Uniform Cauchy and convergence clarification.
$begingroup$
I have a doubt in the definition of uniform Cauchy sequence of functions in the domain $A$.
We say that the sequence $(f_n)_{ninmathbb{N}}$ is uniformly Cauchy when for any positive $epsilon$ we have a $n_1inmathbb{N}$ such that $forall n,minmathbb{N}$ with $n,m>n_1$, $forall xin A$ it has $|f_n(x)-f_m(x)|<epsilon$.
My doubt is whether it is possible to use to different values from the domain in this definition. That is "... $forall x,yin A, |f_n(x)-f_m(y)|<epsilon$ "..
Furthermore I have the same problem for uniform convergence too.
real-analysis uniform-convergence
asked Jan 2 at 20:34
DD90DD90
2648
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add a comment |
$begingroup$
I have a doubt in the definition of uniform Cauchy sequence of functions in the domain $A$.
We say that the sequence $(f_n)_{ninmathbb{N}}$ is uniformly Cauchy when for any positive $epsilon$ we have a $n_1inmathbb{N}$ such that $forall n,minmathbb{N}$ with $n,m>n_1$, $forall xin A$ it has $|f_n(x)-f_m(x)|<epsilon$.
My doubt is whether it is possible to use to different values from the domain in this definition. That is "... $forall x,yin A, |f_n(x)-f_m(y)|<epsilon$ "..
Furthermore I have the same problem for uniform convergence too.
real-analysis uniform-convergence
asked Jan 2 at 20:34
DD90DD90
2648
$endgroup$
add a comment |
$begingroup$
I have a doubt in the definition of uniform Cauchy sequence of functions in the domain $A$.
We say that the sequence $(f_n)_{ninmathbb{N}}$ is uniformly Cauchy when for any positive $epsilon$ we have a $n_1inmathbb{N}$ such that $forall n,minmathbb{N}$ with $n,m>n_1$, $forall xin A$ it has $|f_n(x)-f_m(x)|<epsilon$.
My doubt is whether it is possible to use to different values from the domain in this definition. That is "... $forall x,yin A, |f_n(x)-f_m(y)|<epsilon$ "..
Furthermore I have the same problem for uniform convergence too.
real-analysis uniform-convergence
asked Jan 2 at 20:34
DD90DD90
2648
$endgroup$
I have a doubt in the definition of uniform Cauchy sequence of functions in the domain $A$.
We say that the sequence $(f_n)_{ninmathbb{N}}$ is uniformly Cauchy when for any positive $epsilon$ we have a $n_1inmathbb{N}$ such that $forall n,minmathbb{N}$ with $n,m>n_1$, $forall xin A$ it has $|f_n(x)-f_m(x)|<epsilon$.
My doubt is whether it is possible to use to different values from the domain in this definition. That is "... $forall x,yin A, |f_n(x)-f_m(y)|<epsilon$ "..
Furthermore I have the same problem for uniform convergence too.
real-analysis uniform-convergence
real-analysis uniform-convergence
asked Jan 2 at 20:34
DD90DD90
2648
asked Jan 2 at 20:34
DD90DD90
2648
asked Jan 2 at 20:34
DD90DD90
2648
asked Jan 2 at 20:34
DD90DD90
2648
asked Jan 2 at 20:34
DD90DD90
2648
2648
add a comment |
add a comment |
1 Answer
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Consider $f_n(x)=x$, $|f_n(0)-f_m(1)|=1$ but the sequence of functions converges uniformly since it is constant $f_n(x)$ so in the definition of uniformly convergence, $x$ is fixed.
answered Jan 2 at 20:36
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
$begingroup$
Thanks a lot. Its a good example
$endgroup$
– DD90
Jan 2 at 20:41
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider $f_n(x)=x$, $|f_n(0)-f_m(1)|=1$ but the sequence of functions converges uniformly since it is constant $f_n(x)$ so in the definition of uniformly convergence, $x$ is fixed.
answered Jan 2 at 20:36
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
$begingroup$
Thanks a lot. Its a good example
$endgroup$
– DD90
Jan 2 at 20:41
add a comment |
$begingroup$
Consider $f_n(x)=x$, $|f_n(0)-f_m(1)|=1$ but the sequence of functions converges uniformly since it is constant $f_n(x)$ so in the definition of uniformly convergence, $x$ is fixed.
answered Jan 2 at 20:36
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
$begingroup$
Thanks a lot. Its a good example
$endgroup$
– DD90
Jan 2 at 20:41
add a comment |
$begingroup$
Consider $f_n(x)=x$, $|f_n(0)-f_m(1)|=1$ but the sequence of functions converges uniformly since it is constant $f_n(x)$ so in the definition of uniformly convergence, $x$ is fixed.
answered Jan 2 at 20:36
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
Consider $f_n(x)=x$, $|f_n(0)-f_m(1)|=1$ but the sequence of functions converges uniformly since it is constant $f_n(x)$ so in the definition of uniformly convergence, $x$ is fixed.
answered Jan 2 at 20:36
Tsemo AristideTsemo Aristide
60.1k11446
answered Jan 2 at 20:36
Tsemo AristideTsemo Aristide
60.1k11446
answered Jan 2 at 20:36
Tsemo AristideTsemo Aristide
60.1k11446
answered Jan 2 at 20:36
Tsemo AristideTsemo Aristide
60.1k11446
60.1k11446
$begingroup$
Thanks a lot. Its a good example
$endgroup$
– DD90
Jan 2 at 20:41
add a comment |
$begingroup$
Thanks a lot. Its a good example
$endgroup$
– DD90
Jan 2 at 20:41
$begingroup$
Thanks a lot. Its a good example
$endgroup$
– DD90
Jan 2 at 20:41
$begingroup$
Thanks a lot. Its a good example
$endgroup$
– DD90
Jan 2 at 20:41
add a comment |
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