Metric Given by Riemann integral
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Let $L_p([a,b];mathbbR) = f$ is Riemann-integrable $$. I want to prove that
$$
d_p(f,g) = left(int_a^b |f(x)-g(x)|^p dxright)^1/p
$$
defines a metric in $L_p([a,b],mathbbR).$
Given $f,g$ integrable I easily see that $d_p(f,g)$ exists and that $d_p(f,f)=0$ and $d_p(f,g) = d_p(g,f)$. I also proved the triangle inequality by proving the Hölder inequality for positive real functions. I'm just stuck in the condition $fneq g to d_p(f,g)>0$. For simplicity I'm assuming $f neq 0$ and trying to prove $int |f|^p neq 0$.
I know that $fge 0$ implies $int f ge 0$ but I don't see how to get the strict inequality. Any help or hint will be appreciated.
real-analysis metric-spaces
edited Sep 5 at 9:30
José Carlos Santos
122k16101186
asked Sep 5 at 9:18
AnalyticHarmony
595213
add a comment |Â
up vote
0
down vote
favorite
Let $L_p([a,b];mathbbR) = f$ is Riemann-integrable $$. I want to prove that
$$
d_p(f,g) = left(int_a^b |f(x)-g(x)|^p dxright)^1/p
$$
defines a metric in $L_p([a,b],mathbbR).$
Given $f,g$ integrable I easily see that $d_p(f,g)$ exists and that $d_p(f,f)=0$ and $d_p(f,g) = d_p(g,f)$. I also proved the triangle inequality by proving the Hölder inequality for positive real functions. I'm just stuck in the condition $fneq g to d_p(f,g)>0$. For simplicity I'm assuming $f neq 0$ and trying to prove $int |f|^p neq 0$.
I know that $fge 0$ implies $int f ge 0$ but I don't see how to get the strict inequality. Any help or hint will be appreciated.
real-analysis metric-spaces
edited Sep 5 at 9:30
José Carlos Santos
122k16101186
asked Sep 5 at 9:18
AnalyticHarmony
595213
1
Actually the space should be altered a bit, since $f =0 $ almost everywhere on $[a,b]$ would deduce that $int_a^b |f|^p = 0$.
– xbh
Sep 5 at 9:24
Will continuos functions work?
– AnalyticHarmony
Sep 5 at 13:03
1
Yes, that would avoid the possibility that $int |f - g|^p = 0$ but $f neq g$, since $int |f|^p = 0$ and $f$ be continuous would deduce that $f equiv 0$. This claim could be proved by contradiction.
– xbh
Sep 5 at 13:08
So the usual notation $L_p$ are used for spaces of continuous functions?
– AnalyticHarmony
Sep 5 at 13:09
1
Not exactly. Actually $L$ stands for "Lebesgue" integration. Generally $R$ would be used for Riemann integrals. For continuous functions, try $C$, or $C_p$ to indicate the metric.
– xbh
Sep 5 at 13:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $L_p([a,b];mathbbR) = f$ is Riemann-integrable $$. I want to prove that
$$
d_p(f,g) = left(int_a^b |f(x)-g(x)|^p dxright)^1/p
$$
defines a metric in $L_p([a,b],mathbbR).$
Given $f,g$ integrable I easily see that $d_p(f,g)$ exists and that $d_p(f,f)=0$ and $d_p(f,g) = d_p(g,f)$. I also proved the triangle inequality by proving the Hölder inequality for positive real functions. I'm just stuck in the condition $fneq g to d_p(f,g)>0$. For simplicity I'm assuming $f neq 0$ and trying to prove $int |f|^p neq 0$.
I know that $fge 0$ implies $int f ge 0$ but I don't see how to get the strict inequality. Any help or hint will be appreciated.
real-analysis metric-spaces
edited Sep 5 at 9:30
José Carlos Santos
122k16101186
asked Sep 5 at 9:18
AnalyticHarmony
595213
Let $L_p([a,b];mathbbR) = f$ is Riemann-integrable $$. I want to prove that
$$
d_p(f,g) = left(int_a^b |f(x)-g(x)|^p dxright)^1/p
$$
defines a metric in $L_p([a,b],mathbbR).$
Given $f,g$ integrable I easily see that $d_p(f,g)$ exists and that $d_p(f,f)=0$ and $d_p(f,g) = d_p(g,f)$. I also proved the triangle inequality by proving the Hölder inequality for positive real functions. I'm just stuck in the condition $fneq g to d_p(f,g)>0$. For simplicity I'm assuming $f neq 0$ and trying to prove $int |f|^p neq 0$.
I know that $fge 0$ implies $int f ge 0$ but I don't see how to get the strict inequality. Any help or hint will be appreciated.
real-analysis metric-spaces
real-analysis metric-spaces
edited Sep 5 at 9:30
José Carlos Santos
122k16101186
asked Sep 5 at 9:18
AnalyticHarmony
595213
edited Sep 5 at 9:30
José Carlos Santos
122k16101186
asked Sep 5 at 9:18
AnalyticHarmony
595213
edited Sep 5 at 9:30
José Carlos Santos
122k16101186
edited Sep 5 at 9:30
José Carlos Santos
122k16101186
edited Sep 5 at 9:30
José Carlos Santos
122k16101186
122k16101186
asked Sep 5 at 9:18
AnalyticHarmony
595213
asked Sep 5 at 9:18
AnalyticHarmony
595213
asked Sep 5 at 9:18
AnalyticHarmony
595213
595213
1
Actually the space should be altered a bit, since $f =0 $ almost everywhere on $[a,b]$ would deduce that $int_a^b |f|^p = 0$.
– xbh
Sep 5 at 9:24
Will continuos functions work?
– AnalyticHarmony
Sep 5 at 13:03
1
Yes, that would avoid the possibility that $int |f - g|^p = 0$ but $f neq g$, since $int |f|^p = 0$ and $f$ be continuous would deduce that $f equiv 0$. This claim could be proved by contradiction.
– xbh
Sep 5 at 13:08
So the usual notation $L_p$ are used for spaces of continuous functions?
– AnalyticHarmony
Sep 5 at 13:09
1
Not exactly. Actually $L$ stands for "Lebesgue" integration. Generally $R$ would be used for Riemann integrals. For continuous functions, try $C$, or $C_p$ to indicate the metric.
– xbh
Sep 5 at 13:12
add a comment |Â
1
Actually the space should be altered a bit, since $f =0 $ almost everywhere on $[a,b]$ would deduce that $int_a^b |f|^p = 0$.
– xbh
Sep 5 at 9:24
Will continuos functions work?
– AnalyticHarmony
Sep 5 at 13:03
1
Yes, that would avoid the possibility that $int |f - g|^p = 0$ but $f neq g$, since $int |f|^p = 0$ and $f$ be continuous would deduce that $f equiv 0$. This claim could be proved by contradiction.
– xbh
Sep 5 at 13:08
So the usual notation $L_p$ are used for spaces of continuous functions?
– AnalyticHarmony
Sep 5 at 13:09
1
Not exactly. Actually $L$ stands for "Lebesgue" integration. Generally $R$ would be used for Riemann integrals. For continuous functions, try $C$, or $C_p$ to indicate the metric.
– xbh
Sep 5 at 13:12
1
1
Actually the space should be altered a bit, since $f =0 $ almost everywhere on $[a,b]$ would deduce that $int_a^b |f|^p = 0$.
– xbh
Sep 5 at 9:24
Actually the space should be altered a bit, since $f =0 $ almost everywhere on $[a,b]$ would deduce that $int_a^b |f|^p = 0$.
– xbh
Sep 5 at 9:24
Will continuos functions work?
– AnalyticHarmony
Sep 5 at 13:03
Will continuos functions work?
– AnalyticHarmony
Sep 5 at 13:03
1
1
Yes, that would avoid the possibility that $int |f - g|^p = 0$ but $f neq g$, since $int |f|^p = 0$ and $f$ be continuous would deduce that $f equiv 0$. This claim could be proved by contradiction.
– xbh
Sep 5 at 13:08
Yes, that would avoid the possibility that $int |f - g|^p = 0$ but $f neq g$, since $int |f|^p = 0$ and $f$ be continuous would deduce that $f equiv 0$. This claim could be proved by contradiction.
– xbh
Sep 5 at 13:08
So the usual notation $L_p$ are used for spaces of continuous functions?
– AnalyticHarmony
Sep 5 at 13:09
So the usual notation $L_p$ are used for spaces of continuous functions?
– AnalyticHarmony
Sep 5 at 13:09
1
1
Not exactly. Actually $L$ stands for "Lebesgue" integration. Generally $R$ would be used for Riemann integrals. For continuous functions, try $C$, or $C_p$ to indicate the metric.
– xbh
Sep 5 at 13:12
Not exactly. Actually $L$ stands for "Lebesgue" integration. Generally $R$ would be used for Riemann integrals. For continuous functions, try $C$, or $C_p$ to indicate the metric.
– xbh
Sep 5 at 13:12
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
You can't prove it because it is not true. If$$f(x)=begincases1&text if x=a\0&text otherwise,endcases$$then $d_p(f,0)=0$, but $fneq0$. The function $d_p$ is a pseudometric, but not a metric.
answered Sep 5 at 9:26
José Carlos Santos
122k16101186
If I change the set for continuous functions then it'll be a metric? Or is the problem in the defining function per se?
– AnalyticHarmony
Sep 5 at 13:04
2
@AnalyticHarmony Yes, if you restrict $d_p$ to the space $mathcalC([a,b])$, then it becomes a metric.
– José Carlos Santos
Sep 5 at 16:31
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
You can't prove it because it is not true. If$$f(x)=begincases1&text if x=a\0&text otherwise,endcases$$then $d_p(f,0)=0$, but $fneq0$. The function $d_p$ is a pseudometric, but not a metric.
answered Sep 5 at 9:26
José Carlos Santos
122k16101186
If I change the set for continuous functions then it'll be a metric? Or is the problem in the defining function per se?
– AnalyticHarmony
Sep 5 at 13:04
2
@AnalyticHarmony Yes, if you restrict $d_p$ to the space $mathcalC([a,b])$, then it becomes a metric.
– José Carlos Santos
Sep 5 at 16:31
add a comment |Â
up vote
4
down vote
accepted
You can't prove it because it is not true. If$$f(x)=begincases1&text if x=a\0&text otherwise,endcases$$then $d_p(f,0)=0$, but $fneq0$. The function $d_p$ is a pseudometric, but not a metric.
answered Sep 5 at 9:26
José Carlos Santos
122k16101186
If I change the set for continuous functions then it'll be a metric? Or is the problem in the defining function per se?
– AnalyticHarmony
Sep 5 at 13:04
2
@AnalyticHarmony Yes, if you restrict $d_p$ to the space $mathcalC([a,b])$, then it becomes a metric.
– José Carlos Santos
Sep 5 at 16:31
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
You can't prove it because it is not true. If$$f(x)=begincases1&text if x=a\0&text otherwise,endcases$$then $d_p(f,0)=0$, but $fneq0$. The function $d_p$ is a pseudometric, but not a metric.
answered Sep 5 at 9:26
José Carlos Santos
122k16101186
You can't prove it because it is not true. If$$f(x)=begincases1&text if x=a\0&text otherwise,endcases$$then $d_p(f,0)=0$, but $fneq0$. The function $d_p$ is a pseudometric, but not a metric.
answered Sep 5 at 9:26
José Carlos Santos
122k16101186
answered Sep 5 at 9:26
José Carlos Santos
122k16101186
answered Sep 5 at 9:26
José Carlos Santos
122k16101186
answered Sep 5 at 9:26
José Carlos Santos
122k16101186
122k16101186
If I change the set for continuous functions then it'll be a metric? Or is the problem in the defining function per se?
– AnalyticHarmony
Sep 5 at 13:04
2
@AnalyticHarmony Yes, if you restrict $d_p$ to the space $mathcalC([a,b])$, then it becomes a metric.
– José Carlos Santos
Sep 5 at 16:31
add a comment |Â
If I change the set for continuous functions then it'll be a metric? Or is the problem in the defining function per se?
– AnalyticHarmony
Sep 5 at 13:04
2
@AnalyticHarmony Yes, if you restrict $d_p$ to the space $mathcalC([a,b])$, then it becomes a metric.
– José Carlos Santos
Sep 5 at 16:31
If I change the set for continuous functions then it'll be a metric? Or is the problem in the defining function per se?
– AnalyticHarmony
Sep 5 at 13:04
If I change the set for continuous functions then it'll be a metric? Or is the problem in the defining function per se?
– AnalyticHarmony
Sep 5 at 13:04
2
2
@AnalyticHarmony Yes, if you restrict $d_p$ to the space $mathcalC([a,b])$, then it becomes a metric.
– José Carlos Santos
Sep 5 at 16:31
@AnalyticHarmony Yes, if you restrict $d_p$ to the space $mathcalC([a,b])$, then it becomes a metric.
– José Carlos Santos
Sep 5 at 16:31
add a comment |Â
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1
Actually the space should be altered a bit, since $f =0 $ almost everywhere on $[a,b]$ would deduce that $int_a^b |f|^p = 0$.
– xbh
Sep 5 at 9:24
Will continuos functions work?
– AnalyticHarmony
Sep 5 at 13:03
1
Yes, that would avoid the possibility that $int |f - g|^p = 0$ but $f neq g$, since $int |f|^p = 0$ and $f$ be continuous would deduce that $f equiv 0$. This claim could be proved by contradiction.
– xbh
Sep 5 at 13:08
So the usual notation $L_p$ are used for spaces of continuous functions?
– AnalyticHarmony
Sep 5 at 13:09
1
Not exactly. Actually $L$ stands for "Lebesgue" integration. Generally $R$ would be used for Riemann integrals. For continuous functions, try $C$, or $C_p$ to indicate the metric.
– xbh
Sep 5 at 13:12