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












1














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
























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






  • 1




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










  • 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.
    – Marcos
    Dec 3 '18 at 7:11










  • 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.
    – Marcos
    Dec 3 '18 at 7:13
















1














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
























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






  • 1




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










  • 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.
    – Marcos
    Dec 3 '18 at 7:11










  • 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.
    – Marcos
    Dec 3 '18 at 7:13














1












1








1







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















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

























asked Dec 3 '18 at 6:40









Marcos

1236




1236












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






  • 1




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










  • 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.
    – Marcos
    Dec 3 '18 at 7:11










  • 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.
    – Marcos
    Dec 3 '18 at 7:13


















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






  • 1




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










  • 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.
    – Marcos
    Dec 3 '18 at 7:11










  • 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.
    – Marcos
    Dec 3 '18 at 7:13
















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




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




1




1




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




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












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.
– Marcos
Dec 3 '18 at 7:11




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.
– Marcos
Dec 3 '18 at 7:11












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.
– Marcos
Dec 3 '18 at 7:13




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.
– Marcos
Dec 3 '18 at 7:13










1 Answer
1






active

oldest

votes


















2














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























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










  • 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/…
    – Marcos
    Dec 29 '18 at 6:47













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














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























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










  • 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/…
    – Marcos
    Dec 29 '18 at 6:47


















2














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























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










  • 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/…
    – Marcos
    Dec 29 '18 at 6:47
















2












2








2






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














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









RRL

49k42573




49k42573












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










  • 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/…
    – Marcos
    Dec 29 '18 at 6:47




















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










  • 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/…
    – Marcos
    Dec 29 '18 at 6:47


















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




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












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/…
– Marcos
Dec 29 '18 at 6:47






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/…
– Marcos
Dec 29 '18 at 6:47




















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Type 'String' is not a subtype of type 'int' of 'index'

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up vote 0 down vote favorite My App can authenticate successfully when debugging against the Android emulator, but if I try to authenticate using debugging against the physical device (with same OS version), an error message appears after more than one minute waiting: Exception has occurred. _TypeError (type 'String' is not a subtype of type 'int' of 'index') The error message points to the following code: if (responseJson[AuthUtils.authTokenKey] != null) { AuthUtils.insertDetails( userNameController.text, passwordController.text, responseJson); ... } else { ... }; And in the DEBUG CONSOLE I get the following: I/flutter ( 9531): Auth: SocketException: OS Error: Connection timed out, errno = 110, address = example.com, port = 3897
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