Example of sequence of function that monotone bounded but not uniformly convergent
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My attempt is $f_n(x) = x/n$ but I'm stuck here.
real-analysis functional-analysis
edited Aug 29 at 3:59
David G. Stork
8,10421232
asked Aug 29 at 3:57
zainudin saputra
52
add a comment |Â
up vote
0
down vote
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My attempt is $f_n(x) = x/n$ but I'm stuck here.
real-analysis functional-analysis
edited Aug 29 at 3:59
David G. Stork
8,10421232
asked Aug 29 at 3:57
zainudin saputra
52
A caution on communication: You did not specify whether you mean (i) each $f_n$ is a monotone function, or (ii) each sequence $(f_n(x))_xin Bbb N$ is a monotone sequence or (iii) both (i) and (ii). And you did not specify which possible domain(s) for the $f_n $.... In this subject, when asking a Q it is better to use more words than less, and to divide what you say into separate very short sentences and clauses. .English grammar (what little there is of it) is slippery .
– DanielWainfleet
Aug 29 at 10:28
If the domain of each $f_n$ is $Bbb R$ then your attempt is valid.
– DanielWainfleet
Aug 29 at 10:32
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
My attempt is $f_n(x) = x/n$ but I'm stuck here.
real-analysis functional-analysis
edited Aug 29 at 3:59
David G. Stork
8,10421232
asked Aug 29 at 3:57
zainudin saputra
52
My attempt is $f_n(x) = x/n$ but I'm stuck here.
real-analysis functional-analysis
edited Aug 29 at 3:59
David G. Stork
8,10421232
asked Aug 29 at 3:57
zainudin saputra
52
edited Aug 29 at 3:59
David G. Stork
8,10421232
edited Aug 29 at 3:59
David G. Stork
8,10421232
edited Aug 29 at 3:59
David G. Stork
8,10421232
8,10421232
asked Aug 29 at 3:57
zainudin saputra
52
asked Aug 29 at 3:57
zainudin saputra
52
asked Aug 29 at 3:57
zainudin saputra
52
52
A caution on communication: You did not specify whether you mean (i) each $f_n$ is a monotone function, or (ii) each sequence $(f_n(x))_xin Bbb N$ is a monotone sequence or (iii) both (i) and (ii). And you did not specify which possible domain(s) for the $f_n $.... In this subject, when asking a Q it is better to use more words than less, and to divide what you say into separate very short sentences and clauses. .English grammar (what little there is of it) is slippery .
– DanielWainfleet
Aug 29 at 10:28
If the domain of each $f_n$ is $Bbb R$ then your attempt is valid.
– DanielWainfleet
Aug 29 at 10:32
add a comment |Â
A caution on communication: You did not specify whether you mean (i) each $f_n$ is a monotone function, or (ii) each sequence $(f_n(x))_xin Bbb N$ is a monotone sequence or (iii) both (i) and (ii). And you did not specify which possible domain(s) for the $f_n $.... In this subject, when asking a Q it is better to use more words than less, and to divide what you say into separate very short sentences and clauses. .English grammar (what little there is of it) is slippery .
– DanielWainfleet
Aug 29 at 10:28
If the domain of each $f_n$ is $Bbb R$ then your attempt is valid.
– DanielWainfleet
Aug 29 at 10:32
A caution on communication: You did not specify whether you mean (i) each $f_n$ is a monotone function, or (ii) each sequence $(f_n(x))_xin Bbb N$ is a monotone sequence or (iii) both (i) and (ii). And you did not specify which possible domain(s) for the $f_n $.... In this subject, when asking a Q it is better to use more words than less, and to divide what you say into separate very short sentences and clauses. .English grammar (what little there is of it) is slippery .
– DanielWainfleet
Aug 29 at 10:28
A caution on communication: You did not specify whether you mean (i) each $f_n$ is a monotone function, or (ii) each sequence $(f_n(x))_xin Bbb N$ is a monotone sequence or (iii) both (i) and (ii). And you did not specify which possible domain(s) for the $f_n $.... In this subject, when asking a Q it is better to use more words than less, and to divide what you say into separate very short sentences and clauses. .English grammar (what little there is of it) is slippery .
– DanielWainfleet
Aug 29 at 10:28
If the domain of each $f_n$ is $Bbb R$ then your attempt is valid.
– DanielWainfleet
Aug 29 at 10:32
If the domain of each $f_n$ is $Bbb R$ then your attempt is valid.
– DanielWainfleet
Aug 29 at 10:32
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
0
down vote
accepted
Classic example: on $[0,1]$, let
$$
f_n(x) = x^n,
$$
then each $f_n$ is increasing and bounded above by $1$, but $f_n$ is not uniformly convergent, since $lim f_n = mathbb 1_1(x);[x in [0,1]]$ is not continuous.
P.S. $(f_n(x))_1^infty$ are also monotonic numerical sequences for every $x in [0,1]$.
answered Aug 29 at 4:17
xbh
3,232320
add a comment |Â
up vote
1
down vote
HINT:
Let
$$
f_n(x)=begincases
1-frac1nx&text when $xin[0,1/n]$\
0&text when $x>1/n$.
endcases
$$
Can you prove the desired properties?
answered Aug 29 at 4:09
Przemysław Scherwentke
11.8k52751
Did you mean $f_n(x) = 0$ when $x notin [0, 1/n]$?
– matt stokes
Aug 29 at 4:16
@mattstokes Thank you! Corrected (for functions on $[0,infty)$).
– Przemysław Scherwentke
Aug 29 at 4:18
add a comment |Â
up vote
0
down vote
Similar to Przemyslaw Scherwentke's answer, you can try $(f_n(x))_n in mathbbN$ defined on $[0,infty)$, where
beginalign*
f_n(x) =
begincases
1 & text if x in [0,1/n]
\
0 & text if x > 1/n
endcases.
endalign*
$f_n$ is decreasing and bounded for each $n$, and $f_n to f$ where
beginalign*
f(x) =
begincases
1 & text if x =0
\
0 & text if x >0
endcases
endalign*
($f$ also defined on $[0,infty)$). It should be easy to show that $f_n$ does not converge uniformly to $f$.
answered Aug 29 at 4:40
matt stokes
48029
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Classic example: on $[0,1]$, let
$$
f_n(x) = x^n,
$$
then each $f_n$ is increasing and bounded above by $1$, but $f_n$ is not uniformly convergent, since $lim f_n = mathbb 1_1(x);[x in [0,1]]$ is not continuous.
P.S. $(f_n(x))_1^infty$ are also monotonic numerical sequences for every $x in [0,1]$.
answered Aug 29 at 4:17
xbh
3,232320
add a comment |Â
up vote
0
down vote
accepted
Classic example: on $[0,1]$, let
$$
f_n(x) = x^n,
$$
then each $f_n$ is increasing and bounded above by $1$, but $f_n$ is not uniformly convergent, since $lim f_n = mathbb 1_1(x);[x in [0,1]]$ is not continuous.
P.S. $(f_n(x))_1^infty$ are also monotonic numerical sequences for every $x in [0,1]$.
answered Aug 29 at 4:17
xbh
3,232320
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Classic example: on $[0,1]$, let
$$
f_n(x) = x^n,
$$
then each $f_n$ is increasing and bounded above by $1$, but $f_n$ is not uniformly convergent, since $lim f_n = mathbb 1_1(x);[x in [0,1]]$ is not continuous.
P.S. $(f_n(x))_1^infty$ are also monotonic numerical sequences for every $x in [0,1]$.
answered Aug 29 at 4:17
xbh
3,232320
Classic example: on $[0,1]$, let
$$
f_n(x) = x^n,
$$
then each $f_n$ is increasing and bounded above by $1$, but $f_n$ is not uniformly convergent, since $lim f_n = mathbb 1_1(x);[x in [0,1]]$ is not continuous.
P.S. $(f_n(x))_1^infty$ are also monotonic numerical sequences for every $x in [0,1]$.
answered Aug 29 at 4:17
xbh
3,232320
edited Aug 29 at 6:22
answered Aug 29 at 4:17
xbh
3,232320
answered Aug 29 at 4:17
xbh
3,232320
answered Aug 29 at 4:17
xbh
3,232320
3,232320
add a comment |Â
add a comment |Â
up vote
1
down vote
HINT:
Let
$$
f_n(x)=begincases
1-frac1nx&text when $xin[0,1/n]$\
0&text when $x>1/n$.
endcases
$$
Can you prove the desired properties?
answered Aug 29 at 4:09
Przemysław Scherwentke
11.8k52751
Did you mean $f_n(x) = 0$ when $x notin [0, 1/n]$?
– matt stokes
Aug 29 at 4:16
@mattstokes Thank you! Corrected (for functions on $[0,infty)$).
– Przemysław Scherwentke
Aug 29 at 4:18
add a comment |Â
up vote
1
down vote
HINT:
Let
$$
f_n(x)=begincases
1-frac1nx&text when $xin[0,1/n]$\
0&text when $x>1/n$.
endcases
$$
Can you prove the desired properties?
answered Aug 29 at 4:09
Przemysław Scherwentke
11.8k52751
Did you mean $f_n(x) = 0$ when $x notin [0, 1/n]$?
– matt stokes
Aug 29 at 4:16
@mattstokes Thank you! Corrected (for functions on $[0,infty)$).
– Przemysław Scherwentke
Aug 29 at 4:18
add a comment |Â
up vote
1
down vote
up vote
1
down vote
HINT:
Let
$$
f_n(x)=begincases
1-frac1nx&text when $xin[0,1/n]$\
0&text when $x>1/n$.
endcases
$$
Can you prove the desired properties?
answered Aug 29 at 4:09
Przemysław Scherwentke
11.8k52751
HINT:
Let
$$
f_n(x)=begincases
1-frac1nx&text when $xin[0,1/n]$\
0&text when $x>1/n$.
endcases
$$
Can you prove the desired properties?
answered Aug 29 at 4:09
Przemysław Scherwentke
11.8k52751
edited Aug 29 at 4:17
answered Aug 29 at 4:09
Przemysław Scherwentke
11.8k52751
answered Aug 29 at 4:09
Przemysław Scherwentke
11.8k52751
answered Aug 29 at 4:09
Przemysław Scherwentke
11.8k52751
11.8k52751
Did you mean $f_n(x) = 0$ when $x notin [0, 1/n]$?
– matt stokes
Aug 29 at 4:16
@mattstokes Thank you! Corrected (for functions on $[0,infty)$).
– Przemysław Scherwentke
Aug 29 at 4:18
add a comment |Â
Did you mean $f_n(x) = 0$ when $x notin [0, 1/n]$?
– matt stokes
Aug 29 at 4:16
@mattstokes Thank you! Corrected (for functions on $[0,infty)$).
– Przemysław Scherwentke
Aug 29 at 4:18
Did you mean $f_n(x) = 0$ when $x notin [0, 1/n]$?
– matt stokes
Aug 29 at 4:16
Did you mean $f_n(x) = 0$ when $x notin [0, 1/n]$?
– matt stokes
Aug 29 at 4:16
@mattstokes Thank you! Corrected (for functions on $[0,infty)$).
– Przemysław Scherwentke
Aug 29 at 4:18
@mattstokes Thank you! Corrected (for functions on $[0,infty)$).
– Przemysław Scherwentke
Aug 29 at 4:18
add a comment |Â
up vote
0
down vote
Similar to Przemyslaw Scherwentke's answer, you can try $(f_n(x))_n in mathbbN$ defined on $[0,infty)$, where
beginalign*
f_n(x) =
begincases
1 & text if x in [0,1/n]
\
0 & text if x > 1/n
endcases.
endalign*
$f_n$ is decreasing and bounded for each $n$, and $f_n to f$ where
beginalign*
f(x) =
begincases
1 & text if x =0
\
0 & text if x >0
endcases
endalign*
($f$ also defined on $[0,infty)$). It should be easy to show that $f_n$ does not converge uniformly to $f$.
answered Aug 29 at 4:40
matt stokes
48029
add a comment |Â
up vote
0
down vote
Similar to Przemyslaw Scherwentke's answer, you can try $(f_n(x))_n in mathbbN$ defined on $[0,infty)$, where
beginalign*
f_n(x) =
begincases
1 & text if x in [0,1/n]
\
0 & text if x > 1/n
endcases.
endalign*
$f_n$ is decreasing and bounded for each $n$, and $f_n to f$ where
beginalign*
f(x) =
begincases
1 & text if x =0
\
0 & text if x >0
endcases
endalign*
($f$ also defined on $[0,infty)$). It should be easy to show that $f_n$ does not converge uniformly to $f$.
answered Aug 29 at 4:40
matt stokes
48029
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Similar to Przemyslaw Scherwentke's answer, you can try $(f_n(x))_n in mathbbN$ defined on $[0,infty)$, where
beginalign*
f_n(x) =
begincases
1 & text if x in [0,1/n]
\
0 & text if x > 1/n
endcases.
endalign*
$f_n$ is decreasing and bounded for each $n$, and $f_n to f$ where
beginalign*
f(x) =
begincases
1 & text if x =0
\
0 & text if x >0
endcases
endalign*
($f$ also defined on $[0,infty)$). It should be easy to show that $f_n$ does not converge uniformly to $f$.
answered Aug 29 at 4:40
matt stokes
48029
Similar to Przemyslaw Scherwentke's answer, you can try $(f_n(x))_n in mathbbN$ defined on $[0,infty)$, where
beginalign*
f_n(x) =
begincases
1 & text if x in [0,1/n]
\
0 & text if x > 1/n
endcases.
endalign*
$f_n$ is decreasing and bounded for each $n$, and $f_n to f$ where
beginalign*
f(x) =
begincases
1 & text if x =0
\
0 & text if x >0
endcases
endalign*
($f$ also defined on $[0,infty)$). It should be easy to show that $f_n$ does not converge uniformly to $f$.
answered Aug 29 at 4:40
matt stokes
48029
answered Aug 29 at 4:40
matt stokes
48029
answered Aug 29 at 4:40
matt stokes
48029
answered Aug 29 at 4:40
matt stokes
48029
48029
add a comment |Â
add a comment |Â
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A caution on communication: You did not specify whether you mean (i) each $f_n$ is a monotone function, or (ii) each sequence $(f_n(x))_xin Bbb N$ is a monotone sequence or (iii) both (i) and (ii). And you did not specify which possible domain(s) for the $f_n $.... In this subject, when asking a Q it is better to use more words than less, and to divide what you say into separate very short sentences and clauses. .English grammar (what little there is of it) is slippery .
– DanielWainfleet
Aug 29 at 10:28
If the domain of each $f_n$ is $Bbb R$ then your attempt is valid.
– DanielWainfleet
Aug 29 at 10:32