Mixing
net convergence implies bounded?
$begingroup$
I just saw the following theorem:
Theorem Let $alpha:[a,b] to mathbb{R}$ be a mapping. If the Riemann-Stieltjes integral $$I(f) := int_a^b f(t) , dalpha(t)$$ exists for all continuous functions $f:[a,b] to mathbb{R}$, then $alpha$ is of bounded variation.
in this answer.
but I'm confused by this step in the proof:
Since, by assumption, $I^{Pi}(f) to I(f)$ as $|Pi| to 0$ for all $f in C[a,b]$, we have
$$sup_{Pi} |I^{Pi}(f)| leq c_f < infty$$
Does he use the fact that if $a_n$ converge, then $sup_n a_n<infty$ ? But I have heard from someone that the convergence of Riemann sums is a kind of "net convergence", does this convergence have the same property of ordinary convergence?
real-analysis general-topology convergence riemann-integration
asked Aug 30 '17 at 3:33
LookoutLookout
897818
$endgroup$
add a comment |
$begingroup$
I just saw the following theorem:
Theorem Let $alpha:[a,b] to mathbb{R}$ be a mapping. If the Riemann-Stieltjes integral $$I(f) := int_a^b f(t) , dalpha(t)$$ exists for all continuous functions $f:[a,b] to mathbb{R}$, then $alpha$ is of bounded variation.
in this answer.
but I'm confused by this step in the proof:
Since, by assumption, $I^{Pi}(f) to I(f)$ as $|Pi| to 0$ for all $f in C[a,b]$, we have
$$sup_{Pi} |I^{Pi}(f)| leq c_f < infty$$
Does he use the fact that if $a_n$ converge, then $sup_n a_n<infty$ ? But I have heard from someone that the convergence of Riemann sums is a kind of "net convergence", does this convergence have the same property of ordinary convergence?
real-analysis general-topology convergence riemann-integration
asked Aug 30 '17 at 3:33
LookoutLookout
897818
$endgroup$
$begingroup$
In general, a convergent net need not be bounded. The special situation here may imply the boundedness of $I^{Pi}(f)$, but it's not obvious to me.
$endgroup$
– Daniel Fischer♦
Aug 31 '17 at 15:21
add a comment |
$begingroup$
I just saw the following theorem:
Theorem Let $alpha:[a,b] to mathbb{R}$ be a mapping. If the Riemann-Stieltjes integral $$I(f) := int_a^b f(t) , dalpha(t)$$ exists for all continuous functions $f:[a,b] to mathbb{R}$, then $alpha$ is of bounded variation.
in this answer.
but I'm confused by this step in the proof:
Since, by assumption, $I^{Pi}(f) to I(f)$ as $|Pi| to 0$ for all $f in C[a,b]$, we have
$$sup_{Pi} |I^{Pi}(f)| leq c_f < infty$$
Does he use the fact that if $a_n$ converge, then $sup_n a_n<infty$ ? But I have heard from someone that the convergence of Riemann sums is a kind of "net convergence", does this convergence have the same property of ordinary convergence?
real-analysis general-topology convergence riemann-integration
asked Aug 30 '17 at 3:33
LookoutLookout
897818
$endgroup$
I just saw the following theorem:
Theorem Let $alpha:[a,b] to mathbb{R}$ be a mapping. If the Riemann-Stieltjes integral $$I(f) := int_a^b f(t) , dalpha(t)$$ exists for all continuous functions $f:[a,b] to mathbb{R}$, then $alpha$ is of bounded variation.
in this answer.
but I'm confused by this step in the proof:
Since, by assumption, $I^{Pi}(f) to I(f)$ as $|Pi| to 0$ for all $f in C[a,b]$, we have
$$sup_{Pi} |I^{Pi}(f)| leq c_f < infty$$
Does he use the fact that if $a_n$ converge, then $sup_n a_n<infty$ ? But I have heard from someone that the convergence of Riemann sums is a kind of "net convergence", does this convergence have the same property of ordinary convergence?
real-analysis general-topology convergence riemann-integration
real-analysis general-topology convergence riemann-integration
asked Aug 30 '17 at 3:33
LookoutLookout
897818
asked Aug 30 '17 at 3:33
LookoutLookout
897818
edited Aug 31 '17 at 13:50
Lookout
asked Aug 30 '17 at 3:33
LookoutLookout
897818
asked Aug 30 '17 at 3:33
LookoutLookout
897818
asked Aug 30 '17 at 3:33
LookoutLookout
897818
897818
$begingroup$
In general, a convergent net need not be bounded. The special situation here may imply the boundedness of $I^{Pi}(f)$, but it's not obvious to me.
$endgroup$
– Daniel Fischer♦
Aug 31 '17 at 15:21
add a comment |
$begingroup$
In general, a convergent net need not be bounded. The special situation here may imply the boundedness of $I^{Pi}(f)$, but it's not obvious to me.
$endgroup$
– Daniel Fischer♦
Aug 31 '17 at 15:21
$begingroup$
In general, a convergent net need not be bounded. The special situation here may imply the boundedness of $I^{Pi}(f)$, but it's not obvious to me.
$endgroup$
– Daniel Fischer♦
Aug 31 '17 at 15:21
$begingroup$
In general, a convergent net need not be bounded. The special situation here may imply the boundedness of $I^{Pi}(f)$, but it's not obvious to me.
$endgroup$
– Daniel Fischer♦
Aug 31 '17 at 15:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can take limits along one sequence of partitions with the norms (i.e. the maximum of the lengths of subintervals) tending to zero. There is no need to use nets here since you are not proving the existence of limits of Riemann Steiltje sums.
answered Aug 30 '17 at 6:32
Kavi Rama MurthyKavi Rama Murthy
64.5k42765
$endgroup$
$begingroup$
But I want to apply Banach-Steinhaus theorem, so I need to verify $sup_Pi |I^Pi (f)|<infty$ i.e. every partition is needed.
$endgroup$
– Lookout
Aug 30 '17 at 7:23
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can take limits along one sequence of partitions with the norms (i.e. the maximum of the lengths of subintervals) tending to zero. There is no need to use nets here since you are not proving the existence of limits of Riemann Steiltje sums.
answered Aug 30 '17 at 6:32
Kavi Rama MurthyKavi Rama Murthy
64.5k42765
$endgroup$
$begingroup$
But I want to apply Banach-Steinhaus theorem, so I need to verify $sup_Pi |I^Pi (f)|<infty$ i.e. every partition is needed.
$endgroup$
– Lookout
Aug 30 '17 at 7:23
add a comment |
$begingroup$
You can take limits along one sequence of partitions with the norms (i.e. the maximum of the lengths of subintervals) tending to zero. There is no need to use nets here since you are not proving the existence of limits of Riemann Steiltje sums.
answered Aug 30 '17 at 6:32
Kavi Rama MurthyKavi Rama Murthy
64.5k42765
$endgroup$
$begingroup$
But I want to apply Banach-Steinhaus theorem, so I need to verify $sup_Pi |I^Pi (f)|<infty$ i.e. every partition is needed.
$endgroup$
– Lookout
Aug 30 '17 at 7:23
add a comment |
$begingroup$
You can take limits along one sequence of partitions with the norms (i.e. the maximum of the lengths of subintervals) tending to zero. There is no need to use nets here since you are not proving the existence of limits of Riemann Steiltje sums.
answered Aug 30 '17 at 6:32
Kavi Rama MurthyKavi Rama Murthy
64.5k42765
$endgroup$
You can take limits along one sequence of partitions with the norms (i.e. the maximum of the lengths of subintervals) tending to zero. There is no need to use nets here since you are not proving the existence of limits of Riemann Steiltje sums.
answered Aug 30 '17 at 6:32
Kavi Rama MurthyKavi Rama Murthy
64.5k42765
answered Aug 30 '17 at 6:32
Kavi Rama MurthyKavi Rama Murthy
64.5k42765
answered Aug 30 '17 at 6:32
Kavi Rama MurthyKavi Rama Murthy
64.5k42765
answered Aug 30 '17 at 6:32
Kavi Rama MurthyKavi Rama Murthy
64.5k42765
64.5k42765
$begingroup$
But I want to apply Banach-Steinhaus theorem, so I need to verify $sup_Pi |I^Pi (f)|<infty$ i.e. every partition is needed.
$endgroup$
– Lookout
Aug 30 '17 at 7:23
add a comment |
$begingroup$
But I want to apply Banach-Steinhaus theorem, so I need to verify $sup_Pi |I^Pi (f)|<infty$ i.e. every partition is needed.
$endgroup$
– Lookout
Aug 30 '17 at 7:23
$begingroup$
But I want to apply Banach-Steinhaus theorem, so I need to verify $sup_Pi |I^Pi (f)|<infty$ i.e. every partition is needed.
$endgroup$
– Lookout
Aug 30 '17 at 7:23
$begingroup$
But I want to apply Banach-Steinhaus theorem, so I need to verify $sup_Pi |I^Pi (f)|<infty$ i.e. every partition is needed.
$endgroup$
– Lookout
Aug 30 '17 at 7:23
add a comment |
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$begingroup$
In general, a convergent net need not be bounded. The special situation here may imply the boundedness of $I^{Pi}(f)$, but it's not obvious to me.
$endgroup$
– Daniel Fischer♦
Aug 31 '17 at 15:21