Is the following statement is True/false ? ..
$begingroup$
Is the following statement is True/false ?
Given that $f_n(x) =(-x)^n $ , $x in [0,1] $ then $f_n$ converges pointwise everywhere ?
i thinks it will be true same as $f(x) = 0 $ when x =0 ,$f(x) =1$ when
$0 <x le 1$
Am i right ??
real-analysis
asked Dec 12 '18 at 13:00
jasminejasmine
1,747417
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add a comment |
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Is the following statement is True/false ?
Given that $f_n(x) =(-x)^n $ , $x in [0,1] $ then $f_n$ converges pointwise everywhere ?
i thinks it will be true same as $f(x) = 0 $ when x =0 ,$f(x) =1$ when
$0 <x le 1$
Am i right ??
real-analysis
asked Dec 12 '18 at 13:00
jasminejasmine
1,747417
$endgroup$
add a comment |
$begingroup$
Is the following statement is True/false ?
Given that $f_n(x) =(-x)^n $ , $x in [0,1] $ then $f_n$ converges pointwise everywhere ?
i thinks it will be true same as $f(x) = 0 $ when x =0 ,$f(x) =1$ when
$0 <x le 1$
Am i right ??
real-analysis
asked Dec 12 '18 at 13:00
jasminejasmine
1,747417
$endgroup$
Is the following statement is True/false ?
Given that $f_n(x) =(-x)^n $ , $x in [0,1] $ then $f_n$ converges pointwise everywhere ?
i thinks it will be true same as $f(x) = 0 $ when x =0 ,$f(x) =1$ when
$0 <x le 1$
Am i right ??
real-analysis
real-analysis
asked Dec 12 '18 at 13:00
jasminejasmine
1,747417
asked Dec 12 '18 at 13:00
jasminejasmine
1,747417
asked Dec 12 '18 at 13:00
jasminejasmine
1,747417
asked Dec 12 '18 at 13:00
jasminejasmine
1,747417
asked Dec 12 '18 at 13:00
jasminejasmine
1,747417
1,747417
add a comment |
add a comment |
2 Answers
2
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You are not correct. If you define $$f(x)=begin{cases}0&xin[0,1)\1&x=1end{cases}$$
then $f_n$ converges to $f$ pointwise on $[0,1)$, however, for $x=1$, the sequence $f_n(x)$ becomes $(-1)^n$ which is the sequence $1,-1,1,-1,dots$, a sequence that does not converge.
answered Dec 12 '18 at 13:04
5xum5xum
90.6k394161
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@%xum but im confusing at negative sign $(-x)^n = x^n$
$endgroup$
– jasmine
Dec 12 '18 at 13:06
1
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@jasmine Oh, I missed the negative sign! Sorry, I will edit my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:07
add a comment |
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This is correct except for when $x=1$. When $x=1$ you get a sequence $f_n = (-1)^n$ which does not converge, and therefore does not converge pointwise at that point.
answered Dec 12 '18 at 13:08
dllegenddllegend
163
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2 Answers
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2 Answers
2
active
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$begingroup$
You are not correct. If you define $$f(x)=begin{cases}0&xin[0,1)\1&x=1end{cases}$$
then $f_n$ converges to $f$ pointwise on $[0,1)$, however, for $x=1$, the sequence $f_n(x)$ becomes $(-1)^n$ which is the sequence $1,-1,1,-1,dots$, a sequence that does not converge.
answered Dec 12 '18 at 13:04
5xum5xum
90.6k394161
$endgroup$
$begingroup$
@%xum but im confusing at negative sign $(-x)^n = x^n$
$endgroup$
– jasmine
Dec 12 '18 at 13:06
1
$begingroup$
@jasmine Oh, I missed the negative sign! Sorry, I will edit my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:07
add a comment |
$begingroup$
You are not correct. If you define $$f(x)=begin{cases}0&xin[0,1)\1&x=1end{cases}$$
then $f_n$ converges to $f$ pointwise on $[0,1)$, however, for $x=1$, the sequence $f_n(x)$ becomes $(-1)^n$ which is the sequence $1,-1,1,-1,dots$, a sequence that does not converge.
answered Dec 12 '18 at 13:04
5xum5xum
90.6k394161
$endgroup$
$begingroup$
@%xum but im confusing at negative sign $(-x)^n = x^n$
$endgroup$
– jasmine
Dec 12 '18 at 13:06
1
$begingroup$
@jasmine Oh, I missed the negative sign! Sorry, I will edit my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:07
add a comment |
$begingroup$
You are not correct. If you define $$f(x)=begin{cases}0&xin[0,1)\1&x=1end{cases}$$
then $f_n$ converges to $f$ pointwise on $[0,1)$, however, for $x=1$, the sequence $f_n(x)$ becomes $(-1)^n$ which is the sequence $1,-1,1,-1,dots$, a sequence that does not converge.
answered Dec 12 '18 at 13:04
5xum5xum
90.6k394161
$endgroup$
You are not correct. If you define $$f(x)=begin{cases}0&xin[0,1)\1&x=1end{cases}$$
then $f_n$ converges to $f$ pointwise on $[0,1)$, however, for $x=1$, the sequence $f_n(x)$ becomes $(-1)^n$ which is the sequence $1,-1,1,-1,dots$, a sequence that does not converge.
answered Dec 12 '18 at 13:04
5xum5xum
90.6k394161
edited Dec 12 '18 at 13:08
answered Dec 12 '18 at 13:04
5xum5xum
90.6k394161
answered Dec 12 '18 at 13:04
5xum5xum
90.6k394161
answered Dec 12 '18 at 13:04
5xum5xum
90.6k394161
90.6k394161
$begingroup$
@%xum but im confusing at negative sign $(-x)^n = x^n$
$endgroup$
– jasmine
Dec 12 '18 at 13:06
1
$begingroup$
@jasmine Oh, I missed the negative sign! Sorry, I will edit my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:07
add a comment |
$begingroup$
@%xum but im confusing at negative sign $(-x)^n = x^n$
$endgroup$
– jasmine
Dec 12 '18 at 13:06
1
$begingroup$
@jasmine Oh, I missed the negative sign! Sorry, I will edit my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:07
$begingroup$
@%xum but im confusing at negative sign $(-x)^n = x^n$
$endgroup$
– jasmine
Dec 12 '18 at 13:06
$begingroup$
@%xum but im confusing at negative sign $(-x)^n = x^n$
$endgroup$
– jasmine
Dec 12 '18 at 13:06
1
1
$begingroup$
@jasmine Oh, I missed the negative sign! Sorry, I will edit my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:07
$begingroup$
@jasmine Oh, I missed the negative sign! Sorry, I will edit my answer.
$endgroup$
– 5xum
Dec 12 '18 at 13:07
add a comment |
$begingroup$
This is correct except for when $x=1$. When $x=1$ you get a sequence $f_n = (-1)^n$ which does not converge, and therefore does not converge pointwise at that point.
answered Dec 12 '18 at 13:08
dllegenddllegend
163
$endgroup$
add a comment |
$begingroup$
This is correct except for when $x=1$. When $x=1$ you get a sequence $f_n = (-1)^n$ which does not converge, and therefore does not converge pointwise at that point.
answered Dec 12 '18 at 13:08
dllegenddllegend
163
$endgroup$
add a comment |
$begingroup$
This is correct except for when $x=1$. When $x=1$ you get a sequence $f_n = (-1)^n$ which does not converge, and therefore does not converge pointwise at that point.
answered Dec 12 '18 at 13:08
dllegenddllegend
163
$endgroup$
This is correct except for when $x=1$. When $x=1$ you get a sequence $f_n = (-1)^n$ which does not converge, and therefore does not converge pointwise at that point.
answered Dec 12 '18 at 13:08
dllegenddllegend
163
answered Dec 12 '18 at 13:08
dllegenddllegend
163
answered Dec 12 '18 at 13:08
dllegenddllegend
163
answered Dec 12 '18 at 13:08
dllegenddllegend
163
163
add a comment |
add a comment |
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