Absolute Value inside an integral












0














So I have that $|f(x) - h(x)| le |f(x) - g(x)| + |g(x) - h(x)|$.



What I'm wondering is if this is the same as saying



$$int_a^b |f(x) - h(x)|{rm d}x le int_a^b |f(x) - g(x)|{rm d}x + int_a^b |g(x) - h(x)|{rm d}x.$$



Is this valid?






integration absolute-value







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  • It is valid as long as $g(x)$ is integralble on $[a,b]$.
    – LeB
    Nov 27 '18 at 2:01










  • It's not the same thing, but the first thing implies the second thing.
    – zhw.
    Nov 27 '18 at 2:45
















0














So I have that $|f(x) - h(x)| le |f(x) - g(x)| + |g(x) - h(x)|$.



What I'm wondering is if this is the same as saying



$$int_a^b |f(x) - h(x)|{rm d}x le int_a^b |f(x) - g(x)|{rm d}x + int_a^b |g(x) - h(x)|{rm d}x.$$



Is this valid?






integration absolute-value







share|cite|improve this question
























  • It is valid as long as $g(x)$ is integralble on $[a,b]$.
    – LeB
    Nov 27 '18 at 2:01










  • It's not the same thing, but the first thing implies the second thing.
    – zhw.
    Nov 27 '18 at 2:45














0












0








0







So I have that $|f(x) - h(x)| le |f(x) - g(x)| + |g(x) - h(x)|$.



What I'm wondering is if this is the same as saying



$$int_a^b |f(x) - h(x)|{rm d}x le int_a^b |f(x) - g(x)|{rm d}x + int_a^b |g(x) - h(x)|{rm d}x.$$



Is this valid?






integration absolute-value







share|cite|improve this question















So I have that $|f(x) - h(x)| le |f(x) - g(x)| + |g(x) - h(x)|$.



What I'm wondering is if this is the same as saying



$$int_a^b |f(x) - h(x)|{rm d}x le int_a^b |f(x) - g(x)|{rm d}x + int_a^b |g(x) - h(x)|{rm d}x.$$



Is this valid?






integration absolute-value




integration absolute-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 2:08









caverac

13.9k21130




13.9k21130










asked Nov 27 '18 at 1:58









kendal

337




337












  • It is valid as long as $g(x)$ is integralble on $[a,b]$.
    – LeB
    Nov 27 '18 at 2:01










  • It's not the same thing, but the first thing implies the second thing.
    – zhw.
    Nov 27 '18 at 2:45


















  • It is valid as long as $g(x)$ is integralble on $[a,b]$.
    – LeB
    Nov 27 '18 at 2:01










  • It's not the same thing, but the first thing implies the second thing.
    – zhw.
    Nov 27 '18 at 2:45
















It is valid as long as $g(x)$ is integralble on $[a,b]$.
– LeB
Nov 27 '18 at 2:01




It is valid as long as $g(x)$ is integralble on $[a,b]$.
– LeB
Nov 27 '18 at 2:01












It's not the same thing, but the first thing implies the second thing.
– zhw.
Nov 27 '18 at 2:45




It's not the same thing, but the first thing implies the second thing.
– zhw.
Nov 27 '18 at 2:45










1 Answer
1






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0














Yes, it is triangle inequality evaluated in the integral from a to b.



It is posible thanks to the monotonic property of the integrals.



The only condition is that f,g,h must be integrable on [a,b], indeed continouos in (a,b) so the rest of them would be.



https://es.wikipedia.org/wiki/Desigualdad_triangular#Generalizaci%C3%B3n_de_la_desigualdad_triangular






share|cite|improve this answer





















  • Monotonicity gives you that $f(x) le g(x) Rightarrow int_a^b f(x) dx le int_a^b g(x) dx$, but how do you argue that $int_a^b f(x) dx le int_a^b g(x) dx Rightarrow f(x) le g(x) $?
    – AlkaKadri
    Nov 27 '18 at 2:40










  • I thought he only wanted to solve the first implication.
    – Alberto Torrejon Valenzuela
    Nov 27 '18 at 2:43










  • I see. "Same thing as saying" usually means "if and only if", but if it's just the one direction that OP is looking for then I guess indeed we are done :)
    – AlkaKadri
    Nov 27 '18 at 2:44










  • @AlkaKadri If the second inequality holds for all choices of $a,b$ with $a<b$ then (assuming existence of the integrals) the first equality holds almost everywhere. So we can convert this into and iff statement.
    – Kavi Rama Murthy
    Nov 27 '18 at 8:30










  • @KaviRamaMurthy Definitely. And if $f,g,h$ are continuous then it holds everywhere, but the individual that answered hasn't stated that so I was just trying to prompt him/OP to state why they think the converse is true.
    – AlkaKadri
    Nov 28 '18 at 0:19











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1 Answer
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1 Answer
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active

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0














Yes, it is triangle inequality evaluated in the integral from a to b.



It is posible thanks to the monotonic property of the integrals.



The only condition is that f,g,h must be integrable on [a,b], indeed continouos in (a,b) so the rest of them would be.



https://es.wikipedia.org/wiki/Desigualdad_triangular#Generalizaci%C3%B3n_de_la_desigualdad_triangular






share|cite|improve this answer





















  • Monotonicity gives you that $f(x) le g(x) Rightarrow int_a^b f(x) dx le int_a^b g(x) dx$, but how do you argue that $int_a^b f(x) dx le int_a^b g(x) dx Rightarrow f(x) le g(x) $?
    – AlkaKadri
    Nov 27 '18 at 2:40










  • I thought he only wanted to solve the first implication.
    – Alberto Torrejon Valenzuela
    Nov 27 '18 at 2:43










  • I see. "Same thing as saying" usually means "if and only if", but if it's just the one direction that OP is looking for then I guess indeed we are done :)
    – AlkaKadri
    Nov 27 '18 at 2:44










  • @AlkaKadri If the second inequality holds for all choices of $a,b$ with $a<b$ then (assuming existence of the integrals) the first equality holds almost everywhere. So we can convert this into and iff statement.
    – Kavi Rama Murthy
    Nov 27 '18 at 8:30










  • @KaviRamaMurthy Definitely. And if $f,g,h$ are continuous then it holds everywhere, but the individual that answered hasn't stated that so I was just trying to prompt him/OP to state why they think the converse is true.
    – AlkaKadri
    Nov 28 '18 at 0:19
















0














Yes, it is triangle inequality evaluated in the integral from a to b.



It is posible thanks to the monotonic property of the integrals.



The only condition is that f,g,h must be integrable on [a,b], indeed continouos in (a,b) so the rest of them would be.



https://es.wikipedia.org/wiki/Desigualdad_triangular#Generalizaci%C3%B3n_de_la_desigualdad_triangular






share|cite|improve this answer





















  • Monotonicity gives you that $f(x) le g(x) Rightarrow int_a^b f(x) dx le int_a^b g(x) dx$, but how do you argue that $int_a^b f(x) dx le int_a^b g(x) dx Rightarrow f(x) le g(x) $?
    – AlkaKadri
    Nov 27 '18 at 2:40










  • I thought he only wanted to solve the first implication.
    – Alberto Torrejon Valenzuela
    Nov 27 '18 at 2:43










  • I see. "Same thing as saying" usually means "if and only if", but if it's just the one direction that OP is looking for then I guess indeed we are done :)
    – AlkaKadri
    Nov 27 '18 at 2:44










  • @AlkaKadri If the second inequality holds for all choices of $a,b$ with $a<b$ then (assuming existence of the integrals) the first equality holds almost everywhere. So we can convert this into and iff statement.
    – Kavi Rama Murthy
    Nov 27 '18 at 8:30










  • @KaviRamaMurthy Definitely. And if $f,g,h$ are continuous then it holds everywhere, but the individual that answered hasn't stated that so I was just trying to prompt him/OP to state why they think the converse is true.
    – AlkaKadri
    Nov 28 '18 at 0:19














0












0








0






Yes, it is triangle inequality evaluated in the integral from a to b.



It is posible thanks to the monotonic property of the integrals.



The only condition is that f,g,h must be integrable on [a,b], indeed continouos in (a,b) so the rest of them would be.



https://es.wikipedia.org/wiki/Desigualdad_triangular#Generalizaci%C3%B3n_de_la_desigualdad_triangular






share|cite|improve this answer












Yes, it is triangle inequality evaluated in the integral from a to b.



It is posible thanks to the monotonic property of the integrals.



The only condition is that f,g,h must be integrable on [a,b], indeed continouos in (a,b) so the rest of them would be.



https://es.wikipedia.org/wiki/Desigualdad_triangular#Generalizaci%C3%B3n_de_la_desigualdad_triangular







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 27 '18 at 2:14









Alberto Torrejon Valenzuela

276




276












  • Monotonicity gives you that $f(x) le g(x) Rightarrow int_a^b f(x) dx le int_a^b g(x) dx$, but how do you argue that $int_a^b f(x) dx le int_a^b g(x) dx Rightarrow f(x) le g(x) $?
    – AlkaKadri
    Nov 27 '18 at 2:40










  • I thought he only wanted to solve the first implication.
    – Alberto Torrejon Valenzuela
    Nov 27 '18 at 2:43










  • I see. "Same thing as saying" usually means "if and only if", but if it's just the one direction that OP is looking for then I guess indeed we are done :)
    – AlkaKadri
    Nov 27 '18 at 2:44










  • @AlkaKadri If the second inequality holds for all choices of $a,b$ with $a<b$ then (assuming existence of the integrals) the first equality holds almost everywhere. So we can convert this into and iff statement.
    – Kavi Rama Murthy
    Nov 27 '18 at 8:30










  • @KaviRamaMurthy Definitely. And if $f,g,h$ are continuous then it holds everywhere, but the individual that answered hasn't stated that so I was just trying to prompt him/OP to state why they think the converse is true.
    – AlkaKadri
    Nov 28 '18 at 0:19


















  • Monotonicity gives you that $f(x) le g(x) Rightarrow int_a^b f(x) dx le int_a^b g(x) dx$, but how do you argue that $int_a^b f(x) dx le int_a^b g(x) dx Rightarrow f(x) le g(x) $?
    – AlkaKadri
    Nov 27 '18 at 2:40










  • I thought he only wanted to solve the first implication.
    – Alberto Torrejon Valenzuela
    Nov 27 '18 at 2:43










  • I see. "Same thing as saying" usually means "if and only if", but if it's just the one direction that OP is looking for then I guess indeed we are done :)
    – AlkaKadri
    Nov 27 '18 at 2:44










  • @AlkaKadri If the second inequality holds for all choices of $a,b$ with $a<b$ then (assuming existence of the integrals) the first equality holds almost everywhere. So we can convert this into and iff statement.
    – Kavi Rama Murthy
    Nov 27 '18 at 8:30










  • @KaviRamaMurthy Definitely. And if $f,g,h$ are continuous then it holds everywhere, but the individual that answered hasn't stated that so I was just trying to prompt him/OP to state why they think the converse is true.
    – AlkaKadri
    Nov 28 '18 at 0:19
















Monotonicity gives you that $f(x) le g(x) Rightarrow int_a^b f(x) dx le int_a^b g(x) dx$, but how do you argue that $int_a^b f(x) dx le int_a^b g(x) dx Rightarrow f(x) le g(x) $?
– AlkaKadri
Nov 27 '18 at 2:40




Monotonicity gives you that $f(x) le g(x) Rightarrow int_a^b f(x) dx le int_a^b g(x) dx$, but how do you argue that $int_a^b f(x) dx le int_a^b g(x) dx Rightarrow f(x) le g(x) $?
– AlkaKadri
Nov 27 '18 at 2:40












I thought he only wanted to solve the first implication.
– Alberto Torrejon Valenzuela
Nov 27 '18 at 2:43




I thought he only wanted to solve the first implication.
– Alberto Torrejon Valenzuela
Nov 27 '18 at 2:43












I see. "Same thing as saying" usually means "if and only if", but if it's just the one direction that OP is looking for then I guess indeed we are done :)
– AlkaKadri
Nov 27 '18 at 2:44




I see. "Same thing as saying" usually means "if and only if", but if it's just the one direction that OP is looking for then I guess indeed we are done :)
– AlkaKadri
Nov 27 '18 at 2:44












@AlkaKadri If the second inequality holds for all choices of $a,b$ with $a<b$ then (assuming existence of the integrals) the first equality holds almost everywhere. So we can convert this into and iff statement.
– Kavi Rama Murthy
Nov 27 '18 at 8:30




@AlkaKadri If the second inequality holds for all choices of $a,b$ with $a<b$ then (assuming existence of the integrals) the first equality holds almost everywhere. So we can convert this into and iff statement.
– Kavi Rama Murthy
Nov 27 '18 at 8:30












@KaviRamaMurthy Definitely. And if $f,g,h$ are continuous then it holds everywhere, but the individual that answered hasn't stated that so I was just trying to prompt him/OP to state why they think the converse is true.
– AlkaKadri
Nov 28 '18 at 0:19




@KaviRamaMurthy Definitely. And if $f,g,h$ are continuous then it holds everywhere, but the individual that answered hasn't stated that so I was just trying to prompt him/OP to state why they think the converse is true.
– AlkaKadri
Nov 28 '18 at 0:19


















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