g:X to [0,infty)$ be defined by $g(x)=d(x,f(x))$



Choose $epsilon>0$. Set $delta=epsilon/2$.



Then, let $delta>0$ such that $d(x,y)<delta$,
we have



|$g(x)-g(y)$| = $d(x,f(x))-d(y,f(y))$



=$||x-f(x)|-|y-f(y)||$



$=||x-f(x)+f(x)|-|y-f(y)+f(y)||$



$leq|(|x-f(x)|+f(x))-(|y-f(y)|+f(y))|$



However, I'm stuck in this step,



what I need to do is to somehow make the following inequality



|$g(x)-g(y)| < d(x,y) + d(f(x),f(y))$



which would lead to



$< 2d(x,y) < 2delta leq 2(epsilon/2) leq epsilon$



However, I can't seem to get pass the above step. Any help or suggestions would be greatly appreciated.



For (b) We must show that there exists a unique fixed point $x in X$ s.t. $f(x) = x$, i.e. such that $g(x) = 0$



I know that by compactness there exists a $z in X$ such that $inf_{x in X} g(x) = g(z)$ by Theorem. So existence of this x is given by compactness and continuity, however, I'm stuck following this. I know I still need to prove uniqueness as well, but does the previous statement serve as a proof for existence?



Any help would be greatly appreciated.






real-analysis uniform-continuity fixed-point-theorems fixedpoints







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $f$ is continuous and so does $d$. So $g$ is continuous on compact space $X$, which means that $g$ is uniformly continuous.
    $endgroup$
    – yoyo
    Dec 8 '18 at 11:37






  • 1




    $begingroup$
    If $g(z)=d(z,f(z))>0$, then $g(f(z))=d(f(z),f^2(z))<d(z,f(z))=g(z)$. Then $g(z)$ is not the minimum value, since $g(f(z))<g(z)$.
    $endgroup$
    – yoyo
    Dec 8 '18 at 11:41
















0












0








0





$begingroup$


Let $f:X to X$ be a function on a compact metric space $(X,d)$ such that



$d(f(x),f(y)) < d(x,y)$ for all $xneq y$



a) If $g:X to [0,infty)$ is defined by $g(x)=d(x,f(x))$, prove that $g$ is uniformly continuous on $X$.



b) Prove that $f$ has a unique fixed point in X.



My attempt:



a) We must show that $g$ is uniformly continuous on X.
Let $g:X to [0,infty)$ be defined by $g(x)=d(x,f(x))$



Choose $epsilon>0$. Set $delta=epsilon/2$.



Then, let $delta>0$ such that $d(x,y)<delta$,
we have



|$g(x)-g(y)$| = $d(x,f(x))-d(y,f(y))$



=$||x-f(x)|-|y-f(y)||$



$=||x-f(x)+f(x)|-|y-f(y)+f(y)||$



$leq|(|x-f(x)|+f(x))-(|y-f(y)|+f(y))|$



However, I'm stuck in this step,



what I need to do is to somehow make the following inequality



|$g(x)-g(y)| < d(x,y) + d(f(x),f(y))$



which would lead to



$< 2d(x,y) < 2delta leq 2(epsilon/2) leq epsilon$



However, I can't seem to get pass the above step. Any help or suggestions would be greatly appreciated.



For (b) We must show that there exists a unique fixed point $x in X$ s.t. $f(x) = x$, i.e. such that $g(x) = 0$



I know that by compactness there exists a $z in X$ such that $inf_{x in X} g(x) = g(z)$ by Theorem. So existence of this x is given by compactness and continuity, however, I'm stuck following this. I know I still need to prove uniqueness as well, but does the previous statement serve as a proof for existence?



Any help would be greatly appreciated.






real-analysis uniform-continuity fixed-point-theorems fixedpoints







share|cite|improve this question











$endgroup$




Let $f:X to X$ be a function on a compact metric space $(X,d)$ such that



$d(f(x),f(y)) < d(x,y)$ for all $xneq y$



a) If $g:X to [0,infty)$ is defined by $g(x)=d(x,f(x))$, prove that $g$ is uniformly continuous on $X$.



b) Prove that $f$ has a unique fixed point in X.



My attempt:



a) We must show that $g$ is uniformly continuous on X.
Let $g:X to [0,infty)$ be defined by $g(x)=d(x,f(x))$



Choose $epsilon>0$. Set $delta=epsilon/2$.



Then, let $delta>0$ such that $d(x,y)<delta$,
we have



|$g(x)-g(y)$| = $d(x,f(x))-d(y,f(y))$



=$||x-f(x)|-|y-f(y)||$



$=||x-f(x)+f(x)|-|y-f(y)+f(y)||$



$leq|(|x-f(x)|+f(x))-(|y-f(y)|+f(y))|$



However, I'm stuck in this step,



what I need to do is to somehow make the following inequality



|$g(x)-g(y)| < d(x,y) + d(f(x),f(y))$



which would lead to



$< 2d(x,y) < 2delta leq 2(epsilon/2) leq epsilon$



However, I can't seem to get pass the above step. Any help or suggestions would be greatly appreciated.



For (b) We must show that there exists a unique fixed point $x in X$ s.t. $f(x) = x$, i.e. such that $g(x) = 0$



I know that by compactness there exists a $z in X$ such that $inf_{x in X} g(x) = g(z)$ by Theorem. So existence of this x is given by compactness and continuity, however, I'm stuck following this. I know I still need to prove uniqueness as well, but does the previous statement serve as a proof for existence?



Any help would be greatly appreciated.






real-analysis uniform-continuity fixed-point-theorems fixedpoints




real-analysis uniform-continuity fixed-point-theorems fixedpoints






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 12:21









Namaste

1




1










asked Dec 8 '18 at 11:31









lastgunslingerlastgunslinger

728




728








  • 1




    $begingroup$
    $f$ is continuous and so does $d$. So $g$ is continuous on compact space $X$, which means that $g$ is uniformly continuous.
    $endgroup$
    – yoyo
    Dec 8 '18 at 11:37






  • 1




    $begingroup$
    If $g(z)=d(z,f(z))>0$, then $g(f(z))=d(f(z),f^2(z))<d(z,f(z))=g(z)$. Then $g(z)$ is not the minimum value, since $g(f(z))<g(z)$.
    $endgroup$
    – yoyo
    Dec 8 '18 at 11:41
















  • 1




    $begingroup$
    $f$ is continuous and so does $d$. So $g$ is continuous on compact space $X$, which means that $g$ is uniformly continuous.
    $endgroup$
    – yoyo
    Dec 8 '18 at 11:37






  • 1




    $begingroup$
    If $g(z)=d(z,f(z))>0$, then $g(f(z))=d(f(z),f^2(z))<d(z,f(z))=g(z)$. Then $g(z)$ is not the minimum value, since $g(f(z))<g(z)$.
    $endgroup$
    – yoyo
    Dec 8 '18 at 11:41










1




1




$begingroup$
$f$ is continuous and so does $d$. So $g$ is continuous on compact space $X$, which means that $g$ is uniformly continuous.
$endgroup$
– yoyo
Dec 8 '18 at 11:37




$begingroup$
$f$ is continuous and so does $d$. So $g$ is continuous on compact space $X$, which means that $g$ is uniformly continuous.
$endgroup$
– yoyo
Dec 8 '18 at 11:37




1




1




$begingroup$
If $g(z)=d(z,f(z))>0$, then $g(f(z))=d(f(z),f^2(z))<d(z,f(z))=g(z)$. Then $g(z)$ is not the minimum value, since $g(f(z))<g(z)$.
$endgroup$
– yoyo
Dec 8 '18 at 11:41






$begingroup$
If $g(z)=d(z,f(z))>0$, then $g(f(z))=d(f(z),f^2(z))<d(z,f(z))=g(z)$. Then $g(z)$ is not the minimum value, since $g(f(z))<g(z)$.
$endgroup$
– yoyo
Dec 8 '18 at 11:41












1 Answer
1






active

oldest

votes


















1












$begingroup$

Note that $f$ is continuous as $$x_nrightarrow ximplies d(x_n,x)rightarrow 0implies d(f(x_n),f(x))rightarrow 0$$ as $d(f(x),f(y))≤d(x,y)$. Next note that $d:X×Xrightarrow [0,infty)$ is continuous , therefore $$x_nrightarrow ximplies (x_n,f(x_n))rightarrow (x,f(x))implies d(x_n,f(x_n))rightarrow d(x,f(x)).$$ Hence $g$ is continuous. Compactness of $X$ implies $g$ is uniformly continuous.



Let if possible $$g(a)=inf_{xin X} d(x,f(x))>0implies d(a,f(a))>0implies anot =f(a)implies d(f(a),f(f(a)))<d(a,f(a))implies g(f(a))<g(a)=inf_{xin X} g(x),$$ contradiction. Hence $g(a)=0implies$ $f$ has a fixed point, namely $a$.



Let $b$ be a fixed point of $f$ and $anot= b$ , then $d(a,b)>0$ and $$d(f(a),f(b))=d(a,b)<d(a,b),$$ contradiction. So $f$ has unique fixed point.



Another way of proving uniform continuity of $g$ is given below :--



$$g(x)=d(x,f(x))≤d(x,y)+d(y,f(y))+d(f(y),f(x))≤2d(x,y)+g(y)implies |g(x)-g(y)|≤2d(x,y),forall x,yin X.$$ The last inequality follows from interchanging the role of $x,y$ and $d(x,y)=d(y,x)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you SOO MUCH!! This was a huge help!
    $endgroup$
    – lastgunslinger
    Dec 8 '18 at 20:02











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3030988%2fif-gx-to-0-infty-is-defined-by-gx-dx-fx-prove-that-g-is-unifor%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Note that $f$ is continuous as $$x_nrightarrow ximplies d(x_n,x)rightarrow 0implies d(f(x_n),f(x))rightarrow 0$$ as $d(f(x),f(y))≤d(x,y)$. Next note that $d:X×Xrightarrow [0,infty)$ is continuous , therefore $$x_nrightarrow ximplies (x_n,f(x_n))rightarrow (x,f(x))implies d(x_n,f(x_n))rightarrow d(x,f(x)).$$ Hence $g$ is continuous. Compactness of $X$ implies $g$ is uniformly continuous.



Let if possible $$g(a)=inf_{xin X} d(x,f(x))>0implies d(a,f(a))>0implies anot =f(a)implies d(f(a),f(f(a)))<d(a,f(a))implies g(f(a))<g(a)=inf_{xin X} g(x),$$ contradiction. Hence $g(a)=0implies$ $f$ has a fixed point, namely $a$.



Let $b$ be a fixed point of $f$ and $anot= b$ , then $d(a,b)>0$ and $$d(f(a),f(b))=d(a,b)<d(a,b),$$ contradiction. So $f$ has unique fixed point.



Another way of proving uniform continuity of $g$ is given below :--



$$g(x)=d(x,f(x))≤d(x,y)+d(y,f(y))+d(f(y),f(x))≤2d(x,y)+g(y)implies |g(x)-g(y)|≤2d(x,y),forall x,yin X.$$ The last inequality follows from interchanging the role of $x,y$ and $d(x,y)=d(y,x)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you SOO MUCH!! This was a huge help!
    $endgroup$
    – lastgunslinger
    Dec 8 '18 at 20:02
















1












$begingroup$

Note that $f$ is continuous as $$x_nrightarrow ximplies d(x_n,x)rightarrow 0implies d(f(x_n),f(x))rightarrow 0$$ as $d(f(x),f(y))≤d(x,y)$. Next note that $d:X×Xrightarrow [0,infty)$ is continuous , therefore $$x_nrightarrow ximplies (x_n,f(x_n))rightarrow (x,f(x))implies d(x_n,f(x_n))rightarrow d(x,f(x)).$$ Hence $g$ is continuous. Compactness of $X$ implies $g$ is uniformly continuous.



Let if possible $$g(a)=inf_{xin X} d(x,f(x))>0implies d(a,f(a))>0implies anot =f(a)implies d(f(a),f(f(a)))<d(a,f(a))implies g(f(a))<g(a)=inf_{xin X} g(x),$$ contradiction. Hence $g(a)=0implies$ $f$ has a fixed point, namely $a$.



Let $b$ be a fixed point of $f$ and $anot= b$ , then $d(a,b)>0$ and $$d(f(a),f(b))=d(a,b)<d(a,b),$$ contradiction. So $f$ has unique fixed point.



Another way of proving uniform continuity of $g$ is given below :--



$$g(x)=d(x,f(x))≤d(x,y)+d(y,f(y))+d(f(y),f(x))≤2d(x,y)+g(y)implies |g(x)-g(y)|≤2d(x,y),forall x,yin X.$$ The last inequality follows from interchanging the role of $x,y$ and $d(x,y)=d(y,x)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you SOO MUCH!! This was a huge help!
    $endgroup$
    – lastgunslinger
    Dec 8 '18 at 20:02














1












1








1





$begingroup$

Note that $f$ is continuous as $$x_nrightarrow ximplies d(x_n,x)rightarrow 0implies d(f(x_n),f(x))rightarrow 0$$ as $d(f(x),f(y))≤d(x,y)$. Next note that $d:X×Xrightarrow [0,infty)$ is continuous , therefore $$x_nrightarrow ximplies (x_n,f(x_n))rightarrow (x,f(x))implies d(x_n,f(x_n))rightarrow d(x,f(x)).$$ Hence $g$ is continuous. Compactness of $X$ implies $g$ is uniformly continuous.



Let if possible $$g(a)=inf_{xin X} d(x,f(x))>0implies d(a,f(a))>0implies anot =f(a)implies d(f(a),f(f(a)))<d(a,f(a))implies g(f(a))<g(a)=inf_{xin X} g(x),$$ contradiction. Hence $g(a)=0implies$ $f$ has a fixed point, namely $a$.



Let $b$ be a fixed point of $f$ and $anot= b$ , then $d(a,b)>0$ and $$d(f(a),f(b))=d(a,b)<d(a,b),$$ contradiction. So $f$ has unique fixed point.



Another way of proving uniform continuity of $g$ is given below :--



$$g(x)=d(x,f(x))≤d(x,y)+d(y,f(y))+d(f(y),f(x))≤2d(x,y)+g(y)implies |g(x)-g(y)|≤2d(x,y),forall x,yin X.$$ The last inequality follows from interchanging the role of $x,y$ and $d(x,y)=d(y,x)$.






share|cite|improve this answer











$endgroup$



Note that $f$ is continuous as $$x_nrightarrow ximplies d(x_n,x)rightarrow 0implies d(f(x_n),f(x))rightarrow 0$$ as $d(f(x),f(y))≤d(x,y)$. Next note that $d:X×Xrightarrow [0,infty)$ is continuous , therefore $$x_nrightarrow ximplies (x_n,f(x_n))rightarrow (x,f(x))implies d(x_n,f(x_n))rightarrow d(x,f(x)).$$ Hence $g$ is continuous. Compactness of $X$ implies $g$ is uniformly continuous.



Let if possible $$g(a)=inf_{xin X} d(x,f(x))>0implies d(a,f(a))>0implies anot =f(a)implies d(f(a),f(f(a)))<d(a,f(a))implies g(f(a))<g(a)=inf_{xin X} g(x),$$ contradiction. Hence $g(a)=0implies$ $f$ has a fixed point, namely $a$.



Let $b$ be a fixed point of $f$ and $anot= b$ , then $d(a,b)>0$ and $$d(f(a),f(b))=d(a,b)<d(a,b),$$ contradiction. So $f$ has unique fixed point.



Another way of proving uniform continuity of $g$ is given below :--



$$g(x)=d(x,f(x))≤d(x,y)+d(y,f(y))+d(f(y),f(x))≤2d(x,y)+g(y)implies |g(x)-g(y)|≤2d(x,y),forall x,yin X.$$ The last inequality follows from interchanging the role of $x,y$ and $d(x,y)=d(y,x)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 '18 at 13:36

























answered Dec 8 '18 at 13:19









UserSUserS

1,5541112




1,5541112












  • $begingroup$
    Thank you SOO MUCH!! This was a huge help!
    $endgroup$
    – lastgunslinger
    Dec 8 '18 at 20:02


















  • $begingroup$
    Thank you SOO MUCH!! This was a huge help!
    $endgroup$
    – lastgunslinger
    Dec 8 '18 at 20:02
















$begingroup$
Thank you SOO MUCH!! This was a huge help!
$endgroup$
– lastgunslinger
Dec 8 '18 at 20:02




$begingroup$
Thank you SOO MUCH!! This was a huge help!
$endgroup$
– lastgunslinger
Dec 8 '18 at 20:02


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3030988%2fif-gx-to-0-infty-is-defined-by-gx-dx-fx-prove-that-g-is-unifor%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

Image
0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
Read more

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
Read more