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$f_n:=int _I h(x,y)f_n-1(y)dy$ uniformly convergent
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Proposition:$h(x,y)$ is $C^1$ function on $[0,1]^2$ and $f_0(x)$ is continuous on $I:=[0,1]$.
Let $f_n:=int _I h(x,y)f_n-1(y)dy$ $(n=1,2,cdots)$
Suppose $M:=sup_n max_x |f_n(x)| <infty $ and for all continuous $g(x)$ on $I$, $int_I f_n(y)g(y)dy$ convergent.
Then, $f_n(x)$ uniformly convergent.
My idea: I proved $f_n$ pointwise convergent by $g(y):=h(x,y)$ and there exist subsequence $f_n_k$ uniformly convergent by Ascoli-Arzela.
But I can't prove $f_n$ convergent uniformly.
real-analysis integration
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up vote
1
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Proposition:$h(x,y)$ is $C^1$ function on $[0,1]^2$ and $f_0(x)$ is continuous on $I:=[0,1]$.
Let $f_n:=int _I h(x,y)f_n-1(y)dy$ $(n=1,2,cdots)$
Suppose $M:=sup_n max_x |f_n(x)| <infty $ and for all continuous $g(x)$ on $I$, $int_I f_n(y)g(y)dy$ convergent.
Then, $f_n(x)$ uniformly convergent.
My idea: I proved $f_n$ pointwise convergent by $g(y):=h(x,y)$ and there exist subsequence $f_n_k$ uniformly convergent by Ascoli-Arzela.
But I can't prove $f_n$ convergent uniformly.
real-analysis integration
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Proposition:$h(x,y)$ is $C^1$ function on $[0,1]^2$ and $f_0(x)$ is continuous on $I:=[0,1]$.
Let $f_n:=int _I h(x,y)f_n-1(y)dy$ $(n=1,2,cdots)$
Suppose $M:=sup_n max_x |f_n(x)| <infty $ and for all continuous $g(x)$ on $I$, $int_I f_n(y)g(y)dy$ convergent.
Then, $f_n(x)$ uniformly convergent.
My idea: I proved $f_n$ pointwise convergent by $g(y):=h(x,y)$ and there exist subsequence $f_n_k$ uniformly convergent by Ascoli-Arzela.
But I can't prove $f_n$ convergent uniformly.
real-analysis integration
Proposition:$h(x,y)$ is $C^1$ function on $[0,1]^2$ and $f_0(x)$ is continuous on $I:=[0,1]$.
Let $f_n:=int _I h(x,y)f_n-1(y)dy$ $(n=1,2,cdots)$
Suppose $M:=sup_n max_x |f_n(x)| <infty $ and for all continuous $g(x)$ on $I$, $int_I f_n(y)g(y)dy$ convergent.
Then, $f_n(x)$ uniformly convergent.
My idea: I proved $f_n$ pointwise convergent by $g(y):=h(x,y)$ and there exist subsequence $f_n_k$ uniformly convergent by Ascoli-Arzela.
But I can't prove $f_n$ convergent uniformly.
real-analysis integration
asked Jul 17 at 12:13
user577275
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1 Answer
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In the metric space $C[0,1]$ with the supremum metric the sequence $f_n$ is relatively compact. If a subsequence $f_n_k$ converges to a function $h$ then $int f_n_k g to int hg$. If $h'$ is the limit of another subsequence then we get $int hg =int h'g$ for all $g in C[0,1]$ and this implies $h=h'$. Thus all subsequential limits of $f_n$ are the same. This implies that the entire sequence converges in the metric of $C[0,1]$,i.e. uniformly.
answered Jul 17 at 12:26
Kavi Rama Murthy
21.1k2829
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
In the metric space $C[0,1]$ with the supremum metric the sequence $f_n$ is relatively compact. If a subsequence $f_n_k$ converges to a function $h$ then $int f_n_k g to int hg$. If $h'$ is the limit of another subsequence then we get $int hg =int h'g$ for all $g in C[0,1]$ and this implies $h=h'$. Thus all subsequential limits of $f_n$ are the same. This implies that the entire sequence converges in the metric of $C[0,1]$,i.e. uniformly.
answered Jul 17 at 12:26
Kavi Rama Murthy
21.1k2829
add a comment |Â
up vote
0
down vote
accepted
In the metric space $C[0,1]$ with the supremum metric the sequence $f_n$ is relatively compact. If a subsequence $f_n_k$ converges to a function $h$ then $int f_n_k g to int hg$. If $h'$ is the limit of another subsequence then we get $int hg =int h'g$ for all $g in C[0,1]$ and this implies $h=h'$. Thus all subsequential limits of $f_n$ are the same. This implies that the entire sequence converges in the metric of $C[0,1]$,i.e. uniformly.
answered Jul 17 at 12:26
Kavi Rama Murthy
21.1k2829
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
In the metric space $C[0,1]$ with the supremum metric the sequence $f_n$ is relatively compact. If a subsequence $f_n_k$ converges to a function $h$ then $int f_n_k g to int hg$. If $h'$ is the limit of another subsequence then we get $int hg =int h'g$ for all $g in C[0,1]$ and this implies $h=h'$. Thus all subsequential limits of $f_n$ are the same. This implies that the entire sequence converges in the metric of $C[0,1]$,i.e. uniformly.
answered Jul 17 at 12:26
Kavi Rama Murthy
21.1k2829
In the metric space $C[0,1]$ with the supremum metric the sequence $f_n$ is relatively compact. If a subsequence $f_n_k$ converges to a function $h$ then $int f_n_k g to int hg$. If $h'$ is the limit of another subsequence then we get $int hg =int h'g$ for all $g in C[0,1]$ and this implies $h=h'$. Thus all subsequential limits of $f_n$ are the same. This implies that the entire sequence converges in the metric of $C[0,1]$,i.e. uniformly.
answered Jul 17 at 12:26
Kavi Rama Murthy
21.1k2829
answered Jul 17 at 12:26
Kavi Rama Murthy
21.1k2829
answered Jul 17 at 12:26
Kavi Rama Murthy
21.1k2829
answered Jul 17 at 12:26
Kavi Rama Murthy
21.1k2829
21.1k2829
add a comment |Â
add a comment |Â
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