User:IssaRice/Taking inf and sup separately - Revision history

IssaRice at 22:24, 20 October 2023

2023-10-20T22:24:59Z

← Older revision		Revision as of 22:24, 20 October 2023
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	Now consider the sets <math>A := \{a^+_N : N \geq m\}</math> and <math>B := \{a^-_N : N \geq m\}</math>. If we can show that <math>a^+_j \geq a^-_k</math> for arbitrary <math>j,k\geq m</math>, then we can apply the trick to these sets to conclude that <math>L^+ = \inf(a^+_N)_{N=m} = \inf(A) \geq \sup(B) = \sup(a^-_N)_{N=m} = L^-</math>.		Now consider the sets <math>A := \{a^+_N : N \geq m\}</math> and <math>B := \{a^-_N : N \geq m\}</math>. If we can show that <math>a^+_j \geq a^-_k</math> for arbitrary <math>j,k\geq m</math>, then we can apply the trick to these sets to conclude that <math>L^+ = \inf(a^+_N)_{N=m} = \inf(A) \geq \sup(B) = \sup(a^-_N)_{N=m} = L^-</math>.

	==Comparison principle==		===Comparison principle===

	This technique can also be used to show that if <math>a_n \leq b_n</math> for all <math>n</math>, then <math>\sup(a_n)_{n=0}^\infty \leq \sup(b_n)_{n=0}^\infty</math>, <math>\inf(a_n)_{n=0}^\infty \leq \inf(b_n)_{n=0}^\infty</math>, <math>\limsup_{n\to\infty}(a_n)_{n=0}^\infty \leq \limsup_{n\to\infty}(b_n)_{n=0}^\infty</math>, and <math>\liminf_{n\to\infty}(a_n)_{n=0}^\infty \leq \liminf_{n\to\infty}(b_n)_{n=0}^\infty</math>.		This technique, in modified form where we take two sups separately or two infs separately, can also be used to show that if <math>a_n \leq b_n</math> for all <math>n</math>, then <math>\sup(a_n)_{n=0}^\infty \leq \sup(b_n)_{n=0}^\infty</math>, <math>\inf(a_n)_{n=0}^\infty \leq \inf(b_n)_{n=0}^\infty</math>, <math>\limsup_{n\to\infty}(a_n)_{n=0}^\infty \leq \limsup_{n\to\infty}(b_n)_{n=0}^\infty</math>, and <math>\liminf_{n\to\infty}(a_n)_{n=0}^\infty \leq \liminf_{n\to\infty}(b_n)_{n=0}^\infty</math>.

	===Lower and upper Riemann integral===		===Lower and upper Riemann integral===

IssaRice at 22:23, 20 October 2023

2023-10-20T22:23:45Z

← Older revision		Revision as of 22:23, 20 October 2023
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	Now consider the sets <math>A := \{a^+_N : N \geq m\}</math> and <math>B := \{a^-_N : N \geq m\}</math>. If we can show that <math>a^+_j \geq a^-_k</math> for arbitrary <math>j,k\geq m</math>, then we can apply the trick to these sets to conclude that <math>L^+ = \inf(a^+_N)_{N=m} = \inf(A) \geq \sup(B) = \sup(a^-_N)_{N=m} = L^-</math>.		Now consider the sets <math>A := \{a^+_N : N \geq m\}</math> and <math>B := \{a^-_N : N \geq m\}</math>. If we can show that <math>a^+_j \geq a^-_k</math> for arbitrary <math>j,k\geq m</math>, then we can apply the trick to these sets to conclude that <math>L^+ = \inf(a^+_N)_{N=m} = \inf(A) \geq \sup(B) = \sup(a^-_N)_{N=m} = L^-</math>.

			==Comparison principle==

			This technique can also be used to show that if <math>a_n \leq b_n</math> for all <math>n</math>, then <math>\sup(a_n)_{n=0}^\infty \leq \sup(b_n)_{n=0}^\infty</math>, <math>\inf(a_n)_{n=0}^\infty \leq \inf(b_n)_{n=0}^\infty</math>, <math>\limsup_{n\to\infty}(a_n)_{n=0}^\infty \leq \limsup_{n\to\infty}(b_n)_{n=0}^\infty</math>, and <math>\liminf_{n\to\infty}(a_n)_{n=0}^\infty \leq \liminf_{n\to\infty}(b_n)_{n=0}^\infty</math>.

	===Lower and upper Riemann integral===		===Lower and upper Riemann integral===

IssaRice: /* alternating series test */

2020-01-22T10:01:08Z

alternating series test

← Older revision		Revision as of 10:01, 22 January 2020
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	(this one is more of a failed application)		(this one is more of a failed application)

	each even partial ~~sums~~ is at least as large as each odd partial sum, so the inf over the even partial sums is at least as large as the sup over the odd partial sums. this actually isn't strong enough to prove what we want. we actually need the stronger condition that the even partial sums are a decreasing sequence, and that the odd partial sums are an increasing sequence, and that eventually their difference becomes arbitrarily small.		each even partial sum is at least as large as each odd partial sum, so the inf over the even partial sums is at least as large as the sup over the odd partial sums. this actually isn't strong enough to prove what we want. we actually need the stronger condition that the even partial sums are a decreasing sequence, and that the odd partial sums are an increasing sequence, and that eventually their difference becomes arbitrarily small.

	==References==		==References==

	After I wrote this page, I found the same theorem in Apostol's ''Calculus'' (volume 1, 2nd edition, p. 28) in the section "Fundamental properties of the supremum and infimum".		After I wrote this page, I found the same theorem in Apostol's ''Calculus'' (volume 1, 2nd edition, p. 28) in the section "Fundamental properties of the supremum and infimum".

IssaRice: /* Applications */

2020-01-22T10:00:29Z

Applications

← Older revision		Revision as of 10:00, 22 January 2020
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	Then we have <math>\overline{\int}_I f = \inf(A)</math> and <math>\underline{\int}_I f = \sup(B)</math>. To apply the trick all we need to do is to let <math>g</math> be a p.c. function on <math>I</math> that majorizes <math>f</math>, and let <math>h</math> be a p.c. function on <math>I</math> that minorizes <math>f</math>, and show that <math>p.c.\int_I g\geq p.c.\int_I h</math>.		Then we have <math>\overline{\int}_I f = \inf(A)</math> and <math>\underline{\int}_I f = \sup(B)</math>. To apply the trick all we need to do is to let <math>g</math> be a p.c. function on <math>I</math> that majorizes <math>f</math>, and let <math>h</math> be a p.c. function on <math>I</math> that minorizes <math>f</math>, and show that <math>p.c.\int_I g\geq p.c.\int_I h</math>.

			==alternating series test==

			(this one is more of a failed application)

			each even partial sums is at least as large as each odd partial sum, so the inf over the even partial sums is at least as large as the sup over the odd partial sums. this actually isn't strong enough to prove what we want. we actually need the stronger condition that the even partial sums are a decreasing sequence, and that the odd partial sums are an increasing sequence, and that eventually their difference becomes arbitrarily small.

	==References==		==References==

	After I wrote this page, I found the same theorem in Apostol's ''Calculus'' (volume 1, 2nd edition, p. 28) in the section "Fundamental properties of the supremum and infimum".		After I wrote this page, I found the same theorem in Apostol's ''Calculus'' (volume 1, 2nd edition, p. 28) in the section "Fundamental properties of the supremum and infimum".

IssaRice at 07:07, 21 June 2019

2019-06-21T07:07:49Z

← Older revision		Revision as of 07:07, 21 June 2019
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	Then we have <math>\overline{\int}_I f = \inf(A)</math> and <math>\underline{\int}_I f = \sup(B)</math>. To apply the trick all we need to do is to let <math>g</math> be a p.c. function on <math>I</math> that majorizes <math>f</math>, and let <math>h</math> be a p.c. function on <math>I</math> that minorizes <math>f</math>, and show that <math>p.c.\int_I g\geq p.c.\int_I h</math>.		Then we have <math>\overline{\int}_I f = \inf(A)</math> and <math>\underline{\int}_I f = \sup(B)</math>. To apply the trick all we need to do is to let <math>g</math> be a p.c. function on <math>I</math> that majorizes <math>f</math>, and let <math>h</math> be a p.c. function on <math>I</math> that minorizes <math>f</math>, and show that <math>p.c.\int_I g\geq p.c.\int_I h</math>.

			==References==

			After I wrote this page, I found the same theorem in Apostol's ''Calculus'' (volume 1, 2nd edition, p. 28) in the section "Fundamental properties of the supremum and infimum".

IssaRice: /* Lower and upper Riemann integral */

2018-12-17T23:04:30Z

Lower and upper Riemann integral

← Older revision		Revision as of 23:04, 17 December 2018
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	(Notation from Tao's ''Analysis I''.)		(Notation from Tao's ''Analysis I''.)

			Let <math>I</math> be a bounded interval on the real line, and let <math>f : I \to \mathbf R</math>.

	We have		We have

IssaRice: /* Lower and upper Riemann integral */

2018-12-17T23:02:25Z

Lower and upper Riemann integral

← Older revision		Revision as of 23:02, 17 December 2018
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	<math>\underline{\int}_I f := \sup\left\{p.c.\int_I g : g\text{ is a p.c. function on }I\text{ that minorizes }f\right\}</math>		<math>\underline{\int}_I f := \sup\left\{p.c.\int_I g : g\text{ is a p.c. function on }I\text{ that minorizes }f\right\}</math>

			We want to show <math>\underline{\int}_I f \leq \overline{\int}_I f</math>.

			Define

			<math>A := \left\{p.c.\int_I g : g\text{ is a p.c. function on }I\text{ that majorizes }f\right\}</math>

			<math>B := \left\{p.c.\int_I g : g\text{ is a p.c. function on }I\text{ that minorizes }f\right\}</math>

			Then we have <math>\overline{\int}_I f = \inf(A)</math> and <math>\underline{\int}_I f = \sup(B)</math>. To apply the trick all we need to do is to let <math>g</math> be a p.c. function on <math>I</math> that majorizes <math>f</math>, and let <math>h</math> be a p.c. function on <math>I</math> that minorizes <math>f</math>, and show that <math>p.c.\int_I g\geq p.c.\int_I h</math>.

IssaRice: /* Lower and upper Riemann integral */

2018-12-17T22:58:46Z

Lower and upper Riemann integral

← Older revision		Revision as of 22:58, 17 December 2018
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	===Lower and upper Riemann integral===		===Lower and upper Riemann integral===

			(Notation from Tao's ''Analysis I''.)

			We have

			<math>\overline{\int}_I f := \inf\left\{p.c.\int_I g : g\text{ is a p.c. function on }I\text{ that majorizes }f\right\}</math>

			<math>\underline{\int}_I f := \sup\left\{p.c.\int_I g : g\text{ is a p.c. function on }I\text{ that minorizes }f\right\}</math>

IssaRice: /* liminf vs limsup */

2018-12-17T22:57:10Z

liminf vs limsup

← Older revision		Revision as of 22:57, 17 December 2018
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	===liminf vs limsup===		===liminf vs limsup===

			(Notation from Tao's ''Analysis I''.)

	Let <math>(a_n)_{n=m}^\infty</math> be a sequence of real numbers. Let <math>L^- := \liminf_{n\to\infty} a_n</math> and let <math>L^+ := \limsup_{n\to\infty} a_n</math>. Then we have <math>L^- \leq L^+</math>.		Let <math>(a_n)_{n=m}^\infty</math> be a sequence of real numbers. Let <math>L^- := \liminf_{n\to\infty} a_n</math> and let <math>L^+ := \limsup_{n\to\infty} a_n</math>. Then we have <math>L^- \leq L^+</math>.

IssaRice: /* liminf vs limsup */

2018-12-17T22:54:38Z

liminf vs limsup

← Older revision		Revision as of 22:54, 17 December 2018
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	Consider the sequences <math>(a^-_N)_{N=m}^\infty</math> and <math>(a^+_N)_{N=m}^\infty</math> defined by <math>a^-_N := \inf(a_n)_{n=N}^\infty</math> and <math>a^+_N := \sup(a_n)_{n=N}^\infty</math>.		Consider the sequences <math>(a^-_N)_{N=m}^\infty</math> and <math>(a^+_N)_{N=m}^\infty</math> defined by <math>a^-_N := \inf(a_n)_{n=N}^\infty</math> and <math>a^+_N := \sup(a_n)_{n=N}^\infty</math>.

	Now consider the sets <math>A := \{a^+_N : N \geq m\}</math> and <math>B := \{a^-_N : N \geq m\}</math>. If we can show that <math>a^+_j \geq a^-_k</math> for arbitrary <math>j,k\geq m</math>, then we can apply the trick to these sets to conclude that <math>L^+ = \inf(a^+_N)_{N=m}\inf(A) \geq \sup(B) = \sup(a^-_N)_{N=m} = L^-</math>.		Now consider the sets <math>A := \{a^+_N : N \geq m\}</math> and <math>B := \{a^-_N : N \geq m\}</math>. If we can show that <math>a^+_j \geq a^-_k</math> for arbitrary <math>j,k\geq m</math>, then we can apply the trick to these sets to conclude that <math>L^+ = \inf(a^+_N)_{N=m} = \inf(A) \geq \sup(B) = \sup(a^-_N)_{N=m} = L^-</math>.

	===Lower and upper Riemann integral===		===Lower and upper Riemann integral===