Mixing
Constructing a uniform convergence sequence
Let $f : overline{Omega} subset mathbb{R}^{N} to mathbb{R}$ be a $C^{2}(overline{Omega})$ function. Can we always construct a sequence $f_{n}$ such that $f_{n} to f$ uniformly in $overline{Omega}$? In this case, $Omega$ is a bounded open set. If I weaken the condition to $C(overline{Omega})$, can I also construct such a sequence as well?
Any help will be much appreciated!
uniform-convergence sequence-of-function smooth-functions
asked Nov 20 '18 at 2:25
Evan William Chandra
508313
add a comment |
Let $f : overline{Omega} subset mathbb{R}^{N} to mathbb{R}$ be a $C^{2}(overline{Omega})$ function. Can we always construct a sequence $f_{n}$ such that $f_{n} to f$ uniformly in $overline{Omega}$? In this case, $Omega$ is a bounded open set. If I weaken the condition to $C(overline{Omega})$, can I also construct such a sequence as well?
Any help will be much appreciated!
uniform-convergence sequence-of-function smooth-functions
asked Nov 20 '18 at 2:25
Evan William Chandra
508313
It seems to me that you can just let $f_n$ equal $f$, then the sequence of $f_n$ converge uniformly to $f$.
– irchans
Nov 20 '18 at 2:38
I see, thank you very much... However, I want to know non-constant sequence with respect to $n$.
– Evan William Chandra
Nov 20 '18 at 3:24
1
The image of $Omega$ under $f$ is a bounded set because $Omega$ is compact. So for a non-constant sequence let $f_n=(1+1/n)f$ if $fne 0, $ or $f_n=1/n$ if $f$ is constantly $0$.
– DanielWainfleet
Nov 20 '18 at 4:43
Thank you very much!
– Evan William Chandra
Nov 20 '18 at 5:03
add a comment |
Let $f : overline{Omega} subset mathbb{R}^{N} to mathbb{R}$ be a $C^{2}(overline{Omega})$ function. Can we always construct a sequence $f_{n}$ such that $f_{n} to f$ uniformly in $overline{Omega}$? In this case, $Omega$ is a bounded open set. If I weaken the condition to $C(overline{Omega})$, can I also construct such a sequence as well?
Any help will be much appreciated!
uniform-convergence sequence-of-function smooth-functions
asked Nov 20 '18 at 2:25
Evan William Chandra
508313
Let $f : overline{Omega} subset mathbb{R}^{N} to mathbb{R}$ be a $C^{2}(overline{Omega})$ function. Can we always construct a sequence $f_{n}$ such that $f_{n} to f$ uniformly in $overline{Omega}$? In this case, $Omega$ is a bounded open set. If I weaken the condition to $C(overline{Omega})$, can I also construct such a sequence as well?
Any help will be much appreciated!
uniform-convergence sequence-of-function smooth-functions
uniform-convergence sequence-of-function smooth-functions
asked Nov 20 '18 at 2:25
Evan William Chandra
508313
asked Nov 20 '18 at 2:25
Evan William Chandra
508313
asked Nov 20 '18 at 2:25
Evan William Chandra
508313
asked Nov 20 '18 at 2:25
Evan William Chandra
508313
asked Nov 20 '18 at 2:25
Evan William Chandra
508313
508313
It seems to me that you can just let $f_n$ equal $f$, then the sequence of $f_n$ converge uniformly to $f$.
– irchans
Nov 20 '18 at 2:38
I see, thank you very much... However, I want to know non-constant sequence with respect to $n$.
– Evan William Chandra
Nov 20 '18 at 3:24
1
The image of $Omega$ under $f$ is a bounded set because $Omega$ is compact. So for a non-constant sequence let $f_n=(1+1/n)f$ if $fne 0, $ or $f_n=1/n$ if $f$ is constantly $0$.
– DanielWainfleet
Nov 20 '18 at 4:43
Thank you very much!
– Evan William Chandra
Nov 20 '18 at 5:03
add a comment |
It seems to me that you can just let $f_n$ equal $f$, then the sequence of $f_n$ converge uniformly to $f$.
– irchans
Nov 20 '18 at 2:38
I see, thank you very much... However, I want to know non-constant sequence with respect to $n$.
– Evan William Chandra
Nov 20 '18 at 3:24
1
The image of $Omega$ under $f$ is a bounded set because $Omega$ is compact. So for a non-constant sequence let $f_n=(1+1/n)f$ if $fne 0, $ or $f_n=1/n$ if $f$ is constantly $0$.
– DanielWainfleet
Nov 20 '18 at 4:43
Thank you very much!
– Evan William Chandra
Nov 20 '18 at 5:03
It seems to me that you can just let $f_n$ equal $f$, then the sequence of $f_n$ converge uniformly to $f$.
– irchans
Nov 20 '18 at 2:38
It seems to me that you can just let $f_n$ equal $f$, then the sequence of $f_n$ converge uniformly to $f$.
– irchans
Nov 20 '18 at 2:38
I see, thank you very much... However, I want to know non-constant sequence with respect to $n$.
– Evan William Chandra
Nov 20 '18 at 3:24
I see, thank you very much... However, I want to know non-constant sequence with respect to $n$.
– Evan William Chandra
Nov 20 '18 at 3:24
1
1
The image of $Omega$ under $f$ is a bounded set because $Omega$ is compact. So for a non-constant sequence let $f_n=(1+1/n)f$ if $fne 0, $ or $f_n=1/n$ if $f$ is constantly $0$.
– DanielWainfleet
Nov 20 '18 at 4:43
The image of $Omega$ under $f$ is a bounded set because $Omega$ is compact. So for a non-constant sequence let $f_n=(1+1/n)f$ if $fne 0, $ or $f_n=1/n$ if $f$ is constantly $0$.
– DanielWainfleet
Nov 20 '18 at 4:43
Thank you very much!
– Evan William Chandra
Nov 20 '18 at 5:03
Thank you very much!
– Evan William Chandra
Nov 20 '18 at 5:03
add a comment |
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It seems to me that you can just let $f_n$ equal $f$, then the sequence of $f_n$ converge uniformly to $f$.
– irchans
Nov 20 '18 at 2:38
I see, thank you very much... However, I want to know non-constant sequence with respect to $n$.
– Evan William Chandra
Nov 20 '18 at 3:24
1
The image of $Omega$ under $f$ is a bounded set because $Omega$ is compact. So for a non-constant sequence let $f_n=(1+1/n)f$ if $fne 0, $ or $f_n=1/n$ if $f$ is constantly $0$.
– DanielWainfleet
Nov 20 '18 at 4:43
Thank you very much!
– Evan William Chandra
Nov 20 '18 at 5:03