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












0












$begingroup$


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











$endgroup$












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












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










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










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






  • 1




    $begingroup$
    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 $a-frac{1}{n}le xle a$.
    $endgroup$
    – herb steinberg
    Nov 29 '18 at 22:45


















0












$begingroup$


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











$endgroup$












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












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










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










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






  • 1




    $begingroup$
    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 $a-frac{1}{n}le xle a$.
    $endgroup$
    – herb steinberg
    Nov 29 '18 at 22:45
















0












0








0


0



$begingroup$


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











$endgroup$




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




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







babemcnuggets

















asked Nov 29 '18 at 2:15









babemcnuggetsbabemcnuggets

1129




1129












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












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










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










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






  • 1




    $begingroup$
    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 $a-frac{1}{n}le xle a$.
    $endgroup$
    – herb steinberg
    Nov 29 '18 at 22:45




















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












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










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










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






  • 1




    $begingroup$
    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 $a-frac{1}{n}le xle a$.
    $endgroup$
    – herb steinberg
    Nov 29 '18 at 22:45


















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






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














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




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












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




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












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




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




1




1




$begingroup$
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 $a-frac{1}{n}le xle a$.
$endgroup$
– herb steinberg
Nov 29 '18 at 22:45






$begingroup$
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 $a-frac{1}{n}le xle a$.
$endgroup$
– herb steinberg
Nov 29 '18 at 22:45












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