Find $fin L^p(0,1)$, but $fnotin L^q(0,1)$, $p < q$












-1












$begingroup$


I want to find a measurable function $f in L^p(0,1)$, for $pin [1,+infty)$, but $f notin L^q(0,1)$ for each $qin (p,+infty]$.



I tried to manipulate $f=frac{1}{x^a}$, improving the exponent $a$ s.t. $int_0^1 |frac{1}{x^a}|^p dmu < infty $, but since $p,q$ are real number, I cannot find anything - I don't know how to say it properly, but I cannot find an explicit expression s.t. some algebraic tricks make appear something smaller or bigger than 1, which is the limit point for the divergence and the convegercence of the function- (otherwise, if we deal with the integers I found $ f=frac{1}{x^{frac{q-1}{p^2}}} $, where $q geq p$ )



Can anyone help me? Thanks in advance.






real-analysis functional-analysis examples-counterexamples lp-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    You know that $int_0^1 frac {1}{x^alpha}<infty$ if and only if $alpha<1$. Can you find an $a$ such that $ap<1$ and $aqgeq1$ for $p<q$?
    $endgroup$
    – Kolja
    Jan 2 at 15:32










  • $begingroup$
    sorry, but shouldn't your first condition be also equal? $int_0^1 frac{1}{x^alpha} < infty $ iff $alpha leq 1$. Anyway I'll try to think about it a bit more
    $endgroup$
    – christmas_light
    Jan 2 at 15:39












  • $begingroup$
    No, because we have $int_0^1 frac{1}{x} = infty$. More precisely $int_varepsilon^1 frac{1}{x} = log(1)-log(varepsilon) rightarrow infty$ as $varepsilon$ goes to $0$.
    $endgroup$
    – Kolja
    Jan 2 at 15:41












  • $begingroup$
    sorry, i misread, my fault ;)
    $endgroup$
    – christmas_light
    Jan 2 at 15:46










  • $begingroup$
    You can find an $a$ such that $ap>1$ and $aq>1$ if you want, but there is a very simple $a$ for the case $ap<1$ and $aqgeq1$. I'll let you find it by yourself.
    $endgroup$
    – Kolja
    Jan 2 at 15:56
















-1












$begingroup$


I want to find a measurable function $f in L^p(0,1)$, for $pin [1,+infty)$, but $f notin L^q(0,1)$ for each $qin (p,+infty]$.



I tried to manipulate $f=frac{1}{x^a}$, improving the exponent $a$ s.t. $int_0^1 |frac{1}{x^a}|^p dmu < infty $, but since $p,q$ are real number, I cannot find anything - I don't know how to say it properly, but I cannot find an explicit expression s.t. some algebraic tricks make appear something smaller or bigger than 1, which is the limit point for the divergence and the convegercence of the function- (otherwise, if we deal with the integers I found $ f=frac{1}{x^{frac{q-1}{p^2}}} $, where $q geq p$ )



Can anyone help me? Thanks in advance.






real-analysis functional-analysis examples-counterexamples lp-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    You know that $int_0^1 frac {1}{x^alpha}<infty$ if and only if $alpha<1$. Can you find an $a$ such that $ap<1$ and $aqgeq1$ for $p<q$?
    $endgroup$
    – Kolja
    Jan 2 at 15:32










  • $begingroup$
    sorry, but shouldn't your first condition be also equal? $int_0^1 frac{1}{x^alpha} < infty $ iff $alpha leq 1$. Anyway I'll try to think about it a bit more
    $endgroup$
    – christmas_light
    Jan 2 at 15:39












  • $begingroup$
    No, because we have $int_0^1 frac{1}{x} = infty$. More precisely $int_varepsilon^1 frac{1}{x} = log(1)-log(varepsilon) rightarrow infty$ as $varepsilon$ goes to $0$.
    $endgroup$
    – Kolja
    Jan 2 at 15:41












  • $begingroup$
    sorry, i misread, my fault ;)
    $endgroup$
    – christmas_light
    Jan 2 at 15:46










  • $begingroup$
    You can find an $a$ such that $ap>1$ and $aq>1$ if you want, but there is a very simple $a$ for the case $ap<1$ and $aqgeq1$. I'll let you find it by yourself.
    $endgroup$
    – Kolja
    Jan 2 at 15:56














-1












-1








-1


1



$begingroup$


I want to find a measurable function $f in L^p(0,1)$, for $pin [1,+infty)$, but $f notin L^q(0,1)$ for each $qin (p,+infty]$.



I tried to manipulate $f=frac{1}{x^a}$, improving the exponent $a$ s.t. $int_0^1 |frac{1}{x^a}|^p dmu < infty $, but since $p,q$ are real number, I cannot find anything - I don't know how to say it properly, but I cannot find an explicit expression s.t. some algebraic tricks make appear something smaller or bigger than 1, which is the limit point for the divergence and the convegercence of the function- (otherwise, if we deal with the integers I found $ f=frac{1}{x^{frac{q-1}{p^2}}} $, where $q geq p$ )



Can anyone help me? Thanks in advance.






real-analysis functional-analysis examples-counterexamples lp-spaces







share|cite|improve this question









$endgroup$




I want to find a measurable function $f in L^p(0,1)$, for $pin [1,+infty)$, but $f notin L^q(0,1)$ for each $qin (p,+infty]$.



I tried to manipulate $f=frac{1}{x^a}$, improving the exponent $a$ s.t. $int_0^1 |frac{1}{x^a}|^p dmu < infty $, but since $p,q$ are real number, I cannot find anything - I don't know how to say it properly, but I cannot find an explicit expression s.t. some algebraic tricks make appear something smaller or bigger than 1, which is the limit point for the divergence and the convegercence of the function- (otherwise, if we deal with the integers I found $ f=frac{1}{x^{frac{q-1}{p^2}}} $, where $q geq p$ )



Can anyone help me? Thanks in advance.






real-analysis functional-analysis examples-counterexamples lp-spaces




real-analysis functional-analysis examples-counterexamples lp-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Jan 2 at 15:27









christmas_lightchristmas_light

777




777












  • $begingroup$
    You know that $int_0^1 frac {1}{x^alpha}<infty$ if and only if $alpha<1$. Can you find an $a$ such that $ap<1$ and $aqgeq1$ for $p<q$?
    $endgroup$
    – Kolja
    Jan 2 at 15:32










  • $begingroup$
    sorry, but shouldn't your first condition be also equal? $int_0^1 frac{1}{x^alpha} < infty $ iff $alpha leq 1$. Anyway I'll try to think about it a bit more
    $endgroup$
    – christmas_light
    Jan 2 at 15:39












  • $begingroup$
    No, because we have $int_0^1 frac{1}{x} = infty$. More precisely $int_varepsilon^1 frac{1}{x} = log(1)-log(varepsilon) rightarrow infty$ as $varepsilon$ goes to $0$.
    $endgroup$
    – Kolja
    Jan 2 at 15:41












  • $begingroup$
    sorry, i misread, my fault ;)
    $endgroup$
    – christmas_light
    Jan 2 at 15:46










  • $begingroup$
    You can find an $a$ such that $ap>1$ and $aq>1$ if you want, but there is a very simple $a$ for the case $ap<1$ and $aqgeq1$. I'll let you find it by yourself.
    $endgroup$
    – Kolja
    Jan 2 at 15:56


















  • $begingroup$
    You know that $int_0^1 frac {1}{x^alpha}<infty$ if and only if $alpha<1$. Can you find an $a$ such that $ap<1$ and $aqgeq1$ for $p<q$?
    $endgroup$
    – Kolja
    Jan 2 at 15:32










  • $begingroup$
    sorry, but shouldn't your first condition be also equal? $int_0^1 frac{1}{x^alpha} < infty $ iff $alpha leq 1$. Anyway I'll try to think about it a bit more
    $endgroup$
    – christmas_light
    Jan 2 at 15:39












  • $begingroup$
    No, because we have $int_0^1 frac{1}{x} = infty$. More precisely $int_varepsilon^1 frac{1}{x} = log(1)-log(varepsilon) rightarrow infty$ as $varepsilon$ goes to $0$.
    $endgroup$
    – Kolja
    Jan 2 at 15:41












  • $begingroup$
    sorry, i misread, my fault ;)
    $endgroup$
    – christmas_light
    Jan 2 at 15:46










  • $begingroup$
    You can find an $a$ such that $ap>1$ and $aq>1$ if you want, but there is a very simple $a$ for the case $ap<1$ and $aqgeq1$. I'll let you find it by yourself.
    $endgroup$
    – Kolja
    Jan 2 at 15:56
















$begingroup$
You know that $int_0^1 frac {1}{x^alpha}<infty$ if and only if $alpha<1$. Can you find an $a$ such that $ap<1$ and $aqgeq1$ for $p<q$?
$endgroup$
– Kolja
Jan 2 at 15:32




$begingroup$
You know that $int_0^1 frac {1}{x^alpha}<infty$ if and only if $alpha<1$. Can you find an $a$ such that $ap<1$ and $aqgeq1$ for $p<q$?
$endgroup$
– Kolja
Jan 2 at 15:32












$begingroup$
sorry, but shouldn't your first condition be also equal? $int_0^1 frac{1}{x^alpha} < infty $ iff $alpha leq 1$. Anyway I'll try to think about it a bit more
$endgroup$
– christmas_light
Jan 2 at 15:39






$begingroup$
sorry, but shouldn't your first condition be also equal? $int_0^1 frac{1}{x^alpha} < infty $ iff $alpha leq 1$. Anyway I'll try to think about it a bit more
$endgroup$
– christmas_light
Jan 2 at 15:39














$begingroup$
No, because we have $int_0^1 frac{1}{x} = infty$. More precisely $int_varepsilon^1 frac{1}{x} = log(1)-log(varepsilon) rightarrow infty$ as $varepsilon$ goes to $0$.
$endgroup$
– Kolja
Jan 2 at 15:41






$begingroup$
No, because we have $int_0^1 frac{1}{x} = infty$. More precisely $int_varepsilon^1 frac{1}{x} = log(1)-log(varepsilon) rightarrow infty$ as $varepsilon$ goes to $0$.
$endgroup$
– Kolja
Jan 2 at 15:41














$begingroup$
sorry, i misread, my fault ;)
$endgroup$
– christmas_light
Jan 2 at 15:46




$begingroup$
sorry, i misread, my fault ;)
$endgroup$
– christmas_light
Jan 2 at 15:46












$begingroup$
You can find an $a$ such that $ap>1$ and $aq>1$ if you want, but there is a very simple $a$ for the case $ap<1$ and $aqgeq1$. I'll let you find it by yourself.
$endgroup$
– Kolja
Jan 2 at 15:56




$begingroup$
You can find an $a$ such that $ap>1$ and $aq>1$ if you want, but there is a very simple $a$ for the case $ap<1$ and $aqgeq1$. I'll let you find it by yourself.
$endgroup$
– Kolja
Jan 2 at 15:56










1 Answer
1






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$begingroup$

You can try with the function
$$
f(x) =
begin{cases}
(x log^2 x)^{-1/p}, &text{if} xin (0,1/2),\
0, &text{if} x in (1/2, 1).
end{cases}
$$






share|cite|improve this answer









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    1 Answer
    1






    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    You can try with the function
    $$
    f(x) =
    begin{cases}
    (x log^2 x)^{-1/p}, &text{if} xin (0,1/2),\
    0, &text{if} x in (1/2, 1).
    end{cases}
    $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      You can try with the function
      $$
      f(x) =
      begin{cases}
      (x log^2 x)^{-1/p}, &text{if} xin (0,1/2),\
      0, &text{if} x in (1/2, 1).
      end{cases}
      $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        You can try with the function
        $$
        f(x) =
        begin{cases}
        (x log^2 x)^{-1/p}, &text{if} xin (0,1/2),\
        0, &text{if} x in (1/2, 1).
        end{cases}
        $$






        share|cite|improve this answer









        $endgroup$



        You can try with the function
        $$
        f(x) =
        begin{cases}
        (x log^2 x)^{-1/p}, &text{if} xin (0,1/2),\
        0, &text{if} x in (1/2, 1).
        end{cases}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 15:41









        RigelRigel

        11.4k11320




        11.4k11320






























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