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For every continuous function $g$ does there exists a sequence of functions $f_n$ which converges to $g$?
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Let,
$C(Bbb R)$$=f:Bbb R to Bbb R$.
Let, $displaystyle g(x)=e^-x^2$ for all $xin Bbb R$.
Then prove that there exists a sequence $f_n$ in $C(Bbb R)$ such that $f_n to g$ uniformly.
Here, $g$ is continuous, so by Weierstress approximation theorem we can say that there always exists a sequence of polynomials $p_n(x)$ in $K$ which converges uniformly to $g$ in $K$. But in that case how can I define $f_n(x)$ in such a way that $f_n in C(Bbb R)$ ?
real-analysis sequences-and-series analysis sequence-of-function
asked Aug 3 at 3:45
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up vote
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Let,
$C(Bbb R)$$=f:Bbb R to Bbb R$.
Let, $displaystyle g(x)=e^-x^2$ for all $xin Bbb R$.
Then prove that there exists a sequence $f_n$ in $C(Bbb R)$ such that $f_n to g$ uniformly.
Here, $g$ is continuous, so by Weierstress approximation theorem we can say that there always exists a sequence of polynomials $p_n(x)$ in $K$ which converges uniformly to $g$ in $K$. But in that case how can I define $f_n(x)$ in such a way that $f_n in C(Bbb R)$ ?
real-analysis sequences-and-series analysis sequence-of-function
asked Aug 3 at 3:45
Empty
7,73742154
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let,
$C(Bbb R)$$=f:Bbb R to Bbb R$.
Let, $displaystyle g(x)=e^-x^2$ for all $xin Bbb R$.
Then prove that there exists a sequence $f_n$ in $C(Bbb R)$ such that $f_n to g$ uniformly.
Here, $g$ is continuous, so by Weierstress approximation theorem we can say that there always exists a sequence of polynomials $p_n(x)$ in $K$ which converges uniformly to $g$ in $K$. But in that case how can I define $f_n(x)$ in such a way that $f_n in C(Bbb R)$ ?
real-analysis sequences-and-series analysis sequence-of-function
asked Aug 3 at 3:45
Empty
7,73742154
Let,
$C(Bbb R)$$=f:Bbb R to Bbb R$.
Let, $displaystyle g(x)=e^-x^2$ for all $xin Bbb R$.
Then prove that there exists a sequence $f_n$ in $C(Bbb R)$ such that $f_n to g$ uniformly.
Here, $g$ is continuous, so by Weierstress approximation theorem we can say that there always exists a sequence of polynomials $p_n(x)$ in $K$ which converges uniformly to $g$ in $K$. But in that case how can I define $f_n(x)$ in such a way that $f_n in C(Bbb R)$ ?
real-analysis sequences-and-series analysis sequence-of-function
asked Aug 3 at 3:45
Empty
7,73742154
asked Aug 3 at 3:45
Empty
7,73742154
asked Aug 3 at 3:45
Empty
7,73742154
asked Aug 3 at 3:45
Empty
7,73742154
7,73742154
add a comment |Â
add a comment |Â
2 Answers
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Let $f_n(x) = begincases g(x), & |x| le n \
g(n)(n+1-x), & n<|x| le n+1 \ 0, & textotherwise
endcases$ (I am using the fact that $g$ is even, but it is clear how this would work for any $g$ that satisfies $g(x) to 0$ as $|x| toinfty$.
Then $|g-f_n|_infty le sup_ g(x) = g(n)$, hence the convergence is uniform since $lim_n to infty g(x) = 0$.
answered Aug 3 at 4:23
copper.hat
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Since $g$ tends to $0$ as $xrightarrowinfty$ you can just take $f_n = gphi_n$
where $phi_n$ is a smooth (for your questions continuous is sufficient) cutoff function such
$phi_n(x)=1$ on $[-n,n]$, $phi_n(x)=0$ on $[-n-1,n+1]$, and $0le phi_n(x) le 1$ elsewhere.
answered Aug 3 at 4:23
Thomas
15.7k21429
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $f_n(x) = begincases g(x), & |x| le n \
g(n)(n+1-x), & n<|x| le n+1 \ 0, & textotherwise
endcases$ (I am using the fact that $g$ is even, but it is clear how this would work for any $g$ that satisfies $g(x) to 0$ as $|x| toinfty$.
Then $|g-f_n|_infty le sup_ g(x) = g(n)$, hence the convergence is uniform since $lim_n to infty g(x) = 0$.
answered Aug 3 at 4:23
copper.hat
122k556155
add a comment |Â
up vote
0
down vote
Let $f_n(x) = begincases g(x), & |x| le n \
g(n)(n+1-x), & n<|x| le n+1 \ 0, & textotherwise
endcases$ (I am using the fact that $g$ is even, but it is clear how this would work for any $g$ that satisfies $g(x) to 0$ as $|x| toinfty$.
Then $|g-f_n|_infty le sup_ g(x) = g(n)$, hence the convergence is uniform since $lim_n to infty g(x) = 0$.
answered Aug 3 at 4:23
copper.hat
122k556155
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $f_n(x) = begincases g(x), & |x| le n \
g(n)(n+1-x), & n<|x| le n+1 \ 0, & textotherwise
endcases$ (I am using the fact that $g$ is even, but it is clear how this would work for any $g$ that satisfies $g(x) to 0$ as $|x| toinfty$.
Then $|g-f_n|_infty le sup_ g(x) = g(n)$, hence the convergence is uniform since $lim_n to infty g(x) = 0$.
answered Aug 3 at 4:23
copper.hat
122k556155
Let $f_n(x) = begincases g(x), & |x| le n \
g(n)(n+1-x), & n<|x| le n+1 \ 0, & textotherwise
endcases$ (I am using the fact that $g$ is even, but it is clear how this would work for any $g$ that satisfies $g(x) to 0$ as $|x| toinfty$.
Then $|g-f_n|_infty le sup_ g(x) = g(n)$, hence the convergence is uniform since $lim_n to infty g(x) = 0$.
answered Aug 3 at 4:23
copper.hat
122k556155
answered Aug 3 at 4:23
copper.hat
122k556155
answered Aug 3 at 4:23
copper.hat
122k556155
answered Aug 3 at 4:23
copper.hat
122k556155
122k556155
add a comment |Â
add a comment |Â
up vote
0
down vote
Since $g$ tends to $0$ as $xrightarrowinfty$ you can just take $f_n = gphi_n$
where $phi_n$ is a smooth (for your questions continuous is sufficient) cutoff function such
$phi_n(x)=1$ on $[-n,n]$, $phi_n(x)=0$ on $[-n-1,n+1]$, and $0le phi_n(x) le 1$ elsewhere.
answered Aug 3 at 4:23
Thomas
15.7k21429
add a comment |Â
up vote
0
down vote
Since $g$ tends to $0$ as $xrightarrowinfty$ you can just take $f_n = gphi_n$
where $phi_n$ is a smooth (for your questions continuous is sufficient) cutoff function such
$phi_n(x)=1$ on $[-n,n]$, $phi_n(x)=0$ on $[-n-1,n+1]$, and $0le phi_n(x) le 1$ elsewhere.
answered Aug 3 at 4:23
Thomas
15.7k21429
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since $g$ tends to $0$ as $xrightarrowinfty$ you can just take $f_n = gphi_n$
where $phi_n$ is a smooth (for your questions continuous is sufficient) cutoff function such
$phi_n(x)=1$ on $[-n,n]$, $phi_n(x)=0$ on $[-n-1,n+1]$, and $0le phi_n(x) le 1$ elsewhere.
answered Aug 3 at 4:23
Thomas
15.7k21429
Since $g$ tends to $0$ as $xrightarrowinfty$ you can just take $f_n = gphi_n$
where $phi_n$ is a smooth (for your questions continuous is sufficient) cutoff function such
$phi_n(x)=1$ on $[-n,n]$, $phi_n(x)=0$ on $[-n-1,n+1]$, and $0le phi_n(x) le 1$ elsewhere.
answered Aug 3 at 4:23
Thomas
15.7k21429
answered Aug 3 at 4:23
Thomas
15.7k21429
answered Aug 3 at 4:23
Thomas
15.7k21429
answered Aug 3 at 4:23
Thomas
15.7k21429
15.7k21429
add a comment |Â
add a comment |Â
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