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Proof involving uniform convergence of a sequence and functions.
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Suppose $f_nto f$ uniformly on $[0,1]$ and $g$ is continuous on $[0,1]$. Prove that the sequence of functions $f_ng$ converges uniformly to the product $fg$ on $[0,1]$.
My draft of a proof is the folloing:
Since $g$ is continuous on $[0,1]$, it is bounded by the Extreme Value Theorem. Since $f$ is bounded and $f_n$ is uniformly bounded, there is an $M>0$ such that $textmaxf_n(x)-f(x)leq M$. Given $epsilon>0$ choose $delta>0$ so small that $0<x<0+delta$ or $1>x>1-delta$ inplies $|g(x)|<fracepsilonM$. By hypothesis, $f_nto f$ uniformly on $[0+delta,1-delta]$. Thus choose $N$ so large that $xin[0+delta,1-delta]$ and $ngeq N$ imply $|f_n(x)-f(x)|<fracepsilonC$. If $ngeq N$ and $xin[0,1]$, then $$|f_n(x)g(x)-f(x)g(x)|=|f_n-f(x)||g(x)|<begincases
(fracepsilonC)cdot C=epsilon hspace.5cm textfor hspace.5cm xin[0+delta,1-delta]\
(fracepsilonM)cdot M=epsilon hspace.5cm textfor hspace.5cm xnotin[0+delta,1-delta].endcases$$
Therefore, $f_ngto fg$ uniformly on $[0,1]$
Is there anywhere I can improve or are incorrect?
real-analysis proof-verification
asked Jul 25 at 20:30
Peetrius
358111
 |Â
show 1 more comment
up vote
1
down vote
favorite
Suppose $f_nto f$ uniformly on $[0,1]$ and $g$ is continuous on $[0,1]$. Prove that the sequence of functions $f_ng$ converges uniformly to the product $fg$ on $[0,1]$.
My draft of a proof is the folloing:
Since $g$ is continuous on $[0,1]$, it is bounded by the Extreme Value Theorem. Since $f$ is bounded and $f_n$ is uniformly bounded, there is an $M>0$ such that $textmaxf_n(x)-f(x)leq M$. Given $epsilon>0$ choose $delta>0$ so small that $0<x<0+delta$ or $1>x>1-delta$ inplies $|g(x)|<fracepsilonM$. By hypothesis, $f_nto f$ uniformly on $[0+delta,1-delta]$. Thus choose $N$ so large that $xin[0+delta,1-delta]$ and $ngeq N$ imply $|f_n(x)-f(x)|<fracepsilonC$. If $ngeq N$ and $xin[0,1]$, then $$|f_n(x)g(x)-f(x)g(x)|=|f_n-f(x)||g(x)|<begincases
(fracepsilonC)cdot C=epsilon hspace.5cm textfor hspace.5cm xin[0+delta,1-delta]\
(fracepsilonM)cdot M=epsilon hspace.5cm textfor hspace.5cm xnotin[0+delta,1-delta].endcases$$
Therefore, $f_ngto fg$ uniformly on $[0,1]$
Is there anywhere I can improve or are incorrect?
real-analysis proof-verification
asked Jul 25 at 20:30
Peetrius
358111
1
How do you define $b$?
– Eduardo Longa
Jul 25 at 20:33
those were placeholder values that I forgot to edit in, my bad. Fixing now
– Peetrius
Jul 25 at 20:34
1
Are you given that $f_n$ are continuous or $f$ is continuous?
– B. Mehta
Jul 25 at 20:34
I am not given either those pieces of information, just that $f_nto f$ uniformly on $[0,1]$.
– Peetrius
Jul 25 at 20:35
2
In that case, it may not be true that $f$ is bounded: for instance consider $f(0) = 0$ and $f(x) = 1/x$ for $x > 0$, and $f_n = f$ for all $n$.
– B. Mehta
Jul 25 at 20:38
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose $f_nto f$ uniformly on $[0,1]$ and $g$ is continuous on $[0,1]$. Prove that the sequence of functions $f_ng$ converges uniformly to the product $fg$ on $[0,1]$.
My draft of a proof is the folloing:
Since $g$ is continuous on $[0,1]$, it is bounded by the Extreme Value Theorem. Since $f$ is bounded and $f_n$ is uniformly bounded, there is an $M>0$ such that $textmaxf_n(x)-f(x)leq M$. Given $epsilon>0$ choose $delta>0$ so small that $0<x<0+delta$ or $1>x>1-delta$ inplies $|g(x)|<fracepsilonM$. By hypothesis, $f_nto f$ uniformly on $[0+delta,1-delta]$. Thus choose $N$ so large that $xin[0+delta,1-delta]$ and $ngeq N$ imply $|f_n(x)-f(x)|<fracepsilonC$. If $ngeq N$ and $xin[0,1]$, then $$|f_n(x)g(x)-f(x)g(x)|=|f_n-f(x)||g(x)|<begincases
(fracepsilonC)cdot C=epsilon hspace.5cm textfor hspace.5cm xin[0+delta,1-delta]\
(fracepsilonM)cdot M=epsilon hspace.5cm textfor hspace.5cm xnotin[0+delta,1-delta].endcases$$
Therefore, $f_ngto fg$ uniformly on $[0,1]$
Is there anywhere I can improve or are incorrect?
real-analysis proof-verification
asked Jul 25 at 20:30
Peetrius
358111
Suppose $f_nto f$ uniformly on $[0,1]$ and $g$ is continuous on $[0,1]$. Prove that the sequence of functions $f_ng$ converges uniformly to the product $fg$ on $[0,1]$.
My draft of a proof is the folloing:
Since $g$ is continuous on $[0,1]$, it is bounded by the Extreme Value Theorem. Since $f$ is bounded and $f_n$ is uniformly bounded, there is an $M>0$ such that $textmaxf_n(x)-f(x)leq M$. Given $epsilon>0$ choose $delta>0$ so small that $0<x<0+delta$ or $1>x>1-delta$ inplies $|g(x)|<fracepsilonM$. By hypothesis, $f_nto f$ uniformly on $[0+delta,1-delta]$. Thus choose $N$ so large that $xin[0+delta,1-delta]$ and $ngeq N$ imply $|f_n(x)-f(x)|<fracepsilonC$. If $ngeq N$ and $xin[0,1]$, then $$|f_n(x)g(x)-f(x)g(x)|=|f_n-f(x)||g(x)|<begincases
(fracepsilonC)cdot C=epsilon hspace.5cm textfor hspace.5cm xin[0+delta,1-delta]\
(fracepsilonM)cdot M=epsilon hspace.5cm textfor hspace.5cm xnotin[0+delta,1-delta].endcases$$
Therefore, $f_ngto fg$ uniformly on $[0,1]$
Is there anywhere I can improve or are incorrect?
real-analysis proof-verification
asked Jul 25 at 20:30
Peetrius
358111
edited Jul 25 at 20:34
asked Jul 25 at 20:30
Peetrius
358111
asked Jul 25 at 20:30
Peetrius
358111
asked Jul 25 at 20:30
Peetrius
358111
358111
1
How do you define $b$?
– Eduardo Longa
Jul 25 at 20:33
those were placeholder values that I forgot to edit in, my bad. Fixing now
– Peetrius
Jul 25 at 20:34
1
Are you given that $f_n$ are continuous or $f$ is continuous?
– B. Mehta
Jul 25 at 20:34
I am not given either those pieces of information, just that $f_nto f$ uniformly on $[0,1]$.
– Peetrius
Jul 25 at 20:35
2
In that case, it may not be true that $f$ is bounded: for instance consider $f(0) = 0$ and $f(x) = 1/x$ for $x > 0$, and $f_n = f$ for all $n$.
– B. Mehta
Jul 25 at 20:38
 |Â
show 1 more comment
1
How do you define $b$?
– Eduardo Longa
Jul 25 at 20:33
those were placeholder values that I forgot to edit in, my bad. Fixing now
– Peetrius
Jul 25 at 20:34
1
Are you given that $f_n$ are continuous or $f$ is continuous?
– B. Mehta
Jul 25 at 20:34
I am not given either those pieces of information, just that $f_nto f$ uniformly on $[0,1]$.
– Peetrius
Jul 25 at 20:35
2
In that case, it may not be true that $f$ is bounded: for instance consider $f(0) = 0$ and $f(x) = 1/x$ for $x > 0$, and $f_n = f$ for all $n$.
– B. Mehta
Jul 25 at 20:38
1
1
How do you define $b$?
– Eduardo Longa
Jul 25 at 20:33
How do you define $b$?
– Eduardo Longa
Jul 25 at 20:33
those were placeholder values that I forgot to edit in, my bad. Fixing now
– Peetrius
Jul 25 at 20:34
those were placeholder values that I forgot to edit in, my bad. Fixing now
– Peetrius
Jul 25 at 20:34
1
1
Are you given that $f_n$ are continuous or $f$ is continuous?
– B. Mehta
Jul 25 at 20:34
Are you given that $f_n$ are continuous or $f$ is continuous?
– B. Mehta
Jul 25 at 20:34
I am not given either those pieces of information, just that $f_nto f$ uniformly on $[0,1]$.
– Peetrius
Jul 25 at 20:35
I am not given either those pieces of information, just that $f_nto f$ uniformly on $[0,1]$.
– Peetrius
Jul 25 at 20:35
2
2
In that case, it may not be true that $f$ is bounded: for instance consider $f(0) = 0$ and $f(x) = 1/x$ for $x > 0$, and $f_n = f$ for all $n$.
– B. Mehta
Jul 25 at 20:38
In that case, it may not be true that $f$ is bounded: for instance consider $f(0) = 0$ and $f(x) = 1/x$ for $x > 0$, and $f_n = f$ for all $n$.
– B. Mehta
Jul 25 at 20:38
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
The property holds as soon as $g$ is bounded in $[0,1]$ (and this is true when $g$ is continuous):
$$sup_xin[0,1]|f_n(x)g(x)-f(x)g(x)|leqsup_xin[0,1]|f_n(x)-f(x)|cdot sup_xin[0,1]|g(x)|.$$
Therefore, if $f_nto f$ uniformly in $[0,1]$ then $sup_xin[0,1]|f_n(x)-f(x)|to 0$ and, by the above inequality, $sup_xin[0,1]|f_n(x)g(x)-f(x)g(x)|to 0$ and we may conclude that $f_ngto fg$ uniformly in $[0,1]$.
answered Jul 25 at 20:40
Robert Z
83.9k954122
1
That might be true only for $n$ large enough if you limit yourself to finite quantities.
– mathcounterexamples.net
Jul 25 at 20:42
$g$ is always bounded in $[0,1]$ as it is continuous.
– B. Mehta
Jul 25 at 20:45
1
@mathcounterexamples.net Do you think that the inequality is not true? Why?
– Robert Z
Jul 25 at 20:50
I’m just saying that $sup vert f(x)-f_n(x) vert $ may not be defined for « small » integers.
– mathcounterexamples.net
Jul 25 at 20:58
The $sup$ can be $+infty$. I don't see any problem with that.
– Robert Z
Jul 25 at 21:02
 |Â
show 1 more comment
up vote
0
down vote
Thanks to Robert and B. Mehta I think I am confident in my proof.
Since $g$ is continuous on $[0,1]$, it is bounded by the Extreme Value Theorem.
In other words, there is a $C>0$ such that $|g(x)|leq C$ for all $xin[0,1]$.
Additionally, there is an $M>0$ such that $sup_xin[0,1]leq M$.
Given $epsilon>0$ choose $delta>0$ so small that $0<x<0+delta$ or $1>x>1-delta$ implies $|g(x)|<fracepsilonM$.
By hypothesis, $f_nto f$ uniformly on $[0+delta,1-delta]$.
Thus choose $N$ so large that $xin[0+delta,1-delta]$ and $ngeq N$ imply $|f_n(x)-f(x)|<fracepsilonC$.
If $ngeq N$ and $xin[0,1]$, then $$sup|f_n(x)g(x)-f(x)g(x)|leqsup|f_n-f(x)|cdotsup|g(x)|leqsup|g(x)|fracepsilonC<epsilonhspace.2cmtextforhspace.2cmxin[0+delta,1-delta]$$
Therefore, $f_ngto fg$ uniformly on $[0,1].$
answered Jul 25 at 21:21
Peetrius
358111
1
Your proof shows that $f_ngto fg$ in $[0+delta,1-delta]$ for any $delta>0$ which does not inply that the convergence is uniformly also in $[0,1]$. But why do you need $delta>0$?
– Robert Z
Jul 25 at 21:31
Line 4 of my proof describes $delta$. Isn't $delta$ necessary since this problem has a $epsilon$ for function values, it would need $delta$ for input values?
– Peetrius
Jul 25 at 21:41
1
There is no need of $delta$. Given $epsilon>0$ there is $N$ such that for $ngeq N$, $sup_xin[0,1]|f_n(x)-f(x)|<fracepsilonC$.
– Robert Z
Jul 25 at 21:48
Got it, my book used $delta$ for a similar proof so I tried using that as a baseline for mine.
– Peetrius
Jul 25 at 21:57
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The property holds as soon as $g$ is bounded in $[0,1]$ (and this is true when $g$ is continuous):
$$sup_xin[0,1]|f_n(x)g(x)-f(x)g(x)|leqsup_xin[0,1]|f_n(x)-f(x)|cdot sup_xin[0,1]|g(x)|.$$
Therefore, if $f_nto f$ uniformly in $[0,1]$ then $sup_xin[0,1]|f_n(x)-f(x)|to 0$ and, by the above inequality, $sup_xin[0,1]|f_n(x)g(x)-f(x)g(x)|to 0$ and we may conclude that $f_ngto fg$ uniformly in $[0,1]$.
answered Jul 25 at 20:40
Robert Z
83.9k954122
1
That might be true only for $n$ large enough if you limit yourself to finite quantities.
– mathcounterexamples.net
Jul 25 at 20:42
$g$ is always bounded in $[0,1]$ as it is continuous.
– B. Mehta
Jul 25 at 20:45
1
@mathcounterexamples.net Do you think that the inequality is not true? Why?
– Robert Z
Jul 25 at 20:50
I’m just saying that $sup vert f(x)-f_n(x) vert $ may not be defined for « small » integers.
– mathcounterexamples.net
Jul 25 at 20:58
The $sup$ can be $+infty$. I don't see any problem with that.
– Robert Z
Jul 25 at 21:02
 |Â
show 1 more comment
up vote
2
down vote
accepted
The property holds as soon as $g$ is bounded in $[0,1]$ (and this is true when $g$ is continuous):
$$sup_xin[0,1]|f_n(x)g(x)-f(x)g(x)|leqsup_xin[0,1]|f_n(x)-f(x)|cdot sup_xin[0,1]|g(x)|.$$
Therefore, if $f_nto f$ uniformly in $[0,1]$ then $sup_xin[0,1]|f_n(x)-f(x)|to 0$ and, by the above inequality, $sup_xin[0,1]|f_n(x)g(x)-f(x)g(x)|to 0$ and we may conclude that $f_ngto fg$ uniformly in $[0,1]$.
answered Jul 25 at 20:40
Robert Z
83.9k954122
1
That might be true only for $n$ large enough if you limit yourself to finite quantities.
– mathcounterexamples.net
Jul 25 at 20:42
$g$ is always bounded in $[0,1]$ as it is continuous.
– B. Mehta
Jul 25 at 20:45
1
@mathcounterexamples.net Do you think that the inequality is not true? Why?
– Robert Z
Jul 25 at 20:50
I’m just saying that $sup vert f(x)-f_n(x) vert $ may not be defined for « small » integers.
– mathcounterexamples.net
Jul 25 at 20:58
The $sup$ can be $+infty$. I don't see any problem with that.
– Robert Z
Jul 25 at 21:02
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The property holds as soon as $g$ is bounded in $[0,1]$ (and this is true when $g$ is continuous):
$$sup_xin[0,1]|f_n(x)g(x)-f(x)g(x)|leqsup_xin[0,1]|f_n(x)-f(x)|cdot sup_xin[0,1]|g(x)|.$$
Therefore, if $f_nto f$ uniformly in $[0,1]$ then $sup_xin[0,1]|f_n(x)-f(x)|to 0$ and, by the above inequality, $sup_xin[0,1]|f_n(x)g(x)-f(x)g(x)|to 0$ and we may conclude that $f_ngto fg$ uniformly in $[0,1]$.
answered Jul 25 at 20:40
Robert Z
83.9k954122
The property holds as soon as $g$ is bounded in $[0,1]$ (and this is true when $g$ is continuous):
$$sup_xin[0,1]|f_n(x)g(x)-f(x)g(x)|leqsup_xin[0,1]|f_n(x)-f(x)|cdot sup_xin[0,1]|g(x)|.$$
Therefore, if $f_nto f$ uniformly in $[0,1]$ then $sup_xin[0,1]|f_n(x)-f(x)|to 0$ and, by the above inequality, $sup_xin[0,1]|f_n(x)g(x)-f(x)g(x)|to 0$ and we may conclude that $f_ngto fg$ uniformly in $[0,1]$.
answered Jul 25 at 20:40
Robert Z
83.9k954122
edited Jul 25 at 20:53
answered Jul 25 at 20:40
Robert Z
83.9k954122
answered Jul 25 at 20:40
Robert Z
83.9k954122
answered Jul 25 at 20:40
Robert Z
83.9k954122
83.9k954122
1
That might be true only for $n$ large enough if you limit yourself to finite quantities.
– mathcounterexamples.net
Jul 25 at 20:42
$g$ is always bounded in $[0,1]$ as it is continuous.
– B. Mehta
Jul 25 at 20:45
1
@mathcounterexamples.net Do you think that the inequality is not true? Why?
– Robert Z
Jul 25 at 20:50
I’m just saying that $sup vert f(x)-f_n(x) vert $ may not be defined for « small » integers.
– mathcounterexamples.net
Jul 25 at 20:58
The $sup$ can be $+infty$. I don't see any problem with that.
– Robert Z
Jul 25 at 21:02
 |Â
show 1 more comment
1
That might be true only for $n$ large enough if you limit yourself to finite quantities.
– mathcounterexamples.net
Jul 25 at 20:42
$g$ is always bounded in $[0,1]$ as it is continuous.
– B. Mehta
Jul 25 at 20:45
1
@mathcounterexamples.net Do you think that the inequality is not true? Why?
– Robert Z
Jul 25 at 20:50
I’m just saying that $sup vert f(x)-f_n(x) vert $ may not be defined for « small » integers.
– mathcounterexamples.net
Jul 25 at 20:58
The $sup$ can be $+infty$. I don't see any problem with that.
– Robert Z
Jul 25 at 21:02
1
1
That might be true only for $n$ large enough if you limit yourself to finite quantities.
– mathcounterexamples.net
Jul 25 at 20:42
That might be true only for $n$ large enough if you limit yourself to finite quantities.
– mathcounterexamples.net
Jul 25 at 20:42
$g$ is always bounded in $[0,1]$ as it is continuous.
– B. Mehta
Jul 25 at 20:45
$g$ is always bounded in $[0,1]$ as it is continuous.
– B. Mehta
Jul 25 at 20:45
1
1
@mathcounterexamples.net Do you think that the inequality is not true? Why?
– Robert Z
Jul 25 at 20:50
@mathcounterexamples.net Do you think that the inequality is not true? Why?
– Robert Z
Jul 25 at 20:50
I’m just saying that $sup vert f(x)-f_n(x) vert $ may not be defined for « small » integers.
– mathcounterexamples.net
Jul 25 at 20:58
I’m just saying that $sup vert f(x)-f_n(x) vert $ may not be defined for « small » integers.
– mathcounterexamples.net
Jul 25 at 20:58
The $sup$ can be $+infty$. I don't see any problem with that.
– Robert Z
Jul 25 at 21:02
The $sup$ can be $+infty$. I don't see any problem with that.
– Robert Z
Jul 25 at 21:02
 |Â
show 1 more comment
up vote
0
down vote
Thanks to Robert and B. Mehta I think I am confident in my proof.
Since $g$ is continuous on $[0,1]$, it is bounded by the Extreme Value Theorem.
In other words, there is a $C>0$ such that $|g(x)|leq C$ for all $xin[0,1]$.
Additionally, there is an $M>0$ such that $sup_xin[0,1]leq M$.
Given $epsilon>0$ choose $delta>0$ so small that $0<x<0+delta$ or $1>x>1-delta$ implies $|g(x)|<fracepsilonM$.
By hypothesis, $f_nto f$ uniformly on $[0+delta,1-delta]$.
Thus choose $N$ so large that $xin[0+delta,1-delta]$ and $ngeq N$ imply $|f_n(x)-f(x)|<fracepsilonC$.
If $ngeq N$ and $xin[0,1]$, then $$sup|f_n(x)g(x)-f(x)g(x)|leqsup|f_n-f(x)|cdotsup|g(x)|leqsup|g(x)|fracepsilonC<epsilonhspace.2cmtextforhspace.2cmxin[0+delta,1-delta]$$
Therefore, $f_ngto fg$ uniformly on $[0,1].$
answered Jul 25 at 21:21
Peetrius
358111
1
Your proof shows that $f_ngto fg$ in $[0+delta,1-delta]$ for any $delta>0$ which does not inply that the convergence is uniformly also in $[0,1]$. But why do you need $delta>0$?
– Robert Z
Jul 25 at 21:31
Line 4 of my proof describes $delta$. Isn't $delta$ necessary since this problem has a $epsilon$ for function values, it would need $delta$ for input values?
– Peetrius
Jul 25 at 21:41
1
There is no need of $delta$. Given $epsilon>0$ there is $N$ such that for $ngeq N$, $sup_xin[0,1]|f_n(x)-f(x)|<fracepsilonC$.
– Robert Z
Jul 25 at 21:48
Got it, my book used $delta$ for a similar proof so I tried using that as a baseline for mine.
– Peetrius
Jul 25 at 21:57
add a comment |Â
up vote
0
down vote
Thanks to Robert and B. Mehta I think I am confident in my proof.
Since $g$ is continuous on $[0,1]$, it is bounded by the Extreme Value Theorem.
In other words, there is a $C>0$ such that $|g(x)|leq C$ for all $xin[0,1]$.
Additionally, there is an $M>0$ such that $sup_xin[0,1]leq M$.
Given $epsilon>0$ choose $delta>0$ so small that $0<x<0+delta$ or $1>x>1-delta$ implies $|g(x)|<fracepsilonM$.
By hypothesis, $f_nto f$ uniformly on $[0+delta,1-delta]$.
Thus choose $N$ so large that $xin[0+delta,1-delta]$ and $ngeq N$ imply $|f_n(x)-f(x)|<fracepsilonC$.
If $ngeq N$ and $xin[0,1]$, then $$sup|f_n(x)g(x)-f(x)g(x)|leqsup|f_n-f(x)|cdotsup|g(x)|leqsup|g(x)|fracepsilonC<epsilonhspace.2cmtextforhspace.2cmxin[0+delta,1-delta]$$
Therefore, $f_ngto fg$ uniformly on $[0,1].$
answered Jul 25 at 21:21
Peetrius
358111
1
Your proof shows that $f_ngto fg$ in $[0+delta,1-delta]$ for any $delta>0$ which does not inply that the convergence is uniformly also in $[0,1]$. But why do you need $delta>0$?
– Robert Z
Jul 25 at 21:31
Line 4 of my proof describes $delta$. Isn't $delta$ necessary since this problem has a $epsilon$ for function values, it would need $delta$ for input values?
– Peetrius
Jul 25 at 21:41
1
There is no need of $delta$. Given $epsilon>0$ there is $N$ such that for $ngeq N$, $sup_xin[0,1]|f_n(x)-f(x)|<fracepsilonC$.
– Robert Z
Jul 25 at 21:48
Got it, my book used $delta$ for a similar proof so I tried using that as a baseline for mine.
– Peetrius
Jul 25 at 21:57
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Thanks to Robert and B. Mehta I think I am confident in my proof.
Since $g$ is continuous on $[0,1]$, it is bounded by the Extreme Value Theorem.
In other words, there is a $C>0$ such that $|g(x)|leq C$ for all $xin[0,1]$.
Additionally, there is an $M>0$ such that $sup_xin[0,1]leq M$.
Given $epsilon>0$ choose $delta>0$ so small that $0<x<0+delta$ or $1>x>1-delta$ implies $|g(x)|<fracepsilonM$.
By hypothesis, $f_nto f$ uniformly on $[0+delta,1-delta]$.
Thus choose $N$ so large that $xin[0+delta,1-delta]$ and $ngeq N$ imply $|f_n(x)-f(x)|<fracepsilonC$.
If $ngeq N$ and $xin[0,1]$, then $$sup|f_n(x)g(x)-f(x)g(x)|leqsup|f_n-f(x)|cdotsup|g(x)|leqsup|g(x)|fracepsilonC<epsilonhspace.2cmtextforhspace.2cmxin[0+delta,1-delta]$$
Therefore, $f_ngto fg$ uniformly on $[0,1].$
answered Jul 25 at 21:21
Peetrius
358111
Thanks to Robert and B. Mehta I think I am confident in my proof.
Since $g$ is continuous on $[0,1]$, it is bounded by the Extreme Value Theorem.
In other words, there is a $C>0$ such that $|g(x)|leq C$ for all $xin[0,1]$.
Additionally, there is an $M>0$ such that $sup_xin[0,1]leq M$.
Given $epsilon>0$ choose $delta>0$ so small that $0<x<0+delta$ or $1>x>1-delta$ implies $|g(x)|<fracepsilonM$.
By hypothesis, $f_nto f$ uniformly on $[0+delta,1-delta]$.
Thus choose $N$ so large that $xin[0+delta,1-delta]$ and $ngeq N$ imply $|f_n(x)-f(x)|<fracepsilonC$.
If $ngeq N$ and $xin[0,1]$, then $$sup|f_n(x)g(x)-f(x)g(x)|leqsup|f_n-f(x)|cdotsup|g(x)|leqsup|g(x)|fracepsilonC<epsilonhspace.2cmtextforhspace.2cmxin[0+delta,1-delta]$$
Therefore, $f_ngto fg$ uniformly on $[0,1].$
answered Jul 25 at 21:21
Peetrius
358111
edited Jul 25 at 21:35
answered Jul 25 at 21:21
Peetrius
358111
answered Jul 25 at 21:21
Peetrius
358111
answered Jul 25 at 21:21
Peetrius
358111
358111
1
Your proof shows that $f_ngto fg$ in $[0+delta,1-delta]$ for any $delta>0$ which does not inply that the convergence is uniformly also in $[0,1]$. But why do you need $delta>0$?
– Robert Z
Jul 25 at 21:31
Line 4 of my proof describes $delta$. Isn't $delta$ necessary since this problem has a $epsilon$ for function values, it would need $delta$ for input values?
– Peetrius
Jul 25 at 21:41
1
There is no need of $delta$. Given $epsilon>0$ there is $N$ such that for $ngeq N$, $sup_xin[0,1]|f_n(x)-f(x)|<fracepsilonC$.
– Robert Z
Jul 25 at 21:48
Got it, my book used $delta$ for a similar proof so I tried using that as a baseline for mine.
– Peetrius
Jul 25 at 21:57
add a comment |Â
1
Your proof shows that $f_ngto fg$ in $[0+delta,1-delta]$ for any $delta>0$ which does not inply that the convergence is uniformly also in $[0,1]$. But why do you need $delta>0$?
– Robert Z
Jul 25 at 21:31
Line 4 of my proof describes $delta$. Isn't $delta$ necessary since this problem has a $epsilon$ for function values, it would need $delta$ for input values?
– Peetrius
Jul 25 at 21:41
1
There is no need of $delta$. Given $epsilon>0$ there is $N$ such that for $ngeq N$, $sup_xin[0,1]|f_n(x)-f(x)|<fracepsilonC$.
– Robert Z
Jul 25 at 21:48
Got it, my book used $delta$ for a similar proof so I tried using that as a baseline for mine.
– Peetrius
Jul 25 at 21:57
1
1
Your proof shows that $f_ngto fg$ in $[0+delta,1-delta]$ for any $delta>0$ which does not inply that the convergence is uniformly also in $[0,1]$. But why do you need $delta>0$?
– Robert Z
Jul 25 at 21:31
Your proof shows that $f_ngto fg$ in $[0+delta,1-delta]$ for any $delta>0$ which does not inply that the convergence is uniformly also in $[0,1]$. But why do you need $delta>0$?
– Robert Z
Jul 25 at 21:31
Line 4 of my proof describes $delta$. Isn't $delta$ necessary since this problem has a $epsilon$ for function values, it would need $delta$ for input values?
– Peetrius
Jul 25 at 21:41
Line 4 of my proof describes $delta$. Isn't $delta$ necessary since this problem has a $epsilon$ for function values, it would need $delta$ for input values?
– Peetrius
Jul 25 at 21:41
1
1
There is no need of $delta$. Given $epsilon>0$ there is $N$ such that for $ngeq N$, $sup_xin[0,1]|f_n(x)-f(x)|<fracepsilonC$.
– Robert Z
Jul 25 at 21:48
There is no need of $delta$. Given $epsilon>0$ there is $N$ such that for $ngeq N$, $sup_xin[0,1]|f_n(x)-f(x)|<fracepsilonC$.
– Robert Z
Jul 25 at 21:48
Got it, my book used $delta$ for a similar proof so I tried using that as a baseline for mine.
– Peetrius
Jul 25 at 21:57
Got it, my book used $delta$ for a similar proof so I tried using that as a baseline for mine.
– Peetrius
Jul 25 at 21:57
add a comment |Â
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1
How do you define $b$?
– Eduardo Longa
Jul 25 at 20:33
those were placeholder values that I forgot to edit in, my bad. Fixing now
– Peetrius
Jul 25 at 20:34
1
Are you given that $f_n$ are continuous or $f$ is continuous?
– B. Mehta
Jul 25 at 20:34
I am not given either those pieces of information, just that $f_nto f$ uniformly on $[0,1]$.
– Peetrius
Jul 25 at 20:35
2
In that case, it may not be true that $f$ is bounded: for instance consider $f(0) = 0$ and $f(x) = 1/x$ for $x > 0$, and $f_n = f$ for all $n$.
– B. Mehta
Jul 25 at 20:38