Convergence of sequence of Riemann-Stieltjes integrals to Riemann-Stieltjes integral












1












$begingroup$


In connection with my post Convergence to Riemann-Stieltjes integral of sequence of Riemann-Stieltjes-like sums with changing integrand and integrator, an alternative approach to my main objective would be considering the convergence of the sequence of Riemann-Stieltjes (RS) integrals $int_0^1 , f_N(x) , mathrm{d}F_N(x)$ to the RS integral $int_0^1 , f(x) , mathrm{d}F(x)$. The properties of the functions involved are as defined in the aforementioned post. I repeat them in the following paragraph for convenience, though.



The functions are all real of a single real variable. Those in the sequence $(f_N)_N$ are continuous and bounded in $[0,1]$, and the sequence converges to a continuous function $f$ bounded in the same interval. In turn, those in $(F_N)_N$ are monotonically increasing step functions bounded in $[0,1]$ and the sequence is uniformly convergent to a function $F$ that is a cumulative distribution function.



Any hints on how to prove the above statement of convergence will be welcome.






integration sequence-of-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm rusty. When you indicate $f_n longrightarrow f$, you intend convergence in the sup norm (i.e. uniform convergence)?
    $endgroup$
    – Eric Towers
    Dec 3 '18 at 7:03






  • 1




    $begingroup$
    Have you tried the "obvious thing": see if you can show $int_0^1 ; |f - f_n| , mathrm{d}(|F - F_n|) longrightarrow 0$?
    $endgroup$
    – Eric Towers
    Dec 3 '18 at 7:09










  • $begingroup$
    Not necessarily uniformly, but if it helps with a proof of convergence (or hints, with the details to be worked out by me), you are welcome to assume it is so.
    $endgroup$
    – Marcos
    Dec 3 '18 at 7:11










  • $begingroup$
    No, actually I haven't. This question dawned on me as a rather indirect but actually alternative approach to the linked post. I'll check that.
    $endgroup$
    – Marcos
    Dec 3 '18 at 7:13
















1












$begingroup$


In connection with my post Convergence to Riemann-Stieltjes integral of sequence of Riemann-Stieltjes-like sums with changing integrand and integrator, an alternative approach to my main objective would be considering the convergence of the sequence of Riemann-Stieltjes (RS) integrals $int_0^1 , f_N(x) , mathrm{d}F_N(x)$ to the RS integral $int_0^1 , f(x) , mathrm{d}F(x)$. The properties of the functions involved are as defined in the aforementioned post. I repeat them in the following paragraph for convenience, though.



The functions are all real of a single real variable. Those in the sequence $(f_N)_N$ are continuous and bounded in $[0,1]$, and the sequence converges to a continuous function $f$ bounded in the same interval. In turn, those in $(F_N)_N$ are monotonically increasing step functions bounded in $[0,1]$ and the sequence is uniformly convergent to a function $F$ that is a cumulative distribution function.



Any hints on how to prove the above statement of convergence will be welcome.






integration sequence-of-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm rusty. When you indicate $f_n longrightarrow f$, you intend convergence in the sup norm (i.e. uniform convergence)?
    $endgroup$
    – Eric Towers
    Dec 3 '18 at 7:03






  • 1




    $begingroup$
    Have you tried the "obvious thing": see if you can show $int_0^1 ; |f - f_n| , mathrm{d}(|F - F_n|) longrightarrow 0$?
    $endgroup$
    – Eric Towers
    Dec 3 '18 at 7:09










  • $begingroup$
    Not necessarily uniformly, but if it helps with a proof of convergence (or hints, with the details to be worked out by me), you are welcome to assume it is so.
    $endgroup$
    – Marcos
    Dec 3 '18 at 7:11










  • $begingroup$
    No, actually I haven't. This question dawned on me as a rather indirect but actually alternative approach to the linked post. I'll check that.
    $endgroup$
    – Marcos
    Dec 3 '18 at 7:13














1












1








1





$begingroup$


In connection with my post Convergence to Riemann-Stieltjes integral of sequence of Riemann-Stieltjes-like sums with changing integrand and integrator, an alternative approach to my main objective would be considering the convergence of the sequence of Riemann-Stieltjes (RS) integrals $int_0^1 , f_N(x) , mathrm{d}F_N(x)$ to the RS integral $int_0^1 , f(x) , mathrm{d}F(x)$. The properties of the functions involved are as defined in the aforementioned post. I repeat them in the following paragraph for convenience, though.



The functions are all real of a single real variable. Those in the sequence $(f_N)_N$ are continuous and bounded in $[0,1]$, and the sequence converges to a continuous function $f$ bounded in the same interval. In turn, those in $(F_N)_N$ are monotonically increasing step functions bounded in $[0,1]$ and the sequence is uniformly convergent to a function $F$ that is a cumulative distribution function.



Any hints on how to prove the above statement of convergence will be welcome.






integration sequence-of-function







share|cite|improve this question











$endgroup$




In connection with my post Convergence to Riemann-Stieltjes integral of sequence of Riemann-Stieltjes-like sums with changing integrand and integrator, an alternative approach to my main objective would be considering the convergence of the sequence of Riemann-Stieltjes (RS) integrals $int_0^1 , f_N(x) , mathrm{d}F_N(x)$ to the RS integral $int_0^1 , f(x) , mathrm{d}F(x)$. The properties of the functions involved are as defined in the aforementioned post. I repeat them in the following paragraph for convenience, though.



The functions are all real of a single real variable. Those in the sequence $(f_N)_N$ are continuous and bounded in $[0,1]$, and the sequence converges to a continuous function $f$ bounded in the same interval. In turn, those in $(F_N)_N$ are monotonically increasing step functions bounded in $[0,1]$ and the sequence is uniformly convergent to a function $F$ that is a cumulative distribution function.



Any hints on how to prove the above statement of convergence will be welcome.






integration sequence-of-function




integration sequence-of-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 7:55







Marcos

















asked Dec 3 '18 at 6:40









MarcosMarcos

1236




1236












  • $begingroup$
    I'm rusty. When you indicate $f_n longrightarrow f$, you intend convergence in the sup norm (i.e. uniform convergence)?
    $endgroup$
    – Eric Towers
    Dec 3 '18 at 7:03






  • 1




    $begingroup$
    Have you tried the "obvious thing": see if you can show $int_0^1 ; |f - f_n| , mathrm{d}(|F - F_n|) longrightarrow 0$?
    $endgroup$
    – Eric Towers
    Dec 3 '18 at 7:09










  • $begingroup$
    Not necessarily uniformly, but if it helps with a proof of convergence (or hints, with the details to be worked out by me), you are welcome to assume it is so.
    $endgroup$
    – Marcos
    Dec 3 '18 at 7:11










  • $begingroup$
    No, actually I haven't. This question dawned on me as a rather indirect but actually alternative approach to the linked post. I'll check that.
    $endgroup$
    – Marcos
    Dec 3 '18 at 7:13


















  • $begingroup$
    I'm rusty. When you indicate $f_n longrightarrow f$, you intend convergence in the sup norm (i.e. uniform convergence)?
    $endgroup$
    – Eric Towers
    Dec 3 '18 at 7:03






  • 1




    $begingroup$
    Have you tried the "obvious thing": see if you can show $int_0^1 ; |f - f_n| , mathrm{d}(|F - F_n|) longrightarrow 0$?
    $endgroup$
    – Eric Towers
    Dec 3 '18 at 7:09










  • $begingroup$
    Not necessarily uniformly, but if it helps with a proof of convergence (or hints, with the details to be worked out by me), you are welcome to assume it is so.
    $endgroup$
    – Marcos
    Dec 3 '18 at 7:11










  • $begingroup$
    No, actually I haven't. This question dawned on me as a rather indirect but actually alternative approach to the linked post. I'll check that.
    $endgroup$
    – Marcos
    Dec 3 '18 at 7:13
















$begingroup$
I'm rusty. When you indicate $f_n longrightarrow f$, you intend convergence in the sup norm (i.e. uniform convergence)?
$endgroup$
– Eric Towers
Dec 3 '18 at 7:03




$begingroup$
I'm rusty. When you indicate $f_n longrightarrow f$, you intend convergence in the sup norm (i.e. uniform convergence)?
$endgroup$
– Eric Towers
Dec 3 '18 at 7:03




1




1




$begingroup$
Have you tried the "obvious thing": see if you can show $int_0^1 ; |f - f_n| , mathrm{d}(|F - F_n|) longrightarrow 0$?
$endgroup$
– Eric Towers
Dec 3 '18 at 7:09




$begingroup$
Have you tried the "obvious thing": see if you can show $int_0^1 ; |f - f_n| , mathrm{d}(|F - F_n|) longrightarrow 0$?
$endgroup$
– Eric Towers
Dec 3 '18 at 7:09












$begingroup$
Not necessarily uniformly, but if it helps with a proof of convergence (or hints, with the details to be worked out by me), you are welcome to assume it is so.
$endgroup$
– Marcos
Dec 3 '18 at 7:11




$begingroup$
Not necessarily uniformly, but if it helps with a proof of convergence (or hints, with the details to be worked out by me), you are welcome to assume it is so.
$endgroup$
– Marcos
Dec 3 '18 at 7:11












$begingroup$
No, actually I haven't. This question dawned on me as a rather indirect but actually alternative approach to the linked post. I'll check that.
$endgroup$
– Marcos
Dec 3 '18 at 7:13




$begingroup$
No, actually I haven't. This question dawned on me as a rather indirect but actually alternative approach to the linked post. I'll check that.
$endgroup$
– Marcos
Dec 3 '18 at 7:13










1 Answer
1






active

oldest

votes


















2












$begingroup$

Hints:



Note that



$$left|int_0^1f_N , dF_N - int_0^1 f , dF right| leqslant left|int_0^1f_N , dF_N - int_0^1 f , dF_N right|+ left|int_0^1f , dF_N - int_0^1 f , dF right|$$



(1) We can estimate the first term on the RHS and prove convergence to $0$, if $F_N$ has bounded variation $V_0^1(F_N)$, using



$$left|int_0^1f_N , dF_N - int_0^1 f , dF_N right| leqslant int_0^1|f_N - f|, dV_0^x(F_N), $$
although with the simplification that $F_N$ is montonically increasing we have



$$left|int_0^1f_N , dF_N - int_0^1 f , dF_N right| leqslant int_0^1|f_N - f|, dF_N $$



With uniform convergence of $f_n to f$ and uniform boundedness of $F_N$ it is easy to progress. Given that $|F_N(x)| leqslant M$ uniformly in $N$ and $x$ -- which is true if $F_N$ converges uniformly to a bounded function $F$ -- then for all sufficiently large $N$ we have $|f_N(x) - f(x)| < epsilon/(2M)$ and



$$left|int_0^1f_N , dF_N - int_0^1 f , dF_N right| < frac{epsilon}{2M}[F_N(1) - F_N(0)] < frac{epsilon}{2M}2M = epsilon $$



(2) Estimate the second term on the RHS most easily using Rieman sums.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks so much. As with my linked post, I'll be back as soon as I have worked out the details.
    $endgroup$
    – Marcos
    Dec 3 '18 at 7:52










  • $begingroup$
    Actually, my difficulty is currently with the second term on the RHS. I was thinking of applying the bound $big|int_0^1 f d(F_n-F) big| leq sup_{[0,1]} |f| , V_0^1(F_n-F)$, but I haven't been able to obtainanything useful regarding the limiting total variation of $F_n-F$ on $[0,1]$, whence it came my question math.stackexchange.com/questions/3055573/…
    $endgroup$
    – Marcos
    Dec 29 '18 at 6:47













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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Hints:



Note that



$$left|int_0^1f_N , dF_N - int_0^1 f , dF right| leqslant left|int_0^1f_N , dF_N - int_0^1 f , dF_N right|+ left|int_0^1f , dF_N - int_0^1 f , dF right|$$



(1) We can estimate the first term on the RHS and prove convergence to $0$, if $F_N$ has bounded variation $V_0^1(F_N)$, using



$$left|int_0^1f_N , dF_N - int_0^1 f , dF_N right| leqslant int_0^1|f_N - f|, dV_0^x(F_N), $$
although with the simplification that $F_N$ is montonically increasing we have



$$left|int_0^1f_N , dF_N - int_0^1 f , dF_N right| leqslant int_0^1|f_N - f|, dF_N $$



With uniform convergence of $f_n to f$ and uniform boundedness of $F_N$ it is easy to progress. Given that $|F_N(x)| leqslant M$ uniformly in $N$ and $x$ -- which is true if $F_N$ converges uniformly to a bounded function $F$ -- then for all sufficiently large $N$ we have $|f_N(x) - f(x)| < epsilon/(2M)$ and



$$left|int_0^1f_N , dF_N - int_0^1 f , dF_N right| < frac{epsilon}{2M}[F_N(1) - F_N(0)] < frac{epsilon}{2M}2M = epsilon $$



(2) Estimate the second term on the RHS most easily using Rieman sums.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks so much. As with my linked post, I'll be back as soon as I have worked out the details.
    $endgroup$
    – Marcos
    Dec 3 '18 at 7:52










  • $begingroup$
    Actually, my difficulty is currently with the second term on the RHS. I was thinking of applying the bound $big|int_0^1 f d(F_n-F) big| leq sup_{[0,1]} |f| , V_0^1(F_n-F)$, but I haven't been able to obtainanything useful regarding the limiting total variation of $F_n-F$ on $[0,1]$, whence it came my question math.stackexchange.com/questions/3055573/…
    $endgroup$
    – Marcos
    Dec 29 '18 at 6:47


















2












$begingroup$

Hints:



Note that



$$left|int_0^1f_N , dF_N - int_0^1 f , dF right| leqslant left|int_0^1f_N , dF_N - int_0^1 f , dF_N right|+ left|int_0^1f , dF_N - int_0^1 f , dF right|$$



(1) We can estimate the first term on the RHS and prove convergence to $0$, if $F_N$ has bounded variation $V_0^1(F_N)$, using



$$left|int_0^1f_N , dF_N - int_0^1 f , dF_N right| leqslant int_0^1|f_N - f|, dV_0^x(F_N), $$
although with the simplification that $F_N$ is montonically increasing we have



$$left|int_0^1f_N , dF_N - int_0^1 f , dF_N right| leqslant int_0^1|f_N - f|, dF_N $$



With uniform convergence of $f_n to f$ and uniform boundedness of $F_N$ it is easy to progress. Given that $|F_N(x)| leqslant M$ uniformly in $N$ and $x$ -- which is true if $F_N$ converges uniformly to a bounded function $F$ -- then for all sufficiently large $N$ we have $|f_N(x) - f(x)| < epsilon/(2M)$ and



$$left|int_0^1f_N , dF_N - int_0^1 f , dF_N right| < frac{epsilon}{2M}[F_N(1) - F_N(0)] < frac{epsilon}{2M}2M = epsilon $$



(2) Estimate the second term on the RHS most easily using Rieman sums.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks so much. As with my linked post, I'll be back as soon as I have worked out the details.
    $endgroup$
    – Marcos
    Dec 3 '18 at 7:52










  • $begingroup$
    Actually, my difficulty is currently with the second term on the RHS. I was thinking of applying the bound $big|int_0^1 f d(F_n-F) big| leq sup_{[0,1]} |f| , V_0^1(F_n-F)$, but I haven't been able to obtainanything useful regarding the limiting total variation of $F_n-F$ on $[0,1]$, whence it came my question math.stackexchange.com/questions/3055573/…
    $endgroup$
    – Marcos
    Dec 29 '18 at 6:47
















2












2








2





$begingroup$

Hints:



Note that



$$left|int_0^1f_N , dF_N - int_0^1 f , dF right| leqslant left|int_0^1f_N , dF_N - int_0^1 f , dF_N right|+ left|int_0^1f , dF_N - int_0^1 f , dF right|$$



(1) We can estimate the first term on the RHS and prove convergence to $0$, if $F_N$ has bounded variation $V_0^1(F_N)$, using



$$left|int_0^1f_N , dF_N - int_0^1 f , dF_N right| leqslant int_0^1|f_N - f|, dV_0^x(F_N), $$
although with the simplification that $F_N$ is montonically increasing we have



$$left|int_0^1f_N , dF_N - int_0^1 f , dF_N right| leqslant int_0^1|f_N - f|, dF_N $$



With uniform convergence of $f_n to f$ and uniform boundedness of $F_N$ it is easy to progress. Given that $|F_N(x)| leqslant M$ uniformly in $N$ and $x$ -- which is true if $F_N$ converges uniformly to a bounded function $F$ -- then for all sufficiently large $N$ we have $|f_N(x) - f(x)| < epsilon/(2M)$ and



$$left|int_0^1f_N , dF_N - int_0^1 f , dF_N right| < frac{epsilon}{2M}[F_N(1) - F_N(0)] < frac{epsilon}{2M}2M = epsilon $$



(2) Estimate the second term on the RHS most easily using Rieman sums.






share|cite|improve this answer











$endgroup$



Hints:



Note that



$$left|int_0^1f_N , dF_N - int_0^1 f , dF right| leqslant left|int_0^1f_N , dF_N - int_0^1 f , dF_N right|+ left|int_0^1f , dF_N - int_0^1 f , dF right|$$



(1) We can estimate the first term on the RHS and prove convergence to $0$, if $F_N$ has bounded variation $V_0^1(F_N)$, using



$$left|int_0^1f_N , dF_N - int_0^1 f , dF_N right| leqslant int_0^1|f_N - f|, dV_0^x(F_N), $$
although with the simplification that $F_N$ is montonically increasing we have



$$left|int_0^1f_N , dF_N - int_0^1 f , dF_N right| leqslant int_0^1|f_N - f|, dF_N $$



With uniform convergence of $f_n to f$ and uniform boundedness of $F_N$ it is easy to progress. Given that $|F_N(x)| leqslant M$ uniformly in $N$ and $x$ -- which is true if $F_N$ converges uniformly to a bounded function $F$ -- then for all sufficiently large $N$ we have $|f_N(x) - f(x)| < epsilon/(2M)$ and



$$left|int_0^1f_N , dF_N - int_0^1 f , dF_N right| < frac{epsilon}{2M}[F_N(1) - F_N(0)] < frac{epsilon}{2M}2M = epsilon $$



(2) Estimate the second term on the RHS most easily using Rieman sums.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 8:05

























answered Dec 3 '18 at 7:47









RRLRRL

50.4k42573




50.4k42573












  • $begingroup$
    Thanks so much. As with my linked post, I'll be back as soon as I have worked out the details.
    $endgroup$
    – Marcos
    Dec 3 '18 at 7:52










  • $begingroup$
    Actually, my difficulty is currently with the second term on the RHS. I was thinking of applying the bound $big|int_0^1 f d(F_n-F) big| leq sup_{[0,1]} |f| , V_0^1(F_n-F)$, but I haven't been able to obtainanything useful regarding the limiting total variation of $F_n-F$ on $[0,1]$, whence it came my question math.stackexchange.com/questions/3055573/…
    $endgroup$
    – Marcos
    Dec 29 '18 at 6:47




















  • $begingroup$
    Thanks so much. As with my linked post, I'll be back as soon as I have worked out the details.
    $endgroup$
    – Marcos
    Dec 3 '18 at 7:52










  • $begingroup$
    Actually, my difficulty is currently with the second term on the RHS. I was thinking of applying the bound $big|int_0^1 f d(F_n-F) big| leq sup_{[0,1]} |f| , V_0^1(F_n-F)$, but I haven't been able to obtainanything useful regarding the limiting total variation of $F_n-F$ on $[0,1]$, whence it came my question math.stackexchange.com/questions/3055573/…
    $endgroup$
    – Marcos
    Dec 29 '18 at 6:47


















$begingroup$
Thanks so much. As with my linked post, I'll be back as soon as I have worked out the details.
$endgroup$
– Marcos
Dec 3 '18 at 7:52




$begingroup$
Thanks so much. As with my linked post, I'll be back as soon as I have worked out the details.
$endgroup$
– Marcos
Dec 3 '18 at 7:52












$begingroup$
Actually, my difficulty is currently with the second term on the RHS. I was thinking of applying the bound $big|int_0^1 f d(F_n-F) big| leq sup_{[0,1]} |f| , V_0^1(F_n-F)$, but I haven't been able to obtainanything useful regarding the limiting total variation of $F_n-F$ on $[0,1]$, whence it came my question math.stackexchange.com/questions/3055573/…
$endgroup$
– Marcos
Dec 29 '18 at 6:47






$begingroup$
Actually, my difficulty is currently with the second term on the RHS. I was thinking of applying the bound $big|int_0^1 f d(F_n-F) big| leq sup_{[0,1]} |f| , V_0^1(F_n-F)$, but I haven't been able to obtainanything useful regarding the limiting total variation of $F_n-F$ on $[0,1]$, whence it came my question math.stackexchange.com/questions/3055573/…
$endgroup$
– Marcos
Dec 29 '18 at 6:47




















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Plaza Victoria

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Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
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Puebla de Zaragoza

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Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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