Why is $(f_n)_{ninmathbb{N}}$ with $f_n :[0,1]rightarrowmathbb{R}, f_n(x)=x^n$ not uniformly convergent?...
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This question already has an answer here:
Prove $x^n$ is not uniformly convergent
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If I choose $f:[0,1]rightarrow mathbb{R}$ with $f(1)=1$ and otherwise $f(x)=0$ then $||f_n-f||_{[0,1]}rightarrow0$ therefore it is uniformly convergent and because it is uniformly convergent $f$ would have to be continuous but it is not, where is the mistake?
real-analysis
edited Jan 8 at 21:48
Bernard
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asked Jan 8 at 21:46
RM777RM777
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Jan 8 at 23:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Prove $x^n$ is not uniformly convergent
2 answers
If I choose $f:[0,1]rightarrow mathbb{R}$ with $f(1)=1$ and otherwise $f(x)=0$ then $||f_n-f||_{[0,1]}rightarrow0$ therefore it is uniformly convergent and because it is uniformly convergent $f$ would have to be continuous but it is not, where is the mistake?
real-analysis
edited Jan 8 at 21:48
Bernard
124k741118
asked Jan 8 at 21:46
RM777RM777
38312
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Jan 8 at 23:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Since $lim_n f_n$ is not continuous, it cannot be the uniform limit of any sequence of continuous functions.
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– copper.hat
Jan 8 at 21:58
add a comment |
$begingroup$
This question already has an answer here:
Prove $x^n$ is not uniformly convergent
2 answers
If I choose $f:[0,1]rightarrow mathbb{R}$ with $f(1)=1$ and otherwise $f(x)=0$ then $||f_n-f||_{[0,1]}rightarrow0$ therefore it is uniformly convergent and because it is uniformly convergent $f$ would have to be continuous but it is not, where is the mistake?
real-analysis
edited Jan 8 at 21:48
Bernard
124k741118
asked Jan 8 at 21:46
RM777RM777
38312
$endgroup$
This question already has an answer here:
Prove $x^n$ is not uniformly convergent
2 answers
If I choose $f:[0,1]rightarrow mathbb{R}$ with $f(1)=1$ and otherwise $f(x)=0$ then $||f_n-f||_{[0,1]}rightarrow0$ therefore it is uniformly convergent and because it is uniformly convergent $f$ would have to be continuous but it is not, where is the mistake?
This question already has an answer here:
Prove $x^n$ is not uniformly convergent
2 answers
real-analysis
real-analysis
edited Jan 8 at 21:48
Bernard
124k741118
asked Jan 8 at 21:46
RM777RM777
38312
edited Jan 8 at 21:48
Bernard
124k741118
asked Jan 8 at 21:46
RM777RM777
38312
edited Jan 8 at 21:48
Bernard
124k741118
edited Jan 8 at 21:48
Bernard
124k741118
edited Jan 8 at 21:48
Bernard
124k741118
124k741118
asked Jan 8 at 21:46
RM777RM777
38312
asked Jan 8 at 21:46
RM777RM777
38312
asked Jan 8 at 21:46
RM777RM777
38312
38312
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Jan 8 at 23:19
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Jan 8 at 23:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Since $lim_n f_n$ is not continuous, it cannot be the uniform limit of any sequence of continuous functions.
$endgroup$
– copper.hat
Jan 8 at 21:58
add a comment |
$begingroup$
Since $lim_n f_n$ is not continuous, it cannot be the uniform limit of any sequence of continuous functions.
$endgroup$
– copper.hat
Jan 8 at 21:58
$begingroup$
Since $lim_n f_n$ is not continuous, it cannot be the uniform limit of any sequence of continuous functions.
$endgroup$
– copper.hat
Jan 8 at 21:58
$begingroup$
Since $lim_n f_n$ is not continuous, it cannot be the uniform limit of any sequence of continuous functions.
$endgroup$
– copper.hat
Jan 8 at 21:58
add a comment |
1 Answer
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It is not uniformly convergent. For each $ninmathbb{N}$ you can choose $x=frac{1}{sqrt[n]{2}}$ and get $|f_n(x)-f(x)|geqfrac{1}{2}$. So it doesn't matter how far you go in the sequence, the distance will never be less than $frac{1}{2}$ for all $x$.
answered Jan 8 at 21:52
MarkMark
10.4k1622
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1 Answer
1
active
oldest
votes
1 Answer
1
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active
oldest
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active
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$begingroup$
It is not uniformly convergent. For each $ninmathbb{N}$ you can choose $x=frac{1}{sqrt[n]{2}}$ and get $|f_n(x)-f(x)|geqfrac{1}{2}$. So it doesn't matter how far you go in the sequence, the distance will never be less than $frac{1}{2}$ for all $x$.
answered Jan 8 at 21:52
MarkMark
10.4k1622
$endgroup$
add a comment |
$begingroup$
It is not uniformly convergent. For each $ninmathbb{N}$ you can choose $x=frac{1}{sqrt[n]{2}}$ and get $|f_n(x)-f(x)|geqfrac{1}{2}$. So it doesn't matter how far you go in the sequence, the distance will never be less than $frac{1}{2}$ for all $x$.
answered Jan 8 at 21:52
MarkMark
10.4k1622
$endgroup$
add a comment |
$begingroup$
It is not uniformly convergent. For each $ninmathbb{N}$ you can choose $x=frac{1}{sqrt[n]{2}}$ and get $|f_n(x)-f(x)|geqfrac{1}{2}$. So it doesn't matter how far you go in the sequence, the distance will never be less than $frac{1}{2}$ for all $x$.
answered Jan 8 at 21:52
MarkMark
10.4k1622
$endgroup$
It is not uniformly convergent. For each $ninmathbb{N}$ you can choose $x=frac{1}{sqrt[n]{2}}$ and get $|f_n(x)-f(x)|geqfrac{1}{2}$. So it doesn't matter how far you go in the sequence, the distance will never be less than $frac{1}{2}$ for all $x$.
answered Jan 8 at 21:52
MarkMark
10.4k1622
answered Jan 8 at 21:52
MarkMark
10.4k1622
answered Jan 8 at 21:52
MarkMark
10.4k1622
answered Jan 8 at 21:52
MarkMark
10.4k1622
10.4k1622
add a comment |
add a comment |
$begingroup$
Since $lim_n f_n$ is not continuous, it cannot be the uniform limit of any sequence of continuous functions.
$endgroup$
– copper.hat
Jan 8 at 21:58