Convergence of sequence of Riemann-Stieltjes integrals to Riemann-Stieltjes integral
$begingroup$
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
asked Dec 3 '18 at 6:40
MarcosMarcos
1236
$endgroup$
add a comment |
$begingroup$
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
asked Dec 3 '18 at 6:40
MarcosMarcos
1236
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Dec 3 '18 at 6:40
MarcosMarcos
1236
$endgroup$
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
asked Dec 3 '18 at 6:40
MarcosMarcos
1236
asked Dec 3 '18 at 6:40
MarcosMarcos
1236
edited Dec 3 '18 at 7:55
Marcos
asked Dec 3 '18 at 6:40
MarcosMarcos
1236
asked Dec 3 '18 at 6:40
MarcosMarcos
1236
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 3 '18 at 7:47
RRLRRL
50.4k42573
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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$begingroup$
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.
answered Dec 3 '18 at 7:47
RRLRRL
50.4k42573
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 3 '18 at 7:47
RRLRRL
50.4k42573
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 3 '18 at 7:47
RRLRRL
50.4k42573
$endgroup$
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.
answered Dec 3 '18 at 7:47
RRLRRL
50.4k42573
edited Dec 3 '18 at 8:05
answered Dec 3 '18 at 7:47
RRLRRL
50.4k42573
answered Dec 3 '18 at 7:47
RRLRRL
50.4k42573
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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
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Have you tried the "obvious thing": see if you can show $int_0^1 ; |f - f_n| , mathrm{d}(|F - F_n|) longrightarrow 0$?
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– Eric Towers
Dec 3 '18 at 7:09
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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.
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– Marcos
Dec 3 '18 at 7:11
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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.
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– Marcos
Dec 3 '18 at 7:13