Proof that bounded right continuous functions are integrable.
$begingroup$
I am reading Davie's book "One parameter Semigroups", on page 16 in the proof that "weak semigroups" are also "strong semigroups" it claims that for a right continuous locally bounded function $g:mathbb{R}rightarrowmathbb{R}$, the integral
$$ varepsilon^{-1} int_0^varepsilon g(t) dt $$ converges.
I am assuming that since it is for $varepsilon approx 0$ then the same would be true for a general interval of integration $[a,b]$ if we assume $g$ bounded instead of locally bounded. If it is not, then the main question is why, under the hypotesis in bold text, that integral converges.
I am also assuming it refers to Riemman integrability.
Note: The book works with a specific function but as far as I am concerned the only hypothesis neccesary are those written above. In case the claim is not true, counterexample needed, then I would edit the question to include the particular functions used.
integration definite-integrals semigroup-of-operators
edited Dec 24 '18 at 15:35
Pedro
10.7k23374
asked Dec 16 '18 at 15:40
I.C.I.C.
899
$endgroup$
add a comment |
$begingroup$
I am reading Davie's book "One parameter Semigroups", on page 16 in the proof that "weak semigroups" are also "strong semigroups" it claims that for a right continuous locally bounded function $g:mathbb{R}rightarrowmathbb{R}$, the integral
$$ varepsilon^{-1} int_0^varepsilon g(t) dt $$ converges.
I am assuming that since it is for $varepsilon approx 0$ then the same would be true for a general interval of integration $[a,b]$ if we assume $g$ bounded instead of locally bounded. If it is not, then the main question is why, under the hypotesis in bold text, that integral converges.
I am also assuming it refers to Riemman integrability.
Note: The book works with a specific function but as far as I am concerned the only hypothesis neccesary are those written above. In case the claim is not true, counterexample needed, then I would edit the question to include the particular functions used.
integration definite-integrals semigroup-of-operators
edited Dec 24 '18 at 15:35
Pedro
10.7k23374
asked Dec 16 '18 at 15:40
I.C.I.C.
899
$endgroup$
add a comment |
$begingroup$
I am reading Davie's book "One parameter Semigroups", on page 16 in the proof that "weak semigroups" are also "strong semigroups" it claims that for a right continuous locally bounded function $g:mathbb{R}rightarrowmathbb{R}$, the integral
$$ varepsilon^{-1} int_0^varepsilon g(t) dt $$ converges.
I am assuming that since it is for $varepsilon approx 0$ then the same would be true for a general interval of integration $[a,b]$ if we assume $g$ bounded instead of locally bounded. If it is not, then the main question is why, under the hypotesis in bold text, that integral converges.
I am also assuming it refers to Riemman integrability.
Note: The book works with a specific function but as far as I am concerned the only hypothesis neccesary are those written above. In case the claim is not true, counterexample needed, then I would edit the question to include the particular functions used.
integration definite-integrals semigroup-of-operators
edited Dec 24 '18 at 15:35
Pedro
10.7k23374
asked Dec 16 '18 at 15:40
I.C.I.C.
899
$endgroup$
I am reading Davie's book "One parameter Semigroups", on page 16 in the proof that "weak semigroups" are also "strong semigroups" it claims that for a right continuous locally bounded function $g:mathbb{R}rightarrowmathbb{R}$, the integral
$$ varepsilon^{-1} int_0^varepsilon g(t) dt $$ converges.
I am assuming that since it is for $varepsilon approx 0$ then the same would be true for a general interval of integration $[a,b]$ if we assume $g$ bounded instead of locally bounded. If it is not, then the main question is why, under the hypotesis in bold text, that integral converges.
I am also assuming it refers to Riemman integrability.
Note: The book works with a specific function but as far as I am concerned the only hypothesis neccesary are those written above. In case the claim is not true, counterexample needed, then I would edit the question to include the particular functions used.
integration definite-integrals semigroup-of-operators
integration definite-integrals semigroup-of-operators
edited Dec 24 '18 at 15:35
Pedro
10.7k23374
asked Dec 16 '18 at 15:40
I.C.I.C.
899
edited Dec 24 '18 at 15:35
Pedro
10.7k23374
asked Dec 16 '18 at 15:40
I.C.I.C.
899
edited Dec 24 '18 at 15:35
Pedro
10.7k23374
edited Dec 24 '18 at 15:35
Pedro
10.7k23374
edited Dec 24 '18 at 15:35
Pedro
10.7k23374
10.7k23374
asked Dec 16 '18 at 15:40
I.C.I.C.
899
asked Dec 16 '18 at 15:40
I.C.I.C.
899
asked Dec 16 '18 at 15:40
I.C.I.C.
899
899
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Since $g$ is right continuous at $0$, for every $varepsilon>0$ there exists $delta>0$ such that $|g(y)-g(0)| leq varepsilon$ when $0leq yleq delta$.
Then in particular, for $0leq hleq delta$, the triangle inequality implies
$$
left| h^{-1}int_0^h (g(t) - g(0)),dt right|
leq h^{-1} int_0^h |g(t)-g(0)|,dt
leq h^{-1}int_0^h varepsilon = varepsilon.
$$
This means that, under these hypotheses,
$$
lim_{hto 0^+} h^{-1} int_0^h g(t),dt = g(0).
$$
Does this answer your question?
EDIT: I think I misunderstood. You want to see that bounded right-continuous functions are Riemann-integrable. I don't know immediately how to prove this.
answered Dec 24 '18 at 15:45
felipehfelipeh
1,606619
$endgroup$
$begingroup$
What I was looking for is what you say in your edit. I am not sure if that statement of integrability is true in fact.
$endgroup$
– I.C.
Jan 7 at 22:00
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Since $g$ is right continuous at $0$, for every $varepsilon>0$ there exists $delta>0$ such that $|g(y)-g(0)| leq varepsilon$ when $0leq yleq delta$.
Then in particular, for $0leq hleq delta$, the triangle inequality implies
$$
left| h^{-1}int_0^h (g(t) - g(0)),dt right|
leq h^{-1} int_0^h |g(t)-g(0)|,dt
leq h^{-1}int_0^h varepsilon = varepsilon.
$$
This means that, under these hypotheses,
$$
lim_{hto 0^+} h^{-1} int_0^h g(t),dt = g(0).
$$
Does this answer your question?
EDIT: I think I misunderstood. You want to see that bounded right-continuous functions are Riemann-integrable. I don't know immediately how to prove this.
answered Dec 24 '18 at 15:45
felipehfelipeh
1,606619
$endgroup$
$begingroup$
What I was looking for is what you say in your edit. I am not sure if that statement of integrability is true in fact.
$endgroup$
– I.C.
Jan 7 at 22:00
add a comment |
$begingroup$
Since $g$ is right continuous at $0$, for every $varepsilon>0$ there exists $delta>0$ such that $|g(y)-g(0)| leq varepsilon$ when $0leq yleq delta$.
Then in particular, for $0leq hleq delta$, the triangle inequality implies
$$
left| h^{-1}int_0^h (g(t) - g(0)),dt right|
leq h^{-1} int_0^h |g(t)-g(0)|,dt
leq h^{-1}int_0^h varepsilon = varepsilon.
$$
This means that, under these hypotheses,
$$
lim_{hto 0^+} h^{-1} int_0^h g(t),dt = g(0).
$$
Does this answer your question?
EDIT: I think I misunderstood. You want to see that bounded right-continuous functions are Riemann-integrable. I don't know immediately how to prove this.
answered Dec 24 '18 at 15:45
felipehfelipeh
1,606619
$endgroup$
$begingroup$
What I was looking for is what you say in your edit. I am not sure if that statement of integrability is true in fact.
$endgroup$
– I.C.
Jan 7 at 22:00
add a comment |
$begingroup$
Since $g$ is right continuous at $0$, for every $varepsilon>0$ there exists $delta>0$ such that $|g(y)-g(0)| leq varepsilon$ when $0leq yleq delta$.
Then in particular, for $0leq hleq delta$, the triangle inequality implies
$$
left| h^{-1}int_0^h (g(t) - g(0)),dt right|
leq h^{-1} int_0^h |g(t)-g(0)|,dt
leq h^{-1}int_0^h varepsilon = varepsilon.
$$
This means that, under these hypotheses,
$$
lim_{hto 0^+} h^{-1} int_0^h g(t),dt = g(0).
$$
Does this answer your question?
EDIT: I think I misunderstood. You want to see that bounded right-continuous functions are Riemann-integrable. I don't know immediately how to prove this.
answered Dec 24 '18 at 15:45
felipehfelipeh
1,606619
$endgroup$
Since $g$ is right continuous at $0$, for every $varepsilon>0$ there exists $delta>0$ such that $|g(y)-g(0)| leq varepsilon$ when $0leq yleq delta$.
Then in particular, for $0leq hleq delta$, the triangle inequality implies
$$
left| h^{-1}int_0^h (g(t) - g(0)),dt right|
leq h^{-1} int_0^h |g(t)-g(0)|,dt
leq h^{-1}int_0^h varepsilon = varepsilon.
$$
This means that, under these hypotheses,
$$
lim_{hto 0^+} h^{-1} int_0^h g(t),dt = g(0).
$$
Does this answer your question?
EDIT: I think I misunderstood. You want to see that bounded right-continuous functions are Riemann-integrable. I don't know immediately how to prove this.
answered Dec 24 '18 at 15:45
felipehfelipeh
1,606619
edited Dec 24 '18 at 15:55
answered Dec 24 '18 at 15:45
felipehfelipeh
1,606619
answered Dec 24 '18 at 15:45
felipehfelipeh
1,606619
answered Dec 24 '18 at 15:45
felipehfelipeh
1,606619
1,606619
$begingroup$
What I was looking for is what you say in your edit. I am not sure if that statement of integrability is true in fact.
$endgroup$
– I.C.
Jan 7 at 22:00
add a comment |
$begingroup$
What I was looking for is what you say in your edit. I am not sure if that statement of integrability is true in fact.
$endgroup$
– I.C.
Jan 7 at 22:00
$begingroup$
What I was looking for is what you say in your edit. I am not sure if that statement of integrability is true in fact.
$endgroup$
– I.C.
Jan 7 at 22:00
$begingroup$
What I was looking for is what you say in your edit. I am not sure if that statement of integrability is true in fact.
$endgroup$
– I.C.
Jan 7 at 22:00
add a comment |
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