Uniformly continuous function on uniformly convergent sequence of functions
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Let {$f_{n} , nin mathbb{N}$} be a uniformly convergent sequence of continuous real–valued functions defined on a metric space $M$ and let $g$ be a continuous function on $mathbb{R}$. Define, for each $nin mathbb{N}$, $h_{n}(x)=g(f_{n}(x))$.
We know if $M= mathbb{R}$ then the sequence {$h_{n}:nin mathbb{N}$} may not be uniformly convergent. A very nice example is given here: Composition of a continuous function with functions that converge uniformly
Now if instead, $h$ is a bounded uniformly continuous function, can we say that {$h_{n}:nin mathbb{N}$} is uniformly convergent?
Also, if I relax the condition on {$f_{n}, nin mathbb{N}$} and make it right continuous instead of continuous then will the answer remain same?
real-analysis analysis
asked Jan 10 at 9:53
arnabarnab
184
$endgroup$
add a comment |
$begingroup$
Let {$f_{n} , nin mathbb{N}$} be a uniformly convergent sequence of continuous real–valued functions defined on a metric space $M$ and let $g$ be a continuous function on $mathbb{R}$. Define, for each $nin mathbb{N}$, $h_{n}(x)=g(f_{n}(x))$.
We know if $M= mathbb{R}$ then the sequence {$h_{n}:nin mathbb{N}$} may not be uniformly convergent. A very nice example is given here: Composition of a continuous function with functions that converge uniformly
Now if instead, $h$ is a bounded uniformly continuous function, can we say that {$h_{n}:nin mathbb{N}$} is uniformly convergent?
Also, if I relax the condition on {$f_{n}, nin mathbb{N}$} and make it right continuous instead of continuous then will the answer remain same?
real-analysis analysis
asked Jan 10 at 9:53
arnabarnab
184
$endgroup$
1
$begingroup$
If $g$ is uniformly continuous and $f_n to f$ uniformly then $h_n$ converges uniformly. No continuity assumptions on $f_n$'s are required. If $g$ is just bounded and continuous then $h_n$ need not converge uniformly.
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– Kavi Rama Murthy
Jan 10 at 10:06
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@KaviRamaMurthy Thank you so much. Can you give me a reference? I have to cite a reference for my work. I have searched for it but not able to find a reference.
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– arnab
Jan 10 at 17:28
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This result follows immediately from definition of uniform continuity.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 23:10
add a comment |
$begingroup$
Let {$f_{n} , nin mathbb{N}$} be a uniformly convergent sequence of continuous real–valued functions defined on a metric space $M$ and let $g$ be a continuous function on $mathbb{R}$. Define, for each $nin mathbb{N}$, $h_{n}(x)=g(f_{n}(x))$.
We know if $M= mathbb{R}$ then the sequence {$h_{n}:nin mathbb{N}$} may not be uniformly convergent. A very nice example is given here: Composition of a continuous function with functions that converge uniformly
Now if instead, $h$ is a bounded uniformly continuous function, can we say that {$h_{n}:nin mathbb{N}$} is uniformly convergent?
Also, if I relax the condition on {$f_{n}, nin mathbb{N}$} and make it right continuous instead of continuous then will the answer remain same?
real-analysis analysis
asked Jan 10 at 9:53
arnabarnab
184
$endgroup$
Let {$f_{n} , nin mathbb{N}$} be a uniformly convergent sequence of continuous real–valued functions defined on a metric space $M$ and let $g$ be a continuous function on $mathbb{R}$. Define, for each $nin mathbb{N}$, $h_{n}(x)=g(f_{n}(x))$.
We know if $M= mathbb{R}$ then the sequence {$h_{n}:nin mathbb{N}$} may not be uniformly convergent. A very nice example is given here: Composition of a continuous function with functions that converge uniformly
Now if instead, $h$ is a bounded uniformly continuous function, can we say that {$h_{n}:nin mathbb{N}$} is uniformly convergent?
Also, if I relax the condition on {$f_{n}, nin mathbb{N}$} and make it right continuous instead of continuous then will the answer remain same?
real-analysis analysis
real-analysis analysis
asked Jan 10 at 9:53
arnabarnab
184
asked Jan 10 at 9:53
arnabarnab
184
asked Jan 10 at 9:53
arnabarnab
184
asked Jan 10 at 9:53
arnabarnab
184
asked Jan 10 at 9:53
arnabarnab
184
184
1
$begingroup$
If $g$ is uniformly continuous and $f_n to f$ uniformly then $h_n$ converges uniformly. No continuity assumptions on $f_n$'s are required. If $g$ is just bounded and continuous then $h_n$ need not converge uniformly.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 10:06
$begingroup$
@KaviRamaMurthy Thank you so much. Can you give me a reference? I have to cite a reference for my work. I have searched for it but not able to find a reference.
$endgroup$
– arnab
Jan 10 at 17:28
$begingroup$
This result follows immediately from definition of uniform continuity.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 23:10
add a comment |
1
$begingroup$
If $g$ is uniformly continuous and $f_n to f$ uniformly then $h_n$ converges uniformly. No continuity assumptions on $f_n$'s are required. If $g$ is just bounded and continuous then $h_n$ need not converge uniformly.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 10:06
$begingroup$
@KaviRamaMurthy Thank you so much. Can you give me a reference? I have to cite a reference for my work. I have searched for it but not able to find a reference.
$endgroup$
– arnab
Jan 10 at 17:28
$begingroup$
This result follows immediately from definition of uniform continuity.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 23:10
1
1
$begingroup$
If $g$ is uniformly continuous and $f_n to f$ uniformly then $h_n$ converges uniformly. No continuity assumptions on $f_n$'s are required. If $g$ is just bounded and continuous then $h_n$ need not converge uniformly.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 10:06
$begingroup$
If $g$ is uniformly continuous and $f_n to f$ uniformly then $h_n$ converges uniformly. No continuity assumptions on $f_n$'s are required. If $g$ is just bounded and continuous then $h_n$ need not converge uniformly.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 10:06
$begingroup$
@KaviRamaMurthy Thank you so much. Can you give me a reference? I have to cite a reference for my work. I have searched for it but not able to find a reference.
$endgroup$
– arnab
Jan 10 at 17:28
$begingroup$
@KaviRamaMurthy Thank you so much. Can you give me a reference? I have to cite a reference for my work. I have searched for it but not able to find a reference.
$endgroup$
– arnab
Jan 10 at 17:28
$begingroup$
This result follows immediately from definition of uniform continuity.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 23:10
$begingroup$
This result follows immediately from definition of uniform continuity.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 23:10
add a comment |
1 Answer
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$begingroup$
Suppose $g:mathbb R to mathbb R$ is defined as follows: $g(n)=g(n+frac 2 n)=0$, $g(n+frac 1 n)=1$ and $g$ has a straightline graph on the intervals $(n,n+frac 1 n)$ as well as $(n+frac 1 n,n+frac 2 n)$ for $n=3,4,cdots$ and $0$ everywhere else. Then $g$ is a bounded continuous function. Let $f_n(x)=x+frac 1 n$. Then $f_n$ converges uniformly on $mathbb R$ to $f(x)=x$ but $h_n$ doesn't converge uniformly: $sup_x {|g(f_n(x))-g(f(x))| geq |g(f_n(n))-g(f(n))|=1$ for all $n$. [My comment above has more information about the question].
answered Jan 10 at 10:31
Kavi Rama MurthyKavi Rama Murthy
63.8k42464
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add a comment |
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$begingroup$
Suppose $g:mathbb R to mathbb R$ is defined as follows: $g(n)=g(n+frac 2 n)=0$, $g(n+frac 1 n)=1$ and $g$ has a straightline graph on the intervals $(n,n+frac 1 n)$ as well as $(n+frac 1 n,n+frac 2 n)$ for $n=3,4,cdots$ and $0$ everywhere else. Then $g$ is a bounded continuous function. Let $f_n(x)=x+frac 1 n$. Then $f_n$ converges uniformly on $mathbb R$ to $f(x)=x$ but $h_n$ doesn't converge uniformly: $sup_x {|g(f_n(x))-g(f(x))| geq |g(f_n(n))-g(f(n))|=1$ for all $n$. [My comment above has more information about the question].
answered Jan 10 at 10:31
Kavi Rama MurthyKavi Rama Murthy
63.8k42464
$endgroup$
add a comment |
$begingroup$
Suppose $g:mathbb R to mathbb R$ is defined as follows: $g(n)=g(n+frac 2 n)=0$, $g(n+frac 1 n)=1$ and $g$ has a straightline graph on the intervals $(n,n+frac 1 n)$ as well as $(n+frac 1 n,n+frac 2 n)$ for $n=3,4,cdots$ and $0$ everywhere else. Then $g$ is a bounded continuous function. Let $f_n(x)=x+frac 1 n$. Then $f_n$ converges uniformly on $mathbb R$ to $f(x)=x$ but $h_n$ doesn't converge uniformly: $sup_x {|g(f_n(x))-g(f(x))| geq |g(f_n(n))-g(f(n))|=1$ for all $n$. [My comment above has more information about the question].
answered Jan 10 at 10:31
Kavi Rama MurthyKavi Rama Murthy
63.8k42464
$endgroup$
add a comment |
$begingroup$
Suppose $g:mathbb R to mathbb R$ is defined as follows: $g(n)=g(n+frac 2 n)=0$, $g(n+frac 1 n)=1$ and $g$ has a straightline graph on the intervals $(n,n+frac 1 n)$ as well as $(n+frac 1 n,n+frac 2 n)$ for $n=3,4,cdots$ and $0$ everywhere else. Then $g$ is a bounded continuous function. Let $f_n(x)=x+frac 1 n$. Then $f_n$ converges uniformly on $mathbb R$ to $f(x)=x$ but $h_n$ doesn't converge uniformly: $sup_x {|g(f_n(x))-g(f(x))| geq |g(f_n(n))-g(f(n))|=1$ for all $n$. [My comment above has more information about the question].
answered Jan 10 at 10:31
Kavi Rama MurthyKavi Rama Murthy
63.8k42464
$endgroup$
Suppose $g:mathbb R to mathbb R$ is defined as follows: $g(n)=g(n+frac 2 n)=0$, $g(n+frac 1 n)=1$ and $g$ has a straightline graph on the intervals $(n,n+frac 1 n)$ as well as $(n+frac 1 n,n+frac 2 n)$ for $n=3,4,cdots$ and $0$ everywhere else. Then $g$ is a bounded continuous function. Let $f_n(x)=x+frac 1 n$. Then $f_n$ converges uniformly on $mathbb R$ to $f(x)=x$ but $h_n$ doesn't converge uniformly: $sup_x {|g(f_n(x))-g(f(x))| geq |g(f_n(n))-g(f(n))|=1$ for all $n$. [My comment above has more information about the question].
answered Jan 10 at 10:31
Kavi Rama MurthyKavi Rama Murthy
63.8k42464
edited Jan 10 at 23:09
answered Jan 10 at 10:31
Kavi Rama MurthyKavi Rama Murthy
63.8k42464
answered Jan 10 at 10:31
Kavi Rama MurthyKavi Rama Murthy
63.8k42464
answered Jan 10 at 10:31
Kavi Rama MurthyKavi Rama Murthy
63.8k42464
63.8k42464
add a comment |
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$begingroup$
If $g$ is uniformly continuous and $f_n to f$ uniformly then $h_n$ converges uniformly. No continuity assumptions on $f_n$'s are required. If $g$ is just bounded and continuous then $h_n$ need not converge uniformly.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 10:06
$begingroup$
@KaviRamaMurthy Thank you so much. Can you give me a reference? I have to cite a reference for my work. I have searched for it but not able to find a reference.
$endgroup$
– arnab
Jan 10 at 17:28
$begingroup$
This result follows immediately from definition of uniform continuity.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 23:10