composition of bounded uniformly convergence sequences
I'm hoping to make a generalization of the answer to this question.
Let's say that instead that we're composing two uniformly continuous function sequences, does this composition converge uniformly to $f circ g$.
Here's why I think it does.
Consider $left( f _ { k } right) _ { k = 1 } ^ { infty }$ and $( g_k ) stackrel { infty } { k } = 1$ to be sequences of continuous functions $[ 0,1 ] rightarrow [ 0,1 ]$ converging uniformly to $f$ and $g : [ 0,1 ] rightarrow mathbb { R }$ respectively.
By applying the Weierstrass M-Test, $ exists quad M > 0$ so that $|f_k(x)| leq M quad forall x in [0,1]$ and $k in mathbb{N}$
Now restrict the domain of $(g_k)$ to $[ - M , M ]$. Since the codomain of $(f_k)$ is $[0,1]$, this $[ - M , M ]$ will be within that domain. Therefore, there exists $N in mathbb{N}$ such that $(g_k)$ converges uniformly within this restricted domain.
Therefore
begin{equation}
left| g _ { k } ( f_k(x )) - g(f ( x )) right| < varepsilon text { for all } x in S text { and } k geq N
end{equation}
so $f _ { k } circ g _ { k }$ converge uniformly to $f circ g$.
Is it that simple, or am I missing something?
real-analysis uniform-convergence sequence-of-function
asked Dec 4 '18 at 0:15
Andrew HardyAndrew Hardy
125
add a comment |
I'm hoping to make a generalization of the answer to this question.
Let's say that instead that we're composing two uniformly continuous function sequences, does this composition converge uniformly to $f circ g$.
Here's why I think it does.
Consider $left( f _ { k } right) _ { k = 1 } ^ { infty }$ and $( g_k ) stackrel { infty } { k } = 1$ to be sequences of continuous functions $[ 0,1 ] rightarrow [ 0,1 ]$ converging uniformly to $f$ and $g : [ 0,1 ] rightarrow mathbb { R }$ respectively.
By applying the Weierstrass M-Test, $ exists quad M > 0$ so that $|f_k(x)| leq M quad forall x in [0,1]$ and $k in mathbb{N}$
Now restrict the domain of $(g_k)$ to $[ - M , M ]$. Since the codomain of $(f_k)$ is $[0,1]$, this $[ - M , M ]$ will be within that domain. Therefore, there exists $N in mathbb{N}$ such that $(g_k)$ converges uniformly within this restricted domain.
Therefore
begin{equation}
left| g _ { k } ( f_k(x )) - g(f ( x )) right| < varepsilon text { for all } x in S text { and } k geq N
end{equation}
so $f _ { k } circ g _ { k }$ converge uniformly to $f circ g$.
Is it that simple, or am I missing something?
real-analysis uniform-convergence sequence-of-function
asked Dec 4 '18 at 0:15
Andrew HardyAndrew Hardy
125
It is not that simple: math.stackexchange.com/questions/1680370/…
– Matt A Pelto
Dec 4 '18 at 0:35
How do I show Lipshitz then? Your link doesn't make sense...
– Andrew Hardy
Dec 4 '18 at 0:40
See zhw.'s response, since $f$ is continuous it is uniformly continuous on a compact subset. There are some errors in your post that make it confusing to read.
– Matt A Pelto
Dec 4 '18 at 0:51
add a comment |
I'm hoping to make a generalization of the answer to this question.
Let's say that instead that we're composing two uniformly continuous function sequences, does this composition converge uniformly to $f circ g$.
Here's why I think it does.
Consider $left( f _ { k } right) _ { k = 1 } ^ { infty }$ and $( g_k ) stackrel { infty } { k } = 1$ to be sequences of continuous functions $[ 0,1 ] rightarrow [ 0,1 ]$ converging uniformly to $f$ and $g : [ 0,1 ] rightarrow mathbb { R }$ respectively.
By applying the Weierstrass M-Test, $ exists quad M > 0$ so that $|f_k(x)| leq M quad forall x in [0,1]$ and $k in mathbb{N}$
Now restrict the domain of $(g_k)$ to $[ - M , M ]$. Since the codomain of $(f_k)$ is $[0,1]$, this $[ - M , M ]$ will be within that domain. Therefore, there exists $N in mathbb{N}$ such that $(g_k)$ converges uniformly within this restricted domain.
Therefore
begin{equation}
left| g _ { k } ( f_k(x )) - g(f ( x )) right| < varepsilon text { for all } x in S text { and } k geq N
end{equation}
so $f _ { k } circ g _ { k }$ converge uniformly to $f circ g$.
Is it that simple, or am I missing something?
real-analysis uniform-convergence sequence-of-function
asked Dec 4 '18 at 0:15
Andrew HardyAndrew Hardy
125
I'm hoping to make a generalization of the answer to this question.
Let's say that instead that we're composing two uniformly continuous function sequences, does this composition converge uniformly to $f circ g$.
Here's why I think it does.
Consider $left( f _ { k } right) _ { k = 1 } ^ { infty }$ and $( g_k ) stackrel { infty } { k } = 1$ to be sequences of continuous functions $[ 0,1 ] rightarrow [ 0,1 ]$ converging uniformly to $f$ and $g : [ 0,1 ] rightarrow mathbb { R }$ respectively.
By applying the Weierstrass M-Test, $ exists quad M > 0$ so that $|f_k(x)| leq M quad forall x in [0,1]$ and $k in mathbb{N}$
Now restrict the domain of $(g_k)$ to $[ - M , M ]$. Since the codomain of $(f_k)$ is $[0,1]$, this $[ - M , M ]$ will be within that domain. Therefore, there exists $N in mathbb{N}$ such that $(g_k)$ converges uniformly within this restricted domain.
Therefore
begin{equation}
left| g _ { k } ( f_k(x )) - g(f ( x )) right| < varepsilon text { for all } x in S text { and } k geq N
end{equation}
so $f _ { k } circ g _ { k }$ converge uniformly to $f circ g$.
Is it that simple, or am I missing something?
real-analysis uniform-convergence sequence-of-function
real-analysis uniform-convergence sequence-of-function
asked Dec 4 '18 at 0:15
Andrew HardyAndrew Hardy
125
asked Dec 4 '18 at 0:15
Andrew HardyAndrew Hardy
125
asked Dec 4 '18 at 0:15
Andrew HardyAndrew Hardy
125
asked Dec 4 '18 at 0:15
Andrew HardyAndrew Hardy
125
asked Dec 4 '18 at 0:15
Andrew HardyAndrew Hardy
125
125
It is not that simple: math.stackexchange.com/questions/1680370/…
– Matt A Pelto
Dec 4 '18 at 0:35
How do I show Lipshitz then? Your link doesn't make sense...
– Andrew Hardy
Dec 4 '18 at 0:40
See zhw.'s response, since $f$ is continuous it is uniformly continuous on a compact subset. There are some errors in your post that make it confusing to read.
– Matt A Pelto
Dec 4 '18 at 0:51
add a comment |
It is not that simple: math.stackexchange.com/questions/1680370/…
– Matt A Pelto
Dec 4 '18 at 0:35
How do I show Lipshitz then? Your link doesn't make sense...
– Andrew Hardy
Dec 4 '18 at 0:40
See zhw.'s response, since $f$ is continuous it is uniformly continuous on a compact subset. There are some errors in your post that make it confusing to read.
– Matt A Pelto
Dec 4 '18 at 0:51
It is not that simple: math.stackexchange.com/questions/1680370/…
– Matt A Pelto
Dec 4 '18 at 0:35
It is not that simple: math.stackexchange.com/questions/1680370/…
– Matt A Pelto
Dec 4 '18 at 0:35
How do I show Lipshitz then? Your link doesn't make sense...
– Andrew Hardy
Dec 4 '18 at 0:40
How do I show Lipshitz then? Your link doesn't make sense...
– Andrew Hardy
Dec 4 '18 at 0:40
See zhw.'s response, since $f$ is continuous it is uniformly continuous on a compact subset. There are some errors in your post that make it confusing to read.
– Matt A Pelto
Dec 4 '18 at 0:51
See zhw.'s response, since $f$ is continuous it is uniformly continuous on a compact subset. There are some errors in your post that make it confusing to read.
– Matt A Pelto
Dec 4 '18 at 0:51
add a comment |
1 Answer
1
active
oldest
votes
Suppose $g_n: [0,1] longrightarrow [0,1]$, $,f_n:[0,1] longrightarrow mathbb R$ $(n=1,2,ldots)$ are two sequences of continuous functions that converge uniformly to $g$ and $f$, respectively.
Let $varepsilon>0$ be given.
By the uniform limit theorem and by the Heine-Cantor theorem, $f$ is uniformly continuous on $[0,1]$. Thus there is $delta>0$ so that $|,f(u)-f(v)|<frac12 varepsilon$ whenever $|u-v|<delta$ and $u,v in [0,1]$. Since $g_n to g$ uniformly, there is a positive integer $N_1$ so that for all $x in [0,1]$ we have
$$|g_n(x)-g(x)|<delta text{ whenever } n geq N_1.$$
Since $f_n to f$ uniformly, there is a positive integer $N_2$ so that for all $x in [0,1]$ we have
$$|,f_n(x)-f(x)|<frac{varepsilon}{2} text{ whenever} n geq N_2.$$
Set $N=max{N_1,N_2}$. So if $n geq N$ and $x in [0,1]$, then we have
begin{equation} begin{split} | ,f_n(g_n(x)) - f(g(x))| &= |,f_n(g_n(x)) - f(g_n(x))+f(g_n(x)) - f(g(x)) |\
& leq |,f_n(g_n(x)) - f(g_n(x))|+|,f(g_n(x)) - f(g(x))| \
& <frac{varepsilon}{2} +frac{varepsilon}{2}\
& = varepsilon.
end{split}end{equation}
Therefore, $f_n circ g_n to f circ g$ uniformly on $[0,1]$ (definition of uniform convergence).
answered Dec 4 '18 at 1:19
Matt A PeltoMatt A Pelto
2,397620
1
OK let me paraphrase to see if I follow every step. Boundedness of the domain gives us uniform continuity of f, Then we use uniform converge to restrict g small enough so that it fits into the epsilon requirement of f. Then using an e/2 argument & triangle inequality, you show the condition for uniform converge?
– Andrew Hardy
Dec 4 '18 at 1:57
Precisely, the only thing I would correct from your comment is that we need the domain of $f$ to be closed and bounded. Here is a proof of the proposition that I referred to as the Heine-Cantor theorem: math.stackexchange.com/questions/2757071/…
– Matt A Pelto
Dec 4 '18 at 2:06
For the record, having every function in the sequence of functions ${g_n}_{n=1}^infty$ map into the same codomain $[0,1]$ is an important assumption unless we want to just initially assume that $f$ is uniformly continuous.
– Matt A Pelto
Dec 4 '18 at 3:06
add a comment |
Your Answer
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Suppose $g_n: [0,1] longrightarrow [0,1]$, $,f_n:[0,1] longrightarrow mathbb R$ $(n=1,2,ldots)$ are two sequences of continuous functions that converge uniformly to $g$ and $f$, respectively.
Let $varepsilon>0$ be given.
By the uniform limit theorem and by the Heine-Cantor theorem, $f$ is uniformly continuous on $[0,1]$. Thus there is $delta>0$ so that $|,f(u)-f(v)|<frac12 varepsilon$ whenever $|u-v|<delta$ and $u,v in [0,1]$. Since $g_n to g$ uniformly, there is a positive integer $N_1$ so that for all $x in [0,1]$ we have
$$|g_n(x)-g(x)|<delta text{ whenever } n geq N_1.$$
Since $f_n to f$ uniformly, there is a positive integer $N_2$ so that for all $x in [0,1]$ we have
$$|,f_n(x)-f(x)|<frac{varepsilon}{2} text{ whenever} n geq N_2.$$
Set $N=max{N_1,N_2}$. So if $n geq N$ and $x in [0,1]$, then we have
begin{equation} begin{split} | ,f_n(g_n(x)) - f(g(x))| &= |,f_n(g_n(x)) - f(g_n(x))+f(g_n(x)) - f(g(x)) |\
& leq |,f_n(g_n(x)) - f(g_n(x))|+|,f(g_n(x)) - f(g(x))| \
& <frac{varepsilon}{2} +frac{varepsilon}{2}\
& = varepsilon.
end{split}end{equation}
Therefore, $f_n circ g_n to f circ g$ uniformly on $[0,1]$ (definition of uniform convergence).
answered Dec 4 '18 at 1:19
Matt A PeltoMatt A Pelto
2,397620
1
OK let me paraphrase to see if I follow every step. Boundedness of the domain gives us uniform continuity of f, Then we use uniform converge to restrict g small enough so that it fits into the epsilon requirement of f. Then using an e/2 argument & triangle inequality, you show the condition for uniform converge?
– Andrew Hardy
Dec 4 '18 at 1:57
Precisely, the only thing I would correct from your comment is that we need the domain of $f$ to be closed and bounded. Here is a proof of the proposition that I referred to as the Heine-Cantor theorem: math.stackexchange.com/questions/2757071/…
– Matt A Pelto
Dec 4 '18 at 2:06
For the record, having every function in the sequence of functions ${g_n}_{n=1}^infty$ map into the same codomain $[0,1]$ is an important assumption unless we want to just initially assume that $f$ is uniformly continuous.
– Matt A Pelto
Dec 4 '18 at 3:06
add a comment |
Suppose $g_n: [0,1] longrightarrow [0,1]$, $,f_n:[0,1] longrightarrow mathbb R$ $(n=1,2,ldots)$ are two sequences of continuous functions that converge uniformly to $g$ and $f$, respectively.
Let $varepsilon>0$ be given.
By the uniform limit theorem and by the Heine-Cantor theorem, $f$ is uniformly continuous on $[0,1]$. Thus there is $delta>0$ so that $|,f(u)-f(v)|<frac12 varepsilon$ whenever $|u-v|<delta$ and $u,v in [0,1]$. Since $g_n to g$ uniformly, there is a positive integer $N_1$ so that for all $x in [0,1]$ we have
$$|g_n(x)-g(x)|<delta text{ whenever } n geq N_1.$$
Since $f_n to f$ uniformly, there is a positive integer $N_2$ so that for all $x in [0,1]$ we have
$$|,f_n(x)-f(x)|<frac{varepsilon}{2} text{ whenever} n geq N_2.$$
Set $N=max{N_1,N_2}$. So if $n geq N$ and $x in [0,1]$, then we have
begin{equation} begin{split} | ,f_n(g_n(x)) - f(g(x))| &= |,f_n(g_n(x)) - f(g_n(x))+f(g_n(x)) - f(g(x)) |\
& leq |,f_n(g_n(x)) - f(g_n(x))|+|,f(g_n(x)) - f(g(x))| \
& <frac{varepsilon}{2} +frac{varepsilon}{2}\
& = varepsilon.
end{split}end{equation}
Therefore, $f_n circ g_n to f circ g$ uniformly on $[0,1]$ (definition of uniform convergence).
answered Dec 4 '18 at 1:19
Matt A PeltoMatt A Pelto
2,397620
1
OK let me paraphrase to see if I follow every step. Boundedness of the domain gives us uniform continuity of f, Then we use uniform converge to restrict g small enough so that it fits into the epsilon requirement of f. Then using an e/2 argument & triangle inequality, you show the condition for uniform converge?
– Andrew Hardy
Dec 4 '18 at 1:57
Precisely, the only thing I would correct from your comment is that we need the domain of $f$ to be closed and bounded. Here is a proof of the proposition that I referred to as the Heine-Cantor theorem: math.stackexchange.com/questions/2757071/…
– Matt A Pelto
Dec 4 '18 at 2:06
For the record, having every function in the sequence of functions ${g_n}_{n=1}^infty$ map into the same codomain $[0,1]$ is an important assumption unless we want to just initially assume that $f$ is uniformly continuous.
– Matt A Pelto
Dec 4 '18 at 3:06
add a comment |
Suppose $g_n: [0,1] longrightarrow [0,1]$, $,f_n:[0,1] longrightarrow mathbb R$ $(n=1,2,ldots)$ are two sequences of continuous functions that converge uniformly to $g$ and $f$, respectively.
Let $varepsilon>0$ be given.
By the uniform limit theorem and by the Heine-Cantor theorem, $f$ is uniformly continuous on $[0,1]$. Thus there is $delta>0$ so that $|,f(u)-f(v)|<frac12 varepsilon$ whenever $|u-v|<delta$ and $u,v in [0,1]$. Since $g_n to g$ uniformly, there is a positive integer $N_1$ so that for all $x in [0,1]$ we have
$$|g_n(x)-g(x)|<delta text{ whenever } n geq N_1.$$
Since $f_n to f$ uniformly, there is a positive integer $N_2$ so that for all $x in [0,1]$ we have
$$|,f_n(x)-f(x)|<frac{varepsilon}{2} text{ whenever} n geq N_2.$$
Set $N=max{N_1,N_2}$. So if $n geq N$ and $x in [0,1]$, then we have
begin{equation} begin{split} | ,f_n(g_n(x)) - f(g(x))| &= |,f_n(g_n(x)) - f(g_n(x))+f(g_n(x)) - f(g(x)) |\
& leq |,f_n(g_n(x)) - f(g_n(x))|+|,f(g_n(x)) - f(g(x))| \
& <frac{varepsilon}{2} +frac{varepsilon}{2}\
& = varepsilon.
end{split}end{equation}
Therefore, $f_n circ g_n to f circ g$ uniformly on $[0,1]$ (definition of uniform convergence).
answered Dec 4 '18 at 1:19
Matt A PeltoMatt A Pelto
2,397620
Suppose $g_n: [0,1] longrightarrow [0,1]$, $,f_n:[0,1] longrightarrow mathbb R$ $(n=1,2,ldots)$ are two sequences of continuous functions that converge uniformly to $g$ and $f$, respectively.
Let $varepsilon>0$ be given.
By the uniform limit theorem and by the Heine-Cantor theorem, $f$ is uniformly continuous on $[0,1]$. Thus there is $delta>0$ so that $|,f(u)-f(v)|<frac12 varepsilon$ whenever $|u-v|<delta$ and $u,v in [0,1]$. Since $g_n to g$ uniformly, there is a positive integer $N_1$ so that for all $x in [0,1]$ we have
$$|g_n(x)-g(x)|<delta text{ whenever } n geq N_1.$$
Since $f_n to f$ uniformly, there is a positive integer $N_2$ so that for all $x in [0,1]$ we have
$$|,f_n(x)-f(x)|<frac{varepsilon}{2} text{ whenever} n geq N_2.$$
Set $N=max{N_1,N_2}$. So if $n geq N$ and $x in [0,1]$, then we have
begin{equation} begin{split} | ,f_n(g_n(x)) - f(g(x))| &= |,f_n(g_n(x)) - f(g_n(x))+f(g_n(x)) - f(g(x)) |\
& leq |,f_n(g_n(x)) - f(g_n(x))|+|,f(g_n(x)) - f(g(x))| \
& <frac{varepsilon}{2} +frac{varepsilon}{2}\
& = varepsilon.
end{split}end{equation}
Therefore, $f_n circ g_n to f circ g$ uniformly on $[0,1]$ (definition of uniform convergence).
answered Dec 4 '18 at 1:19
Matt A PeltoMatt A Pelto
2,397620
edited Dec 4 '18 at 2:19
answered Dec 4 '18 at 1:19
Matt A PeltoMatt A Pelto
2,397620
answered Dec 4 '18 at 1:19
Matt A PeltoMatt A Pelto
2,397620
answered Dec 4 '18 at 1:19
Matt A PeltoMatt A Pelto
2,397620
2,397620
1
OK let me paraphrase to see if I follow every step. Boundedness of the domain gives us uniform continuity of f, Then we use uniform converge to restrict g small enough so that it fits into the epsilon requirement of f. Then using an e/2 argument & triangle inequality, you show the condition for uniform converge?
– Andrew Hardy
Dec 4 '18 at 1:57
Precisely, the only thing I would correct from your comment is that we need the domain of $f$ to be closed and bounded. Here is a proof of the proposition that I referred to as the Heine-Cantor theorem: math.stackexchange.com/questions/2757071/…
– Matt A Pelto
Dec 4 '18 at 2:06
For the record, having every function in the sequence of functions ${g_n}_{n=1}^infty$ map into the same codomain $[0,1]$ is an important assumption unless we want to just initially assume that $f$ is uniformly continuous.
– Matt A Pelto
Dec 4 '18 at 3:06
add a comment |
1
OK let me paraphrase to see if I follow every step. Boundedness of the domain gives us uniform continuity of f, Then we use uniform converge to restrict g small enough so that it fits into the epsilon requirement of f. Then using an e/2 argument & triangle inequality, you show the condition for uniform converge?
– Andrew Hardy
Dec 4 '18 at 1:57
Precisely, the only thing I would correct from your comment is that we need the domain of $f$ to be closed and bounded. Here is a proof of the proposition that I referred to as the Heine-Cantor theorem: math.stackexchange.com/questions/2757071/…
– Matt A Pelto
Dec 4 '18 at 2:06
For the record, having every function in the sequence of functions ${g_n}_{n=1}^infty$ map into the same codomain $[0,1]$ is an important assumption unless we want to just initially assume that $f$ is uniformly continuous.
– Matt A Pelto
Dec 4 '18 at 3:06
1
1
OK let me paraphrase to see if I follow every step. Boundedness of the domain gives us uniform continuity of f, Then we use uniform converge to restrict g small enough so that it fits into the epsilon requirement of f. Then using an e/2 argument & triangle inequality, you show the condition for uniform converge?
– Andrew Hardy
Dec 4 '18 at 1:57
OK let me paraphrase to see if I follow every step. Boundedness of the domain gives us uniform continuity of f, Then we use uniform converge to restrict g small enough so that it fits into the epsilon requirement of f. Then using an e/2 argument & triangle inequality, you show the condition for uniform converge?
– Andrew Hardy
Dec 4 '18 at 1:57
Precisely, the only thing I would correct from your comment is that we need the domain of $f$ to be closed and bounded. Here is a proof of the proposition that I referred to as the Heine-Cantor theorem: math.stackexchange.com/questions/2757071/…
– Matt A Pelto
Dec 4 '18 at 2:06
Precisely, the only thing I would correct from your comment is that we need the domain of $f$ to be closed and bounded. Here is a proof of the proposition that I referred to as the Heine-Cantor theorem: math.stackexchange.com/questions/2757071/…
– Matt A Pelto
Dec 4 '18 at 2:06
For the record, having every function in the sequence of functions ${g_n}_{n=1}^infty$ map into the same codomain $[0,1]$ is an important assumption unless we want to just initially assume that $f$ is uniformly continuous.
– Matt A Pelto
Dec 4 '18 at 3:06
For the record, having every function in the sequence of functions ${g_n}_{n=1}^infty$ map into the same codomain $[0,1]$ is an important assumption unless we want to just initially assume that $f$ is uniformly continuous.
– Matt A Pelto
Dec 4 '18 at 3:06
add a comment |
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It is not that simple: math.stackexchange.com/questions/1680370/…
– Matt A Pelto
Dec 4 '18 at 0:35
How do I show Lipshitz then? Your link doesn't make sense...
– Andrew Hardy
Dec 4 '18 at 0:40
See zhw.'s response, since $f$ is continuous it is uniformly continuous on a compact subset. There are some errors in your post that make it confusing to read.
– Matt A Pelto
Dec 4 '18 at 0:51