If $int_a^b f, dalpha $ exists for all continuous $f$ then $alpha$ is of bounded variation Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Why do we unavoidably (or not) use Riemann integral to define Itô integral?Proof of a change of variables formula in integralsHow to prove $int_a^b f,dalpha =int_a^b g,dalpha$?Riemann-Stieltjes integral of unbounded functionDoes the Riemann-Stieltjes integral $int_0 ^1 cos x , d(alpha (x))$ exists if $alpha$ is not of bounded variation?$alpha: [a,b]tomathbbR$ constant, $f:[a,b]tomathbbR$ continuous, then $int_a^b f dalpha = f(c)cdot [a(c+)-a(c-)]$ (Stieltjes)$alpha:(0, +infty) to (0, +infty)$ continuous and increasing, $alpha(1) = 1$. Show $int_1^x dalpha/alpha = log alpha(x)$$F:[a,b]tomathbbR$, $F(x) = int_a^x f dalpha$, then $F$ has bounded variation and is total variation $lesupf(x)cdot V$$lambda_k(t) = (cos kt, sin kt)$, $w$ is continuous $1$-form, then $int_lambda_kw = kint_lambda_1w$Pairs $(f,g)$ for which the Riemann-Stiltjes integral $int_a^bf(x),dg(x)$ existsRiemann-Stieltjes Integral with respect to total variation
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If $int_a^b f, dalpha $ exists for all continuous $f$ then $alpha$ is of bounded variation
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Why do we unavoidably (or not) use Riemann integral to define Itô integral?Proof of a change of variables formula in integralsHow to prove $int_a^b f,dalpha =int_a^b g,dalpha$?Riemann-Stieltjes integral of unbounded functionDoes the Riemann-Stieltjes integral $int_0 ^1 cos x , d(alpha (x))$ exists if $alpha$ is not of bounded variation?$alpha: [a,b]tomathbbR$ constant, $f:[a,b]tomathbbR$ continuous, then $int_a^b f dalpha = f(c)cdot [a(c+)-a(c-)]$ (Stieltjes)$alpha:(0, +infty) to (0, +infty)$ continuous and increasing, $alpha(1) = 1$. Show $int_1^x dalpha/alpha = log alpha(x)$$F:[a,b]tomathbbR$, $F(x) = int_a^x f dalpha$, then $F$ has bounded variation and is total variation $lesupf(x)cdot V$$lambda_k(t) = (cos kt, sin kt)$, $w$ is continuous $1$-form, then $int_lambda_kw = kint_lambda_1w$Pairs $(f,g)$ for which the Riemann-Stiltjes integral $int_a^bf(x),dg(x)$ existsRiemann-Stieltjes Integral with respect to total variation
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In this answer the following theorem is proved:
Theorem: Let $alpha:[a, b] tomathbb R $ be a function such that the Riemann-Stieltjes integral $$int_a ^b f(x) , dalpha(x) $$ exists for all continuous functions $f:[a, b] tomathbb R $. Then $alpha $ is of bounded variation on $[a, b] $.
However the linked answer uses Banach Steinhaus theorem to prove it.
Is there any proof which does not use theorems from functional analysis and is rather based on the usual theorems dealing with Riemann-Stieltjes integration?
real-analysis alternative-proof riemann-integration
asked Apr 2 at 10:56
Paramanand SinghParamanand Singh
51.6k560170
$endgroup$
add a comment |
$begingroup$
In this answer the following theorem is proved:
Theorem: Let $alpha:[a, b] tomathbb R $ be a function such that the Riemann-Stieltjes integral $$int_a ^b f(x) , dalpha(x) $$ exists for all continuous functions $f:[a, b] tomathbb R $. Then $alpha $ is of bounded variation on $[a, b] $.
However the linked answer uses Banach Steinhaus theorem to prove it.
Is there any proof which does not use theorems from functional analysis and is rather based on the usual theorems dealing with Riemann-Stieltjes integration?
real-analysis alternative-proof riemann-integration
asked Apr 2 at 10:56
Paramanand SinghParamanand Singh
51.6k560170
$endgroup$
2
$begingroup$
I doubt such a proof exists. The setting is the natural habitat of functional analysis: a linear functional (integrating against some measure) defined on a normed space (the space of all continuous functions on $[a,b]$).
$endgroup$
– uniquesolution
Apr 2 at 11:18
add a comment |
$begingroup$
In this answer the following theorem is proved:
Theorem: Let $alpha:[a, b] tomathbb R $ be a function such that the Riemann-Stieltjes integral $$int_a ^b f(x) , dalpha(x) $$ exists for all continuous functions $f:[a, b] tomathbb R $. Then $alpha $ is of bounded variation on $[a, b] $.
However the linked answer uses Banach Steinhaus theorem to prove it.
Is there any proof which does not use theorems from functional analysis and is rather based on the usual theorems dealing with Riemann-Stieltjes integration?
real-analysis alternative-proof riemann-integration
asked Apr 2 at 10:56
Paramanand SinghParamanand Singh
51.6k560170
$endgroup$
In this answer the following theorem is proved:
Theorem: Let $alpha:[a, b] tomathbb R $ be a function such that the Riemann-Stieltjes integral $$int_a ^b f(x) , dalpha(x) $$ exists for all continuous functions $f:[a, b] tomathbb R $. Then $alpha $ is of bounded variation on $[a, b] $.
However the linked answer uses Banach Steinhaus theorem to prove it.
Is there any proof which does not use theorems from functional analysis and is rather based on the usual theorems dealing with Riemann-Stieltjes integration?
real-analysis alternative-proof riemann-integration
real-analysis alternative-proof riemann-integration
asked Apr 2 at 10:56
Paramanand SinghParamanand Singh
51.6k560170
asked Apr 2 at 10:56
Paramanand SinghParamanand Singh
51.6k560170
edited Apr 2 at 14:18
Paramanand Singh
asked Apr 2 at 10:56
Paramanand SinghParamanand Singh
51.6k560170
asked Apr 2 at 10:56
Paramanand SinghParamanand Singh
51.6k560170
asked Apr 2 at 10:56
Paramanand SinghParamanand Singh
51.6k560170
51.6k560170
2
$begingroup$
I doubt such a proof exists. The setting is the natural habitat of functional analysis: a linear functional (integrating against some measure) defined on a normed space (the space of all continuous functions on $[a,b]$).
$endgroup$
– uniquesolution
Apr 2 at 11:18
add a comment |
2
$begingroup$
I doubt such a proof exists. The setting is the natural habitat of functional analysis: a linear functional (integrating against some measure) defined on a normed space (the space of all continuous functions on $[a,b]$).
$endgroup$
– uniquesolution
Apr 2 at 11:18
2
2
$begingroup$
I doubt such a proof exists. The setting is the natural habitat of functional analysis: a linear functional (integrating against some measure) defined on a normed space (the space of all continuous functions on $[a,b]$).
$endgroup$
– uniquesolution
Apr 2 at 11:18
$begingroup$
I doubt such a proof exists. The setting is the natural habitat of functional analysis: a linear functional (integrating against some measure) defined on a normed space (the space of all continuous functions on $[a,b]$).
$endgroup$
– uniquesolution
Apr 2 at 11:18
add a comment |
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I doubt such a proof exists. The setting is the natural habitat of functional analysis: a linear functional (integrating against some measure) defined on a normed space (the space of all continuous functions on $[a,b]$).
$endgroup$
– uniquesolution
Apr 2 at 11:18