Looking for a certain function with compact support
$begingroup$
can someone help me with the following doubt?
I know that if we consider a function $f$ with compact support and $C^infty(mathbb{R^2})$ such that $f(x)=1$ if $|x|leq{1}$ and $f(x)=0$ if $|x|geq{3}$. Then $f_s(x)=f(frac{x}{s})$ with $s>0$ verifies $$lim_{srightarrowinfty}{f_s(x)}=1$$
How can we define $f_s(x)$ to get $lim_{srightarrowinfty}{f_s(x)}=|x|^2$?
I thought of considering $f(x)=|x|^2$ if $|x|leq{1}$ and $f(x)=0$ if $|x|geq{3}$ and using $f_s$ as before, but it doesn't work.
Thanks.
real-analysis functional-analysis analysis
edited Dec 26 '18 at 11:34
Julián Aguirre
69.4k24096
asked Dec 26 '18 at 9:06
mathlifemathlife
609
$endgroup$
add a comment |
$begingroup$
can someone help me with the following doubt?
I know that if we consider a function $f$ with compact support and $C^infty(mathbb{R^2})$ such that $f(x)=1$ if $|x|leq{1}$ and $f(x)=0$ if $|x|geq{3}$. Then $f_s(x)=f(frac{x}{s})$ with $s>0$ verifies $$lim_{srightarrowinfty}{f_s(x)}=1$$
How can we define $f_s(x)$ to get $lim_{srightarrowinfty}{f_s(x)}=|x|^2$?
I thought of considering $f(x)=|x|^2$ if $|x|leq{1}$ and $f(x)=0$ if $|x|geq{3}$ and using $f_s$ as before, but it doesn't work.
Thanks.
real-analysis functional-analysis analysis
edited Dec 26 '18 at 11:34
Julián Aguirre
69.4k24096
asked Dec 26 '18 at 9:06
mathlifemathlife
609
$endgroup$
add a comment |
$begingroup$
can someone help me with the following doubt?
I know that if we consider a function $f$ with compact support and $C^infty(mathbb{R^2})$ such that $f(x)=1$ if $|x|leq{1}$ and $f(x)=0$ if $|x|geq{3}$. Then $f_s(x)=f(frac{x}{s})$ with $s>0$ verifies $$lim_{srightarrowinfty}{f_s(x)}=1$$
How can we define $f_s(x)$ to get $lim_{srightarrowinfty}{f_s(x)}=|x|^2$?
I thought of considering $f(x)=|x|^2$ if $|x|leq{1}$ and $f(x)=0$ if $|x|geq{3}$ and using $f_s$ as before, but it doesn't work.
Thanks.
real-analysis functional-analysis analysis
edited Dec 26 '18 at 11:34
Julián Aguirre
69.4k24096
asked Dec 26 '18 at 9:06
mathlifemathlife
609
$endgroup$
can someone help me with the following doubt?
I know that if we consider a function $f$ with compact support and $C^infty(mathbb{R^2})$ such that $f(x)=1$ if $|x|leq{1}$ and $f(x)=0$ if $|x|geq{3}$. Then $f_s(x)=f(frac{x}{s})$ with $s>0$ verifies $$lim_{srightarrowinfty}{f_s(x)}=1$$
How can we define $f_s(x)$ to get $lim_{srightarrowinfty}{f_s(x)}=|x|^2$?
I thought of considering $f(x)=|x|^2$ if $|x|leq{1}$ and $f(x)=0$ if $|x|geq{3}$ and using $f_s$ as before, but it doesn't work.
Thanks.
real-analysis functional-analysis analysis
real-analysis functional-analysis analysis
edited Dec 26 '18 at 11:34
Julián Aguirre
69.4k24096
asked Dec 26 '18 at 9:06
mathlifemathlife
609
edited Dec 26 '18 at 11:34
Julián Aguirre
69.4k24096
asked Dec 26 '18 at 9:06
mathlifemathlife
609
edited Dec 26 '18 at 11:34
Julián Aguirre
69.4k24096
edited Dec 26 '18 at 11:34
Julián Aguirre
69.4k24096
edited Dec 26 '18 at 11:34
Julián Aguirre
69.4k24096
69.4k24096
asked Dec 26 '18 at 9:06
mathlifemathlife
609
asked Dec 26 '18 at 9:06
mathlifemathlife
609
asked Dec 26 '18 at 9:06
mathlifemathlife
609
609
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Nothing will work.
$$
lim_{stoinfty}f_s(x)=f(0)
$$
is a constant.
answered Dec 26 '18 at 11:35
Julián AguirreJulián Aguirre
69.4k24096
$endgroup$
$begingroup$
And $f_s(x)=f(frac{x}{s})|x|^2$?
$endgroup$
– mathlife
Dec 26 '18 at 13:53
$begingroup$
The limit will be $f(0),|x|^2$.
$endgroup$
– Julián Aguirre
Dec 26 '18 at 13:54
$begingroup$
But un muy cuestión $f(0)=1$
$endgroup$
– mathlife
Dec 26 '18 at 13:57
$begingroup$
Then $f(0),|x|^2=|x|^2$.
$endgroup$
– Julián Aguirre
Dec 26 '18 at 13:58
$begingroup$
In that definition $f_s(x)$ has also compact support and is infinite derivable?
$endgroup$
– mathlife
Dec 26 '18 at 14:02
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Nothing will work.
$$
lim_{stoinfty}f_s(x)=f(0)
$$
is a constant.
answered Dec 26 '18 at 11:35
Julián AguirreJulián Aguirre
69.4k24096
$endgroup$
$begingroup$
And $f_s(x)=f(frac{x}{s})|x|^2$?
$endgroup$
– mathlife
Dec 26 '18 at 13:53
$begingroup$
The limit will be $f(0),|x|^2$.
$endgroup$
– Julián Aguirre
Dec 26 '18 at 13:54
$begingroup$
But un muy cuestión $f(0)=1$
$endgroup$
– mathlife
Dec 26 '18 at 13:57
$begingroup$
Then $f(0),|x|^2=|x|^2$.
$endgroup$
– Julián Aguirre
Dec 26 '18 at 13:58
$begingroup$
In that definition $f_s(x)$ has also compact support and is infinite derivable?
$endgroup$
– mathlife
Dec 26 '18 at 14:02
add a comment |
$begingroup$
Nothing will work.
$$
lim_{stoinfty}f_s(x)=f(0)
$$
is a constant.
answered Dec 26 '18 at 11:35
Julián AguirreJulián Aguirre
69.4k24096
$endgroup$
$begingroup$
And $f_s(x)=f(frac{x}{s})|x|^2$?
$endgroup$
– mathlife
Dec 26 '18 at 13:53
$begingroup$
The limit will be $f(0),|x|^2$.
$endgroup$
– Julián Aguirre
Dec 26 '18 at 13:54
$begingroup$
But un muy cuestión $f(0)=1$
$endgroup$
– mathlife
Dec 26 '18 at 13:57
$begingroup$
Then $f(0),|x|^2=|x|^2$.
$endgroup$
– Julián Aguirre
Dec 26 '18 at 13:58
$begingroup$
In that definition $f_s(x)$ has also compact support and is infinite derivable?
$endgroup$
– mathlife
Dec 26 '18 at 14:02
add a comment |
$begingroup$
Nothing will work.
$$
lim_{stoinfty}f_s(x)=f(0)
$$
is a constant.
answered Dec 26 '18 at 11:35
Julián AguirreJulián Aguirre
69.4k24096
$endgroup$
Nothing will work.
$$
lim_{stoinfty}f_s(x)=f(0)
$$
is a constant.
answered Dec 26 '18 at 11:35
Julián AguirreJulián Aguirre
69.4k24096
edited Dec 26 '18 at 13:56
answered Dec 26 '18 at 11:35
Julián AguirreJulián Aguirre
69.4k24096
answered Dec 26 '18 at 11:35
Julián AguirreJulián Aguirre
69.4k24096
answered Dec 26 '18 at 11:35
Julián AguirreJulián Aguirre
69.4k24096
69.4k24096
$begingroup$
And $f_s(x)=f(frac{x}{s})|x|^2$?
$endgroup$
– mathlife
Dec 26 '18 at 13:53
$begingroup$
The limit will be $f(0),|x|^2$.
$endgroup$
– Julián Aguirre
Dec 26 '18 at 13:54
$begingroup$
But un muy cuestión $f(0)=1$
$endgroup$
– mathlife
Dec 26 '18 at 13:57
$begingroup$
Then $f(0),|x|^2=|x|^2$.
$endgroup$
– Julián Aguirre
Dec 26 '18 at 13:58
$begingroup$
In that definition $f_s(x)$ has also compact support and is infinite derivable?
$endgroup$
– mathlife
Dec 26 '18 at 14:02
add a comment |
$begingroup$
And $f_s(x)=f(frac{x}{s})|x|^2$?
$endgroup$
– mathlife
Dec 26 '18 at 13:53
$begingroup$
The limit will be $f(0),|x|^2$.
$endgroup$
– Julián Aguirre
Dec 26 '18 at 13:54
$begingroup$
But un muy cuestión $f(0)=1$
$endgroup$
– mathlife
Dec 26 '18 at 13:57
$begingroup$
Then $f(0),|x|^2=|x|^2$.
$endgroup$
– Julián Aguirre
Dec 26 '18 at 13:58
$begingroup$
In that definition $f_s(x)$ has also compact support and is infinite derivable?
$endgroup$
– mathlife
Dec 26 '18 at 14:02
$begingroup$
And $f_s(x)=f(frac{x}{s})|x|^2$?
$endgroup$
– mathlife
Dec 26 '18 at 13:53
$begingroup$
And $f_s(x)=f(frac{x}{s})|x|^2$?
$endgroup$
– mathlife
Dec 26 '18 at 13:53
$begingroup$
The limit will be $f(0),|x|^2$.
$endgroup$
– Julián Aguirre
Dec 26 '18 at 13:54
$begingroup$
The limit will be $f(0),|x|^2$.
$endgroup$
– Julián Aguirre
Dec 26 '18 at 13:54
$begingroup$
But un muy cuestión $f(0)=1$
$endgroup$
– mathlife
Dec 26 '18 at 13:57
$begingroup$
But un muy cuestión $f(0)=1$
$endgroup$
– mathlife
Dec 26 '18 at 13:57
$begingroup$
Then $f(0),|x|^2=|x|^2$.
$endgroup$
– Julián Aguirre
Dec 26 '18 at 13:58
$begingroup$
Then $f(0),|x|^2=|x|^2$.
$endgroup$
– Julián Aguirre
Dec 26 '18 at 13:58
$begingroup$
In that definition $f_s(x)$ has also compact support and is infinite derivable?
$endgroup$
– mathlife
Dec 26 '18 at 14:02
$begingroup$
In that definition $f_s(x)$ has also compact support and is infinite derivable?
$endgroup$
– mathlife
Dec 26 '18 at 14:02
add a comment |
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