Finding sequence of functions with compact support for Integral of given function












0














Let $a>0$ and $f(x):mathbb{R}rightarrowmathbb{R}$ with



$$f(x):begin{cases}frac{1}{sqrt{a^2-x^2}},& xin(-a,a)\
0,&text{else} end{cases}$$

I now have to construct sequences of functions $(f_n)_{ninmathbb{N}}in C_c(mathbb{R})$ with $f_nuparrow f$ (meaning $f_n leq f_{n+1}$) to later show, that the integral of $f$ is equal to the improper integral of $fin (-a,a)$. I think I'll be able to solve the integral part, but I'm insecure building this sequence of functions.
I think I do know how it must look like :



Let $$f_n:begin{cases}f,& xin(-a+frac{1}{n},a-frac{1}{n})\
text{at $x=a-frac{1}{n}$}text{is} f(x)=frac{1}{sqrt{2afrac{1}{n}-frac{1}{n^2}}},&text{so it should go straight down to 0 in $[a-frac{1}{n},a)$}\text{at $x=-a+frac{1}{n}$}text{is $f(x)=frac{1}{sqrt{2afrac{1}{n}-frac{1}{n^2}}}$},& text{the same reasoning}\0,& xnotin]-a,a[ end{cases}$$

I think that is pretty much correct, as $nrightarrowinf$ we get that $f_nrightarrow f$. However I am not sure how to build the last part.



EDIT: While writing all this I thought of the simple function $f(x)=mx$ where $m=frac{y_2-y_1}{x_2-x_1}$. For the two points $(-a+frac{1}{n},f(-a+frac{1}{n}))(=(x_1,y_1)$ and ($-a,0)(=x_2,y_2)$ and got $m=frac{frac{1}{sqrt{2afrac{1}{n}-frac{1}{n^2}}}}{frac{1}{n}}$. Then I would have my compact sequence of function. Is this correct or is it $-m$ ?










share|cite|improve this question
























  • How about $f_n(x)=f(x)$ for $-a+frac{1}{n}le xle a-frac{1}{n}$.and $=0$ otherwise?
    – herb steinberg
    Nov 29 '18 at 2:46












  • Doesn't it approach infinity going near its borders? Correct me if I am wrong.
    – babemcnuggets
    Nov 29 '18 at 2:51










  • @babenuggets Each$f_n(x)$ is bounded, so each has compact support.
    – herb steinberg
    Nov 29 '18 at 3:38










  • You're right, however they are not continuous.
    – babemcnuggets
    Nov 29 '18 at 3:40






  • 1




    They can be made continuous trivially. Just extend the function by its end point value to $pm a$. That is let $c_n=f(a-frac{1}{n})$ and $f_n(x)=c_n$ for $-ale xle -a+frac{1}{n}$ and $a-frac{1}{n}le xle a$.
    – herb steinberg
    Nov 29 '18 at 22:45


















0














Let $a>0$ and $f(x):mathbb{R}rightarrowmathbb{R}$ with



$$f(x):begin{cases}frac{1}{sqrt{a^2-x^2}},& xin(-a,a)\
0,&text{else} end{cases}$$

I now have to construct sequences of functions $(f_n)_{ninmathbb{N}}in C_c(mathbb{R})$ with $f_nuparrow f$ (meaning $f_n leq f_{n+1}$) to later show, that the integral of $f$ is equal to the improper integral of $fin (-a,a)$. I think I'll be able to solve the integral part, but I'm insecure building this sequence of functions.
I think I do know how it must look like :



Let $$f_n:begin{cases}f,& xin(-a+frac{1}{n},a-frac{1}{n})\
text{at $x=a-frac{1}{n}$}text{is} f(x)=frac{1}{sqrt{2afrac{1}{n}-frac{1}{n^2}}},&text{so it should go straight down to 0 in $[a-frac{1}{n},a)$}\text{at $x=-a+frac{1}{n}$}text{is $f(x)=frac{1}{sqrt{2afrac{1}{n}-frac{1}{n^2}}}$},& text{the same reasoning}\0,& xnotin]-a,a[ end{cases}$$

I think that is pretty much correct, as $nrightarrowinf$ we get that $f_nrightarrow f$. However I am not sure how to build the last part.



EDIT: While writing all this I thought of the simple function $f(x)=mx$ where $m=frac{y_2-y_1}{x_2-x_1}$. For the two points $(-a+frac{1}{n},f(-a+frac{1}{n}))(=(x_1,y_1)$ and ($-a,0)(=x_2,y_2)$ and got $m=frac{frac{1}{sqrt{2afrac{1}{n}-frac{1}{n^2}}}}{frac{1}{n}}$. Then I would have my compact sequence of function. Is this correct or is it $-m$ ?










share|cite|improve this question
























  • How about $f_n(x)=f(x)$ for $-a+frac{1}{n}le xle a-frac{1}{n}$.and $=0$ otherwise?
    – herb steinberg
    Nov 29 '18 at 2:46












  • Doesn't it approach infinity going near its borders? Correct me if I am wrong.
    – babemcnuggets
    Nov 29 '18 at 2:51










  • @babenuggets Each$f_n(x)$ is bounded, so each has compact support.
    – herb steinberg
    Nov 29 '18 at 3:38










  • You're right, however they are not continuous.
    – babemcnuggets
    Nov 29 '18 at 3:40






  • 1




    They can be made continuous trivially. Just extend the function by its end point value to $pm a$. That is let $c_n=f(a-frac{1}{n})$ and $f_n(x)=c_n$ for $-ale xle -a+frac{1}{n}$ and $a-frac{1}{n}le xle a$.
    – herb steinberg
    Nov 29 '18 at 22:45
















0












0








0


0





Let $a>0$ and $f(x):mathbb{R}rightarrowmathbb{R}$ with



$$f(x):begin{cases}frac{1}{sqrt{a^2-x^2}},& xin(-a,a)\
0,&text{else} end{cases}$$

I now have to construct sequences of functions $(f_n)_{ninmathbb{N}}in C_c(mathbb{R})$ with $f_nuparrow f$ (meaning $f_n leq f_{n+1}$) to later show, that the integral of $f$ is equal to the improper integral of $fin (-a,a)$. I think I'll be able to solve the integral part, but I'm insecure building this sequence of functions.
I think I do know how it must look like :



Let $$f_n:begin{cases}f,& xin(-a+frac{1}{n},a-frac{1}{n})\
text{at $x=a-frac{1}{n}$}text{is} f(x)=frac{1}{sqrt{2afrac{1}{n}-frac{1}{n^2}}},&text{so it should go straight down to 0 in $[a-frac{1}{n},a)$}\text{at $x=-a+frac{1}{n}$}text{is $f(x)=frac{1}{sqrt{2afrac{1}{n}-frac{1}{n^2}}}$},& text{the same reasoning}\0,& xnotin]-a,a[ end{cases}$$

I think that is pretty much correct, as $nrightarrowinf$ we get that $f_nrightarrow f$. However I am not sure how to build the last part.



EDIT: While writing all this I thought of the simple function $f(x)=mx$ where $m=frac{y_2-y_1}{x_2-x_1}$. For the two points $(-a+frac{1}{n},f(-a+frac{1}{n}))(=(x_1,y_1)$ and ($-a,0)(=x_2,y_2)$ and got $m=frac{frac{1}{sqrt{2afrac{1}{n}-frac{1}{n^2}}}}{frac{1}{n}}$. Then I would have my compact sequence of function. Is this correct or is it $-m$ ?










share|cite|improve this question















Let $a>0$ and $f(x):mathbb{R}rightarrowmathbb{R}$ with



$$f(x):begin{cases}frac{1}{sqrt{a^2-x^2}},& xin(-a,a)\
0,&text{else} end{cases}$$

I now have to construct sequences of functions $(f_n)_{ninmathbb{N}}in C_c(mathbb{R})$ with $f_nuparrow f$ (meaning $f_n leq f_{n+1}$) to later show, that the integral of $f$ is equal to the improper integral of $fin (-a,a)$. I think I'll be able to solve the integral part, but I'm insecure building this sequence of functions.
I think I do know how it must look like :



Let $$f_n:begin{cases}f,& xin(-a+frac{1}{n},a-frac{1}{n})\
text{at $x=a-frac{1}{n}$}text{is} f(x)=frac{1}{sqrt{2afrac{1}{n}-frac{1}{n^2}}},&text{so it should go straight down to 0 in $[a-frac{1}{n},a)$}\text{at $x=-a+frac{1}{n}$}text{is $f(x)=frac{1}{sqrt{2afrac{1}{n}-frac{1}{n^2}}}$},& text{the same reasoning}\0,& xnotin]-a,a[ end{cases}$$

I think that is pretty much correct, as $nrightarrowinf$ we get that $f_nrightarrow f$. However I am not sure how to build the last part.



EDIT: While writing all this I thought of the simple function $f(x)=mx$ where $m=frac{y_2-y_1}{x_2-x_1}$. For the two points $(-a+frac{1}{n},f(-a+frac{1}{n}))(=(x_1,y_1)$ and ($-a,0)(=x_2,y_2)$ and got $m=frac{frac{1}{sqrt{2afrac{1}{n}-frac{1}{n^2}}}}{frac{1}{n}}$. Then I would have my compact sequence of function. Is this correct or is it $-m$ ?







calculus measure-theory sequence-of-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 5:25

























asked Nov 29 '18 at 2:15









babemcnuggets

396




396












  • How about $f_n(x)=f(x)$ for $-a+frac{1}{n}le xle a-frac{1}{n}$.and $=0$ otherwise?
    – herb steinberg
    Nov 29 '18 at 2:46












  • Doesn't it approach infinity going near its borders? Correct me if I am wrong.
    – babemcnuggets
    Nov 29 '18 at 2:51










  • @babenuggets Each$f_n(x)$ is bounded, so each has compact support.
    – herb steinberg
    Nov 29 '18 at 3:38










  • You're right, however they are not continuous.
    – babemcnuggets
    Nov 29 '18 at 3:40






  • 1




    They can be made continuous trivially. Just extend the function by its end point value to $pm a$. That is let $c_n=f(a-frac{1}{n})$ and $f_n(x)=c_n$ for $-ale xle -a+frac{1}{n}$ and $a-frac{1}{n}le xle a$.
    – herb steinberg
    Nov 29 '18 at 22:45




















  • How about $f_n(x)=f(x)$ for $-a+frac{1}{n}le xle a-frac{1}{n}$.and $=0$ otherwise?
    – herb steinberg
    Nov 29 '18 at 2:46












  • Doesn't it approach infinity going near its borders? Correct me if I am wrong.
    – babemcnuggets
    Nov 29 '18 at 2:51










  • @babenuggets Each$f_n(x)$ is bounded, so each has compact support.
    – herb steinberg
    Nov 29 '18 at 3:38










  • You're right, however they are not continuous.
    – babemcnuggets
    Nov 29 '18 at 3:40






  • 1




    They can be made continuous trivially. Just extend the function by its end point value to $pm a$. That is let $c_n=f(a-frac{1}{n})$ and $f_n(x)=c_n$ for $-ale xle -a+frac{1}{n}$ and $a-frac{1}{n}le xle a$.
    – herb steinberg
    Nov 29 '18 at 22:45


















How about $f_n(x)=f(x)$ for $-a+frac{1}{n}le xle a-frac{1}{n}$.and $=0$ otherwise?
– herb steinberg
Nov 29 '18 at 2:46






How about $f_n(x)=f(x)$ for $-a+frac{1}{n}le xle a-frac{1}{n}$.and $=0$ otherwise?
– herb steinberg
Nov 29 '18 at 2:46














Doesn't it approach infinity going near its borders? Correct me if I am wrong.
– babemcnuggets
Nov 29 '18 at 2:51




Doesn't it approach infinity going near its borders? Correct me if I am wrong.
– babemcnuggets
Nov 29 '18 at 2:51












@babenuggets Each$f_n(x)$ is bounded, so each has compact support.
– herb steinberg
Nov 29 '18 at 3:38




@babenuggets Each$f_n(x)$ is bounded, so each has compact support.
– herb steinberg
Nov 29 '18 at 3:38












You're right, however they are not continuous.
– babemcnuggets
Nov 29 '18 at 3:40




You're right, however they are not continuous.
– babemcnuggets
Nov 29 '18 at 3:40




1




1




They can be made continuous trivially. Just extend the function by its end point value to $pm a$. That is let $c_n=f(a-frac{1}{n})$ and $f_n(x)=c_n$ for $-ale xle -a+frac{1}{n}$ and $a-frac{1}{n}le xle a$.
– herb steinberg
Nov 29 '18 at 22:45






They can be made continuous trivially. Just extend the function by its end point value to $pm a$. That is let $c_n=f(a-frac{1}{n})$ and $f_n(x)=c_n$ for $-ale xle -a+frac{1}{n}$ and $a-frac{1}{n}le xle a$.
– herb steinberg
Nov 29 '18 at 22:45












0






active

oldest

votes











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%2f3018064%2ffinding-sequence-of-functions-with-compact-support-for-integral-of-given-functio%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes
















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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


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%2f3018064%2ffinding-sequence-of-functions-with-compact-support-for-integral-of-given-functio%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

Probability when a professor distributes a quiz and homework assignment to a class of n students.

Aardman Animations

Are they similar matrix