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







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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














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







share|cite|improve this question

















  • 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












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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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












  • 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















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