Exercise on $L^p([0,1])$
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From now on: $L^p := L^p([0,1],m)$, where $m$ is the Lebesgue measure.
Let $f in L^p$, and let $T(f):= int_0^1 f(x^2)dx$. Show that $T$ is in $(L^4)^ast$ (so it is well defined e bounded) and find $g_T in L^frac43$ such that $T$ is represented by $g_T$ as $int f(x)g_T(x)dx$.
Attempt: What I've tried is this, let $t=x^2$ and so $dt=2x dx$, now, we can make the substitution in the integral and let:
$T(f)=int fracf(t)2sqrttdt$. So to complete proof we have to prove that $frac1sqrtt$ is in $L^frac43$, but: $int t^-frac12 frac43dt=int t^-frac23dt < infty $.
Is my proof correct?
functional-analysis lp-spaces
asked Sep 10 at 10:41
HaroldF
525415
add a comment |Â
up vote
3
down vote
favorite
From now on: $L^p := L^p([0,1],m)$, where $m$ is the Lebesgue measure.
Let $f in L^p$, and let $T(f):= int_0^1 f(x^2)dx$. Show that $T$ is in $(L^4)^ast$ (so it is well defined e bounded) and find $g_T in L^frac43$ such that $T$ is represented by $g_T$ as $int f(x)g_T(x)dx$.
Attempt: What I've tried is this, let $t=x^2$ and so $dt=2x dx$, now, we can make the substitution in the integral and let:
$T(f)=int fracf(t)2sqrttdt$. So to complete proof we have to prove that $frac1sqrtt$ is in $L^frac43$, but: $int t^-frac12 frac43dt=int t^-frac23dt < infty $.
Is my proof correct?
functional-analysis lp-spaces
asked Sep 10 at 10:41
HaroldF
525415
2
Yes, your proof looks good to me. Maybe a sentence stating that in hindsight $T$ is well defined and bounded as a functional on $L^4$ could be added, but that's not crucial.
– Joseph Adams
Sep 10 at 11:17
Thank you. I was concerned about the main idea in this type of exercises. Now I have somehow a different thought about $L^p$ spaces: when one wants to solve a "calculus I" integral by change of variables, multiplication by the suitable function to make sense of the substitution would be a great thing, unfortunately it cannot be done generally. But if you can locate your function in the world of $L^p$'s and everything is good, then you can do it.
– HaroldF
Sep 10 at 12:20
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
From now on: $L^p := L^p([0,1],m)$, where $m$ is the Lebesgue measure.
Let $f in L^p$, and let $T(f):= int_0^1 f(x^2)dx$. Show that $T$ is in $(L^4)^ast$ (so it is well defined e bounded) and find $g_T in L^frac43$ such that $T$ is represented by $g_T$ as $int f(x)g_T(x)dx$.
Attempt: What I've tried is this, let $t=x^2$ and so $dt=2x dx$, now, we can make the substitution in the integral and let:
$T(f)=int fracf(t)2sqrttdt$. So to complete proof we have to prove that $frac1sqrtt$ is in $L^frac43$, but: $int t^-frac12 frac43dt=int t^-frac23dt < infty $.
Is my proof correct?
functional-analysis lp-spaces
asked Sep 10 at 10:41
HaroldF
525415
From now on: $L^p := L^p([0,1],m)$, where $m$ is the Lebesgue measure.
Let $f in L^p$, and let $T(f):= int_0^1 f(x^2)dx$. Show that $T$ is in $(L^4)^ast$ (so it is well defined e bounded) and find $g_T in L^frac43$ such that $T$ is represented by $g_T$ as $int f(x)g_T(x)dx$.
Attempt: What I've tried is this, let $t=x^2$ and so $dt=2x dx$, now, we can make the substitution in the integral and let:
$T(f)=int fracf(t)2sqrttdt$. So to complete proof we have to prove that $frac1sqrtt$ is in $L^frac43$, but: $int t^-frac12 frac43dt=int t^-frac23dt < infty $.
Is my proof correct?
functional-analysis lp-spaces
functional-analysis lp-spaces
asked Sep 10 at 10:41
HaroldF
525415
asked Sep 10 at 10:41
HaroldF
525415
asked Sep 10 at 10:41
HaroldF
525415
asked Sep 10 at 10:41
HaroldF
525415
asked Sep 10 at 10:41
HaroldF
525415
525415
2
Yes, your proof looks good to me. Maybe a sentence stating that in hindsight $T$ is well defined and bounded as a functional on $L^4$ could be added, but that's not crucial.
– Joseph Adams
Sep 10 at 11:17
Thank you. I was concerned about the main idea in this type of exercises. Now I have somehow a different thought about $L^p$ spaces: when one wants to solve a "calculus I" integral by change of variables, multiplication by the suitable function to make sense of the substitution would be a great thing, unfortunately it cannot be done generally. But if you can locate your function in the world of $L^p$'s and everything is good, then you can do it.
– HaroldF
Sep 10 at 12:20
add a comment |Â
2
Yes, your proof looks good to me. Maybe a sentence stating that in hindsight $T$ is well defined and bounded as a functional on $L^4$ could be added, but that's not crucial.
– Joseph Adams
Sep 10 at 11:17
Thank you. I was concerned about the main idea in this type of exercises. Now I have somehow a different thought about $L^p$ spaces: when one wants to solve a "calculus I" integral by change of variables, multiplication by the suitable function to make sense of the substitution would be a great thing, unfortunately it cannot be done generally. But if you can locate your function in the world of $L^p$'s and everything is good, then you can do it.
– HaroldF
Sep 10 at 12:20
2
2
Yes, your proof looks good to me. Maybe a sentence stating that in hindsight $T$ is well defined and bounded as a functional on $L^4$ could be added, but that's not crucial.
– Joseph Adams
Sep 10 at 11:17
Yes, your proof looks good to me. Maybe a sentence stating that in hindsight $T$ is well defined and bounded as a functional on $L^4$ could be added, but that's not crucial.
– Joseph Adams
Sep 10 at 11:17
Thank you. I was concerned about the main idea in this type of exercises. Now I have somehow a different thought about $L^p$ spaces: when one wants to solve a "calculus I" integral by change of variables, multiplication by the suitable function to make sense of the substitution would be a great thing, unfortunately it cannot be done generally. But if you can locate your function in the world of $L^p$'s and everything is good, then you can do it.
– HaroldF
Sep 10 at 12:20
Thank you. I was concerned about the main idea in this type of exercises. Now I have somehow a different thought about $L^p$ spaces: when one wants to solve a "calculus I" integral by change of variables, multiplication by the suitable function to make sense of the substitution would be a great thing, unfortunately it cannot be done generally. But if you can locate your function in the world of $L^p$'s and everything is good, then you can do it.
– HaroldF
Sep 10 at 12:20
add a comment |Â
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2
Yes, your proof looks good to me. Maybe a sentence stating that in hindsight $T$ is well defined and bounded as a functional on $L^4$ could be added, but that's not crucial.
– Joseph Adams
Sep 10 at 11:17
Thank you. I was concerned about the main idea in this type of exercises. Now I have somehow a different thought about $L^p$ spaces: when one wants to solve a "calculus I" integral by change of variables, multiplication by the suitable function to make sense of the substitution would be a great thing, unfortunately it cannot be done generally. But if you can locate your function in the world of $L^p$'s and everything is good, then you can do it.
– HaroldF
Sep 10 at 12:20