Lebesgue space inclusion on half real line











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Assume that $f in L^1 ([0,infty))$ is a real continuous function on $[0,infty)$.



Can we then conclude that $f in L^p ([0,infty))$ for $p in [1, infty)$?






real-analysis integration lebesgue-integral lebesgue-measure







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  • I guess $f(x) = 1$ is a real continuous function on $[0,infty)$. Is it in $L^p$?
    – GEdgar
    Nov 22 at 16:45















up vote
0
down vote

favorite












Assume that $f in L^1 ([0,infty))$ is a real continuous function on $[0,infty)$.



Can we then conclude that $f in L^p ([0,infty))$ for $p in [1, infty)$?






real-analysis integration lebesgue-integral lebesgue-measure







share|cite|improve this question






















  • I guess $f(x) = 1$ is a real continuous function on $[0,infty)$. Is it in $L^p$?
    – GEdgar
    Nov 22 at 16:45













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Assume that $f in L^1 ([0,infty))$ is a real continuous function on $[0,infty)$.



Can we then conclude that $f in L^p ([0,infty))$ for $p in [1, infty)$?






real-analysis integration lebesgue-integral lebesgue-measure







share|cite|improve this question













Assume that $f in L^1 ([0,infty))$ is a real continuous function on $[0,infty)$.



Can we then conclude that $f in L^p ([0,infty))$ for $p in [1, infty)$?






real-analysis integration lebesgue-integral lebesgue-measure




real-analysis integration lebesgue-integral lebesgue-measure






share|cite|improve this question













share|cite|improve this question











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asked Nov 22 at 16:36









sleepingrabbit

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  • I guess $f(x) = 1$ is a real continuous function on $[0,infty)$. Is it in $L^p$?
    – GEdgar
    Nov 22 at 16:45


















  • I guess $f(x) = 1$ is a real continuous function on $[0,infty)$. Is it in $L^p$?
    – GEdgar
    Nov 22 at 16:45
















I guess $f(x) = 1$ is a real continuous function on $[0,infty)$. Is it in $L^p$?
– GEdgar
Nov 22 at 16:45




I guess $f(x) = 1$ is a real continuous function on $[0,infty)$. Is it in $L^p$?
– GEdgar
Nov 22 at 16:45










1 Answer
1






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This is false. To build a counterexample, consider functions $g_n(x)$ that is non-zero only on $(n, n + 2n^{-3})$, piecewise linear, with $g_n(n) = g_n(n+2n^{-3}) = 0$ and $g_n(n+n^{-3}) = n$. Then,
$$
int g_n(x) dx = frac12 n cdot 2n^{-3} = n^{-2}.
$$
Since for $n geq 2$ functions $g_n$ have disjoint support, if we define $f = sum_{n geq 2} g_n$ then $f in L^1([0, infty))$. However, if we take $p = 2$, we have that
$$
int g_n(x)^2 dx geq int_{n + frac12 n^{-3}}^{n + frac32 n^{-3}} (frac12 n)^2 dx = frac14 n^{-1}.
$$
This implies $int f^2 geq frac14 sum_{n geq 2} n^{-1} = infty$.






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    accepted










    This is false. To build a counterexample, consider functions $g_n(x)$ that is non-zero only on $(n, n + 2n^{-3})$, piecewise linear, with $g_n(n) = g_n(n+2n^{-3}) = 0$ and $g_n(n+n^{-3}) = n$. Then,
    $$
    int g_n(x) dx = frac12 n cdot 2n^{-3} = n^{-2}.
    $$
    Since for $n geq 2$ functions $g_n$ have disjoint support, if we define $f = sum_{n geq 2} g_n$ then $f in L^1([0, infty))$. However, if we take $p = 2$, we have that
    $$
    int g_n(x)^2 dx geq int_{n + frac12 n^{-3}}^{n + frac32 n^{-3}} (frac12 n)^2 dx = frac14 n^{-1}.
    $$
    This implies $int f^2 geq frac14 sum_{n geq 2} n^{-1} = infty$.






    share|cite|improve this answer



























      up vote
      2
      down vote



      accepted










      This is false. To build a counterexample, consider functions $g_n(x)$ that is non-zero only on $(n, n + 2n^{-3})$, piecewise linear, with $g_n(n) = g_n(n+2n^{-3}) = 0$ and $g_n(n+n^{-3}) = n$. Then,
      $$
      int g_n(x) dx = frac12 n cdot 2n^{-3} = n^{-2}.
      $$
      Since for $n geq 2$ functions $g_n$ have disjoint support, if we define $f = sum_{n geq 2} g_n$ then $f in L^1([0, infty))$. However, if we take $p = 2$, we have that
      $$
      int g_n(x)^2 dx geq int_{n + frac12 n^{-3}}^{n + frac32 n^{-3}} (frac12 n)^2 dx = frac14 n^{-1}.
      $$
      This implies $int f^2 geq frac14 sum_{n geq 2} n^{-1} = infty$.






      share|cite|improve this answer

























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        This is false. To build a counterexample, consider functions $g_n(x)$ that is non-zero only on $(n, n + 2n^{-3})$, piecewise linear, with $g_n(n) = g_n(n+2n^{-3}) = 0$ and $g_n(n+n^{-3}) = n$. Then,
        $$
        int g_n(x) dx = frac12 n cdot 2n^{-3} = n^{-2}.
        $$
        Since for $n geq 2$ functions $g_n$ have disjoint support, if we define $f = sum_{n geq 2} g_n$ then $f in L^1([0, infty))$. However, if we take $p = 2$, we have that
        $$
        int g_n(x)^2 dx geq int_{n + frac12 n^{-3}}^{n + frac32 n^{-3}} (frac12 n)^2 dx = frac14 n^{-1}.
        $$
        This implies $int f^2 geq frac14 sum_{n geq 2} n^{-1} = infty$.






        share|cite|improve this answer














        This is false. To build a counterexample, consider functions $g_n(x)$ that is non-zero only on $(n, n + 2n^{-3})$, piecewise linear, with $g_n(n) = g_n(n+2n^{-3}) = 0$ and $g_n(n+n^{-3}) = n$. Then,
        $$
        int g_n(x) dx = frac12 n cdot 2n^{-3} = n^{-2}.
        $$
        Since for $n geq 2$ functions $g_n$ have disjoint support, if we define $f = sum_{n geq 2} g_n$ then $f in L^1([0, infty))$. However, if we take $p = 2$, we have that
        $$
        int g_n(x)^2 dx geq int_{n + frac12 n^{-3}}^{n + frac32 n^{-3}} (frac12 n)^2 dx = frac14 n^{-1}.
        $$
        This implies $int f^2 geq frac14 sum_{n geq 2} n^{-1} = infty$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 22 at 17:29

























        answered Nov 22 at 16:55









        Daniel

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        1,516210






























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