Continuous function as pointwise limit but not as uniform limit of a sequence of continuous functions on...
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I recentaly find an article where it is said that there is a sequence of continuous functions ${f_n:[0,1]rightarrowBbb R}_{nin Bbb N}$ that converges pointqise almost everywhere to zero function , but not converges uniformly.
My question is the following :--
Let $f:[0,1]rightarrow Bbb R$ be continuous , does there always exists a sequence of continuous functions ${f_n:[0,1]rightarrowBbb R}_{nin Bbb N}$ such that $f_nrightarrow f$ pointwise, but not uniformly on $[0,1]$.
real-analysis general-topology metric-spaces uniform-convergence pointwise-convergence
asked Dec 12 '18 at 13:30
MathloverMathlover
1447
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add a comment |
$begingroup$
I recentaly find an article where it is said that there is a sequence of continuous functions ${f_n:[0,1]rightarrowBbb R}_{nin Bbb N}$ that converges pointqise almost everywhere to zero function , but not converges uniformly.
My question is the following :--
Let $f:[0,1]rightarrow Bbb R$ be continuous , does there always exists a sequence of continuous functions ${f_n:[0,1]rightarrowBbb R}_{nin Bbb N}$ such that $f_nrightarrow f$ pointwise, but not uniformly on $[0,1]$.
real-analysis general-topology metric-spaces uniform-convergence pointwise-convergence
asked Dec 12 '18 at 13:30
MathloverMathlover
1447
$endgroup$
add a comment |
$begingroup$
I recentaly find an article where it is said that there is a sequence of continuous functions ${f_n:[0,1]rightarrowBbb R}_{nin Bbb N}$ that converges pointqise almost everywhere to zero function , but not converges uniformly.
My question is the following :--
Let $f:[0,1]rightarrow Bbb R$ be continuous , does there always exists a sequence of continuous functions ${f_n:[0,1]rightarrowBbb R}_{nin Bbb N}$ such that $f_nrightarrow f$ pointwise, but not uniformly on $[0,1]$.
real-analysis general-topology metric-spaces uniform-convergence pointwise-convergence
asked Dec 12 '18 at 13:30
MathloverMathlover
1447
$endgroup$
I recentaly find an article where it is said that there is a sequence of continuous functions ${f_n:[0,1]rightarrowBbb R}_{nin Bbb N}$ that converges pointqise almost everywhere to zero function , but not converges uniformly.
My question is the following :--
Let $f:[0,1]rightarrow Bbb R$ be continuous , does there always exists a sequence of continuous functions ${f_n:[0,1]rightarrowBbb R}_{nin Bbb N}$ such that $f_nrightarrow f$ pointwise, but not uniformly on $[0,1]$.
real-analysis general-topology metric-spaces uniform-convergence pointwise-convergence
real-analysis general-topology metric-spaces uniform-convergence pointwise-convergence
asked Dec 12 '18 at 13:30
MathloverMathlover
1447
asked Dec 12 '18 at 13:30
MathloverMathlover
1447
asked Dec 12 '18 at 13:30
MathloverMathlover
1447
asked Dec 12 '18 at 13:30
MathloverMathlover
1447
asked Dec 12 '18 at 13:30
MathloverMathlover
1447
1447
add a comment |
add a comment |
2 Answers
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Yes, there is always such a function.
Take a sequence of continuous functions $g_n$ which converges pointwise, but not uniformly, to $0$. Then $f_n=f + g_n$ converges pointwise, but not uniformly to $f$.
For an example of such $g_n$, you can take the following:
$$
g_n(x) = cases{
nx & if $x< frac1n$\
2-nx & if $frac1n leq x < frac2n$\
0& otherwise}
$$
The graph of $g_n$ will be a triangle starting at $(0,0)$, going up to $(1, frac1n)$, then down to $(frac2n, 0)$, and then flat horizontal from there. This does converge pointwise to $0$ (as at any non-zero point $ain [0,1]$, eventually $frac2n<a$, and $g_n(a) = 0$), but not uniformly, as it always has a maximum value of $1$.
answered Dec 12 '18 at 13:31
ArthurArthur
116k7116198
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1
$begingroup$
Heh, the exact same answer up to the letter used in order to avoid duplication of $f_n$ :)
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– 5xum
Dec 12 '18 at 13:33
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@5xum I gues it's pretty standardized how one makes the "trivial" examples.
$endgroup$
– Arthur
Dec 12 '18 at 13:38
add a comment |
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Let $g_n$ be the sequence that converges almost everywhere to $0$, but does not converge uniformly (we already know that exists).
Let $f$ be a continuous function.
Then, $f_n=f+g_n$ converges almost everywhere to $f$, but does not converge uniformly.
answered Dec 12 '18 at 13:32
5xum5xum
91.1k394161
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add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
oldest
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active
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votes
$begingroup$
Yes, there is always such a function.
Take a sequence of continuous functions $g_n$ which converges pointwise, but not uniformly, to $0$. Then $f_n=f + g_n$ converges pointwise, but not uniformly to $f$.
For an example of such $g_n$, you can take the following:
$$
g_n(x) = cases{
nx & if $x< frac1n$\
2-nx & if $frac1n leq x < frac2n$\
0& otherwise}
$$
The graph of $g_n$ will be a triangle starting at $(0,0)$, going up to $(1, frac1n)$, then down to $(frac2n, 0)$, and then flat horizontal from there. This does converge pointwise to $0$ (as at any non-zero point $ain [0,1]$, eventually $frac2n<a$, and $g_n(a) = 0$), but not uniformly, as it always has a maximum value of $1$.
answered Dec 12 '18 at 13:31
ArthurArthur
116k7116198
$endgroup$
1
$begingroup$
Heh, the exact same answer up to the letter used in order to avoid duplication of $f_n$ :)
$endgroup$
– 5xum
Dec 12 '18 at 13:33
$begingroup$
@5xum I gues it's pretty standardized how one makes the "trivial" examples.
$endgroup$
– Arthur
Dec 12 '18 at 13:38
add a comment |
$begingroup$
Yes, there is always such a function.
Take a sequence of continuous functions $g_n$ which converges pointwise, but not uniformly, to $0$. Then $f_n=f + g_n$ converges pointwise, but not uniformly to $f$.
For an example of such $g_n$, you can take the following:
$$
g_n(x) = cases{
nx & if $x< frac1n$\
2-nx & if $frac1n leq x < frac2n$\
0& otherwise}
$$
The graph of $g_n$ will be a triangle starting at $(0,0)$, going up to $(1, frac1n)$, then down to $(frac2n, 0)$, and then flat horizontal from there. This does converge pointwise to $0$ (as at any non-zero point $ain [0,1]$, eventually $frac2n<a$, and $g_n(a) = 0$), but not uniformly, as it always has a maximum value of $1$.
answered Dec 12 '18 at 13:31
ArthurArthur
116k7116198
$endgroup$
1
$begingroup$
Heh, the exact same answer up to the letter used in order to avoid duplication of $f_n$ :)
$endgroup$
– 5xum
Dec 12 '18 at 13:33
$begingroup$
@5xum I gues it's pretty standardized how one makes the "trivial" examples.
$endgroup$
– Arthur
Dec 12 '18 at 13:38
add a comment |
$begingroup$
Yes, there is always such a function.
Take a sequence of continuous functions $g_n$ which converges pointwise, but not uniformly, to $0$. Then $f_n=f + g_n$ converges pointwise, but not uniformly to $f$.
For an example of such $g_n$, you can take the following:
$$
g_n(x) = cases{
nx & if $x< frac1n$\
2-nx & if $frac1n leq x < frac2n$\
0& otherwise}
$$
The graph of $g_n$ will be a triangle starting at $(0,0)$, going up to $(1, frac1n)$, then down to $(frac2n, 0)$, and then flat horizontal from there. This does converge pointwise to $0$ (as at any non-zero point $ain [0,1]$, eventually $frac2n<a$, and $g_n(a) = 0$), but not uniformly, as it always has a maximum value of $1$.
answered Dec 12 '18 at 13:31
ArthurArthur
116k7116198
$endgroup$
Yes, there is always such a function.
Take a sequence of continuous functions $g_n$ which converges pointwise, but not uniformly, to $0$. Then $f_n=f + g_n$ converges pointwise, but not uniformly to $f$.
For an example of such $g_n$, you can take the following:
$$
g_n(x) = cases{
nx & if $x< frac1n$\
2-nx & if $frac1n leq x < frac2n$\
0& otherwise}
$$
The graph of $g_n$ will be a triangle starting at $(0,0)$, going up to $(1, frac1n)$, then down to $(frac2n, 0)$, and then flat horizontal from there. This does converge pointwise to $0$ (as at any non-zero point $ain [0,1]$, eventually $frac2n<a$, and $g_n(a) = 0$), but not uniformly, as it always has a maximum value of $1$.
answered Dec 12 '18 at 13:31
ArthurArthur
116k7116198
edited Dec 12 '18 at 13:36
answered Dec 12 '18 at 13:31
ArthurArthur
116k7116198
answered Dec 12 '18 at 13:31
ArthurArthur
116k7116198
answered Dec 12 '18 at 13:31
ArthurArthur
116k7116198
116k7116198
1
$begingroup$
Heh, the exact same answer up to the letter used in order to avoid duplication of $f_n$ :)
$endgroup$
– 5xum
Dec 12 '18 at 13:33
$begingroup$
@5xum I gues it's pretty standardized how one makes the "trivial" examples.
$endgroup$
– Arthur
Dec 12 '18 at 13:38
add a comment |
1
$begingroup$
Heh, the exact same answer up to the letter used in order to avoid duplication of $f_n$ :)
$endgroup$
– 5xum
Dec 12 '18 at 13:33
$begingroup$
@5xum I gues it's pretty standardized how one makes the "trivial" examples.
$endgroup$
– Arthur
Dec 12 '18 at 13:38
1
1
$begingroup$
Heh, the exact same answer up to the letter used in order to avoid duplication of $f_n$ :)
$endgroup$
– 5xum
Dec 12 '18 at 13:33
$begingroup$
Heh, the exact same answer up to the letter used in order to avoid duplication of $f_n$ :)
$endgroup$
– 5xum
Dec 12 '18 at 13:33
$begingroup$
@5xum I gues it's pretty standardized how one makes the "trivial" examples.
$endgroup$
– Arthur
Dec 12 '18 at 13:38
$begingroup$
@5xum I gues it's pretty standardized how one makes the "trivial" examples.
$endgroup$
– Arthur
Dec 12 '18 at 13:38
add a comment |
$begingroup$
Let $g_n$ be the sequence that converges almost everywhere to $0$, but does not converge uniformly (we already know that exists).
Let $f$ be a continuous function.
Then, $f_n=f+g_n$ converges almost everywhere to $f$, but does not converge uniformly.
answered Dec 12 '18 at 13:32
5xum5xum
91.1k394161
$endgroup$
add a comment |
$begingroup$
Let $g_n$ be the sequence that converges almost everywhere to $0$, but does not converge uniformly (we already know that exists).
Let $f$ be a continuous function.
Then, $f_n=f+g_n$ converges almost everywhere to $f$, but does not converge uniformly.
answered Dec 12 '18 at 13:32
5xum5xum
91.1k394161
$endgroup$
add a comment |
$begingroup$
Let $g_n$ be the sequence that converges almost everywhere to $0$, but does not converge uniformly (we already know that exists).
Let $f$ be a continuous function.
Then, $f_n=f+g_n$ converges almost everywhere to $f$, but does not converge uniformly.
answered Dec 12 '18 at 13:32
5xum5xum
91.1k394161
$endgroup$
Let $g_n$ be the sequence that converges almost everywhere to $0$, but does not converge uniformly (we already know that exists).
Let $f$ be a continuous function.
Then, $f_n=f+g_n$ converges almost everywhere to $f$, but does not converge uniformly.
answered Dec 12 '18 at 13:32
5xum5xum
91.1k394161
answered Dec 12 '18 at 13:32
5xum5xum
91.1k394161
answered Dec 12 '18 at 13:32
5xum5xum
91.1k394161
answered Dec 12 '18 at 13:32
5xum5xum
91.1k394161
91.1k394161
add a comment |
add a comment |
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