Limit of $L^p$ norm












101












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Could someone help me prove that given a finite measure space $(X, mathcal{M}, sigma)$ and a measurable function $f:Xtomathbb{R}$ in $L^infty$ and some $L^q$, $displaystylelim_{ptoinfty}|f|_p=|f|_infty$?



I don't know where to start.






real-analysis functional-analysis measure-theory lp-spaces







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$endgroup$








  • 1




    $begingroup$
    Why are you taking the limit as p goes to infinity? (i.e. what is the motivation?) I've often seen people use the limit as p goes to 1, since certain optimizations aren't unique in taxicab space.
    $endgroup$
    – Ryan
    Nov 22 '12 at 20:10






  • 9




    $begingroup$
    Its an exercise in a book I'm reading. I don't have any real motivation, except maybe to justify the definition of the $L^{infty}$ norm.
    $endgroup$
    – Parakee
    Nov 22 '12 at 20:19
















101












$begingroup$


Could someone help me prove that given a finite measure space $(X, mathcal{M}, sigma)$ and a measurable function $f:Xtomathbb{R}$ in $L^infty$ and some $L^q$, $displaystylelim_{ptoinfty}|f|_p=|f|_infty$?



I don't know where to start.






real-analysis functional-analysis measure-theory lp-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Why are you taking the limit as p goes to infinity? (i.e. what is the motivation?) I've often seen people use the limit as p goes to 1, since certain optimizations aren't unique in taxicab space.
    $endgroup$
    – Ryan
    Nov 22 '12 at 20:10






  • 9




    $begingroup$
    Its an exercise in a book I'm reading. I don't have any real motivation, except maybe to justify the definition of the $L^{infty}$ norm.
    $endgroup$
    – Parakee
    Nov 22 '12 at 20:19














101












101








101


89



$begingroup$


Could someone help me prove that given a finite measure space $(X, mathcal{M}, sigma)$ and a measurable function $f:Xtomathbb{R}$ in $L^infty$ and some $L^q$, $displaystylelim_{ptoinfty}|f|_p=|f|_infty$?



I don't know where to start.






real-analysis functional-analysis measure-theory lp-spaces







share|cite|improve this question











$endgroup$




Could someone help me prove that given a finite measure space $(X, mathcal{M}, sigma)$ and a measurable function $f:Xtomathbb{R}$ in $L^infty$ and some $L^q$, $displaystylelim_{ptoinfty}|f|_p=|f|_infty$?



I don't know where to start.






real-analysis functional-analysis measure-theory lp-spaces




real-analysis functional-analysis measure-theory lp-spaces






share|cite|improve this question















share|cite|improve this question













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edited Jan 2 at 16:54









klirk

1




1










asked Nov 22 '12 at 19:51









ParakeeParakee

1,04321022




1,04321022








  • 1




    $begingroup$
    Why are you taking the limit as p goes to infinity? (i.e. what is the motivation?) I've often seen people use the limit as p goes to 1, since certain optimizations aren't unique in taxicab space.
    $endgroup$
    – Ryan
    Nov 22 '12 at 20:10






  • 9




    $begingroup$
    Its an exercise in a book I'm reading. I don't have any real motivation, except maybe to justify the definition of the $L^{infty}$ norm.
    $endgroup$
    – Parakee
    Nov 22 '12 at 20:19














  • 1




    $begingroup$
    Why are you taking the limit as p goes to infinity? (i.e. what is the motivation?) I've often seen people use the limit as p goes to 1, since certain optimizations aren't unique in taxicab space.
    $endgroup$
    – Ryan
    Nov 22 '12 at 20:10






  • 9




    $begingroup$
    Its an exercise in a book I'm reading. I don't have any real motivation, except maybe to justify the definition of the $L^{infty}$ norm.
    $endgroup$
    – Parakee
    Nov 22 '12 at 20:19








1




1




$begingroup$
Why are you taking the limit as p goes to infinity? (i.e. what is the motivation?) I've often seen people use the limit as p goes to 1, since certain optimizations aren't unique in taxicab space.
$endgroup$
– Ryan
Nov 22 '12 at 20:10




$begingroup$
Why are you taking the limit as p goes to infinity? (i.e. what is the motivation?) I've often seen people use the limit as p goes to 1, since certain optimizations aren't unique in taxicab space.
$endgroup$
– Ryan
Nov 22 '12 at 20:10




9




9




$begingroup$
Its an exercise in a book I'm reading. I don't have any real motivation, except maybe to justify the definition of the $L^{infty}$ norm.
$endgroup$
– Parakee
Nov 22 '12 at 20:19




$begingroup$
Its an exercise in a book I'm reading. I don't have any real motivation, except maybe to justify the definition of the $L^{infty}$ norm.
$endgroup$
– Parakee
Nov 22 '12 at 20:19










2 Answers
2






active

oldest

votes


















112












$begingroup$

Fix $delta>0$ and let $S_delta:={x,|f(x)|geqslant lVert frVert_infty-delta}$ for $delta<lVert frVert_infty$. We have
$$lVert frVert_pgeqslant left(int_{S_delta}(lVert frVert_infty-delta)^pdmuright)^{1/p}=(lVert frVert_infty-delta)mu(S_delta)^{1/p},$$
since $mu(S_delta)$ is finite and positive.
This gives
$$liminf_{pto +infty}lVert frVert_pgeqslantlVert frVert_infty.$$
As $|f(x)|leqslantlVert frVert_infty$ for almost every $x$, we have for $p>q$, $$
lVert frVert_pleqslantleft(int_X|f(x)|^{p-q}|f(x)|^qdmuright)^{1/p}leqslant lVert frVert_infty^{frac{p-q}p}lVert frVert_q^{q/p},$$
giving the reverse inequality.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    How does that last step give us the reverse inequality?
    $endgroup$
    – Parakee
    Nov 22 '12 at 20:27








  • 9




    $begingroup$
    Take this time $limsup_{pto infty}$.
    $endgroup$
    – Davide Giraudo
    Nov 22 '12 at 20:28






  • 1




    $begingroup$
    Assume that we work on a probability space. Then $lVert frVert_pleqslant lVert frVert_infty$, and we are interested about a control of $lVert frVert_infty$ in terms of the $L^p$ norm. We thus fix an arbitrary small real number and look at the set where $|f|$ is greater than the supremum norm minus the small number.
    $endgroup$
    – Davide Giraudo
    Jan 23 '14 at 21:40






  • 8




    $begingroup$
    Doesn't your proof assume $mu(X)<infty$?
    $endgroup$
    – Eric Auld
    Apr 22 '14 at 22:24






  • 6




    $begingroup$
    Yes. However, I think we can get rid of this assumption: we only have to consider the case where the measure space is $sigma$-finite, hence $X=bigcup_n A_n$ where $A_nuparrow X$ and each $A_n$ is of finite measure. Then we pick $n$ such that $mu(A_ncap S_{delta})$ is a positive real number.
    $endgroup$
    – Davide Giraudo
    Apr 23 '14 at 8:17



















22












$begingroup$

Let $f:Xto mathbb{R}$. Assume that $f$ is measurable, and that $|f|_p<infty$ for all large $p$. Suppose for convenience that $fgeq 0$. (If not, just work with $f^*:=|f|$.) We define
$$
|f|_{infty}:=sup {rin mathbb{R}: muleft( {x:|f(x)|geq r} right)>0}.
$$



I claim without proof that $|f|_p < infty$ for large $p$ implies that $|f|_infty < infty$.



If $|f|_{infty}=0$, we can see that the proposition holds trivially. If $|f|_{infty}neq 0$, let $M:=|f|_{infty}$.



Fix $epsilon$ such that $0< epsilon < M$. Define $D:={x:f(x)geq M-epsilon}$. Observe that $mu(D)>0$ by definition of $|f|_{infty}$. Also, $mu(D)<infty$ since $f$ is integrable for all large $p$. Now we can establish $liminf_{ptoinfty }|f|_pgeq M-epsilon$ by
$$
left( int_{X}f(x)^p dx right)^{1/p} geq left( int_D (M-epsilon)^pdx right)^{1/p} = (M-epsilon)mu(D)^{1/p} xrightarrow{ptoinfty}(M-epsilon)
$$



Now we show $limsup_{ptoinfty}|f|_{p} leq M+epsilon$. Let $tilde{f}(x) := dfrac{f(x)}{M+epsilon}$. Observe that $0leq tilde{f}(x)leq M/(M+epsilon)<1$, and that
$$
left( int_{X} f(x)^p dx right)^{1/p} = (M+epsilon)left( int_{X} tilde{f}(x)^p dx right)^{1/p}.
$$



Now it suffices to show that $int_X tilde{f}(x)^p dx$ is bounded above by $1$ as $pto infty$, since then we have



$$
left( int_{X} f(x)^p dx right)^{1/p} = (M+epsilon)left( int_{X} tilde{f}(x)^p dx right)^{1/p} leq M+epsilon.
$$



But observe that
$$
int_{X} f(x)^{a+b} dx = int_{X} f(x)^{a}f(x)^b dx
$$
$$
leq int_{X} f(x)^{a} left(frac{M}{M+epsilon}right) ^b dx = left(frac{M}{M+epsilon}right)^b int_{X} f(x)^{a} dx.
$$

Therefore $int_{X} f(x)^{p} dx$ will eventually be less than one. This shows $displaystylelimsup_{ptoinfty}|f|_{p} leq M+epsilon$ and completes the proof.






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    2 Answers
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    2 Answers
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    active

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    112












    $begingroup$

    Fix $delta>0$ and let $S_delta:={x,|f(x)|geqslant lVert frVert_infty-delta}$ for $delta<lVert frVert_infty$. We have
    $$lVert frVert_pgeqslant left(int_{S_delta}(lVert frVert_infty-delta)^pdmuright)^{1/p}=(lVert frVert_infty-delta)mu(S_delta)^{1/p},$$
    since $mu(S_delta)$ is finite and positive.
    This gives
    $$liminf_{pto +infty}lVert frVert_pgeqslantlVert frVert_infty.$$
    As $|f(x)|leqslantlVert frVert_infty$ for almost every $x$, we have for $p>q$, $$
    lVert frVert_pleqslantleft(int_X|f(x)|^{p-q}|f(x)|^qdmuright)^{1/p}leqslant lVert frVert_infty^{frac{p-q}p}lVert frVert_q^{q/p},$$
    giving the reverse inequality.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      How does that last step give us the reverse inequality?
      $endgroup$
      – Parakee
      Nov 22 '12 at 20:27








    • 9




      $begingroup$
      Take this time $limsup_{pto infty}$.
      $endgroup$
      – Davide Giraudo
      Nov 22 '12 at 20:28






    • 1




      $begingroup$
      Assume that we work on a probability space. Then $lVert frVert_pleqslant lVert frVert_infty$, and we are interested about a control of $lVert frVert_infty$ in terms of the $L^p$ norm. We thus fix an arbitrary small real number and look at the set where $|f|$ is greater than the supremum norm minus the small number.
      $endgroup$
      – Davide Giraudo
      Jan 23 '14 at 21:40






    • 8




      $begingroup$
      Doesn't your proof assume $mu(X)<infty$?
      $endgroup$
      – Eric Auld
      Apr 22 '14 at 22:24






    • 6




      $begingroup$
      Yes. However, I think we can get rid of this assumption: we only have to consider the case where the measure space is $sigma$-finite, hence $X=bigcup_n A_n$ where $A_nuparrow X$ and each $A_n$ is of finite measure. Then we pick $n$ such that $mu(A_ncap S_{delta})$ is a positive real number.
      $endgroup$
      – Davide Giraudo
      Apr 23 '14 at 8:17
















    112












    $begingroup$

    Fix $delta>0$ and let $S_delta:={x,|f(x)|geqslant lVert frVert_infty-delta}$ for $delta<lVert frVert_infty$. We have
    $$lVert frVert_pgeqslant left(int_{S_delta}(lVert frVert_infty-delta)^pdmuright)^{1/p}=(lVert frVert_infty-delta)mu(S_delta)^{1/p},$$
    since $mu(S_delta)$ is finite and positive.
    This gives
    $$liminf_{pto +infty}lVert frVert_pgeqslantlVert frVert_infty.$$
    As $|f(x)|leqslantlVert frVert_infty$ for almost every $x$, we have for $p>q$, $$
    lVert frVert_pleqslantleft(int_X|f(x)|^{p-q}|f(x)|^qdmuright)^{1/p}leqslant lVert frVert_infty^{frac{p-q}p}lVert frVert_q^{q/p},$$
    giving the reverse inequality.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      How does that last step give us the reverse inequality?
      $endgroup$
      – Parakee
      Nov 22 '12 at 20:27








    • 9




      $begingroup$
      Take this time $limsup_{pto infty}$.
      $endgroup$
      – Davide Giraudo
      Nov 22 '12 at 20:28






    • 1




      $begingroup$
      Assume that we work on a probability space. Then $lVert frVert_pleqslant lVert frVert_infty$, and we are interested about a control of $lVert frVert_infty$ in terms of the $L^p$ norm. We thus fix an arbitrary small real number and look at the set where $|f|$ is greater than the supremum norm minus the small number.
      $endgroup$
      – Davide Giraudo
      Jan 23 '14 at 21:40






    • 8




      $begingroup$
      Doesn't your proof assume $mu(X)<infty$?
      $endgroup$
      – Eric Auld
      Apr 22 '14 at 22:24






    • 6




      $begingroup$
      Yes. However, I think we can get rid of this assumption: we only have to consider the case where the measure space is $sigma$-finite, hence $X=bigcup_n A_n$ where $A_nuparrow X$ and each $A_n$ is of finite measure. Then we pick $n$ such that $mu(A_ncap S_{delta})$ is a positive real number.
      $endgroup$
      – Davide Giraudo
      Apr 23 '14 at 8:17














    112












    112








    112





    $begingroup$

    Fix $delta>0$ and let $S_delta:={x,|f(x)|geqslant lVert frVert_infty-delta}$ for $delta<lVert frVert_infty$. We have
    $$lVert frVert_pgeqslant left(int_{S_delta}(lVert frVert_infty-delta)^pdmuright)^{1/p}=(lVert frVert_infty-delta)mu(S_delta)^{1/p},$$
    since $mu(S_delta)$ is finite and positive.
    This gives
    $$liminf_{pto +infty}lVert frVert_pgeqslantlVert frVert_infty.$$
    As $|f(x)|leqslantlVert frVert_infty$ for almost every $x$, we have for $p>q$, $$
    lVert frVert_pleqslantleft(int_X|f(x)|^{p-q}|f(x)|^qdmuright)^{1/p}leqslant lVert frVert_infty^{frac{p-q}p}lVert frVert_q^{q/p},$$
    giving the reverse inequality.






    share|cite|improve this answer











    $endgroup$



    Fix $delta>0$ and let $S_delta:={x,|f(x)|geqslant lVert frVert_infty-delta}$ for $delta<lVert frVert_infty$. We have
    $$lVert frVert_pgeqslant left(int_{S_delta}(lVert frVert_infty-delta)^pdmuright)^{1/p}=(lVert frVert_infty-delta)mu(S_delta)^{1/p},$$
    since $mu(S_delta)$ is finite and positive.
    This gives
    $$liminf_{pto +infty}lVert frVert_pgeqslantlVert frVert_infty.$$
    As $|f(x)|leqslantlVert frVert_infty$ for almost every $x$, we have for $p>q$, $$
    lVert frVert_pleqslantleft(int_X|f(x)|^{p-q}|f(x)|^qdmuright)^{1/p}leqslant lVert frVert_infty^{frac{p-q}p}lVert frVert_q^{q/p},$$
    giving the reverse inequality.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Apr 22 '13 at 18:08









    Community♦

    1




    1










    answered Nov 22 '12 at 20:21









    Davide GiraudoDavide Giraudo

    127k16154268




    127k16154268








    • 1




      $begingroup$
      How does that last step give us the reverse inequality?
      $endgroup$
      – Parakee
      Nov 22 '12 at 20:27








    • 9




      $begingroup$
      Take this time $limsup_{pto infty}$.
      $endgroup$
      – Davide Giraudo
      Nov 22 '12 at 20:28






    • 1




      $begingroup$
      Assume that we work on a probability space. Then $lVert frVert_pleqslant lVert frVert_infty$, and we are interested about a control of $lVert frVert_infty$ in terms of the $L^p$ norm. We thus fix an arbitrary small real number and look at the set where $|f|$ is greater than the supremum norm minus the small number.
      $endgroup$
      – Davide Giraudo
      Jan 23 '14 at 21:40






    • 8




      $begingroup$
      Doesn't your proof assume $mu(X)<infty$?
      $endgroup$
      – Eric Auld
      Apr 22 '14 at 22:24






    • 6




      $begingroup$
      Yes. However, I think we can get rid of this assumption: we only have to consider the case where the measure space is $sigma$-finite, hence $X=bigcup_n A_n$ where $A_nuparrow X$ and each $A_n$ is of finite measure. Then we pick $n$ such that $mu(A_ncap S_{delta})$ is a positive real number.
      $endgroup$
      – Davide Giraudo
      Apr 23 '14 at 8:17














    • 1




      $begingroup$
      How does that last step give us the reverse inequality?
      $endgroup$
      – Parakee
      Nov 22 '12 at 20:27








    • 9




      $begingroup$
      Take this time $limsup_{pto infty}$.
      $endgroup$
      – Davide Giraudo
      Nov 22 '12 at 20:28






    • 1




      $begingroup$
      Assume that we work on a probability space. Then $lVert frVert_pleqslant lVert frVert_infty$, and we are interested about a control of $lVert frVert_infty$ in terms of the $L^p$ norm. We thus fix an arbitrary small real number and look at the set where $|f|$ is greater than the supremum norm minus the small number.
      $endgroup$
      – Davide Giraudo
      Jan 23 '14 at 21:40






    • 8




      $begingroup$
      Doesn't your proof assume $mu(X)<infty$?
      $endgroup$
      – Eric Auld
      Apr 22 '14 at 22:24






    • 6




      $begingroup$
      Yes. However, I think we can get rid of this assumption: we only have to consider the case where the measure space is $sigma$-finite, hence $X=bigcup_n A_n$ where $A_nuparrow X$ and each $A_n$ is of finite measure. Then we pick $n$ such that $mu(A_ncap S_{delta})$ is a positive real number.
      $endgroup$
      – Davide Giraudo
      Apr 23 '14 at 8:17








    1




    1




    $begingroup$
    How does that last step give us the reverse inequality?
    $endgroup$
    – Parakee
    Nov 22 '12 at 20:27






    $begingroup$
    How does that last step give us the reverse inequality?
    $endgroup$
    – Parakee
    Nov 22 '12 at 20:27






    9




    9




    $begingroup$
    Take this time $limsup_{pto infty}$.
    $endgroup$
    – Davide Giraudo
    Nov 22 '12 at 20:28




    $begingroup$
    Take this time $limsup_{pto infty}$.
    $endgroup$
    – Davide Giraudo
    Nov 22 '12 at 20:28




    1




    1




    $begingroup$
    Assume that we work on a probability space. Then $lVert frVert_pleqslant lVert frVert_infty$, and we are interested about a control of $lVert frVert_infty$ in terms of the $L^p$ norm. We thus fix an arbitrary small real number and look at the set where $|f|$ is greater than the supremum norm minus the small number.
    $endgroup$
    – Davide Giraudo
    Jan 23 '14 at 21:40




    $begingroup$
    Assume that we work on a probability space. Then $lVert frVert_pleqslant lVert frVert_infty$, and we are interested about a control of $lVert frVert_infty$ in terms of the $L^p$ norm. We thus fix an arbitrary small real number and look at the set where $|f|$ is greater than the supremum norm minus the small number.
    $endgroup$
    – Davide Giraudo
    Jan 23 '14 at 21:40




    8




    8




    $begingroup$
    Doesn't your proof assume $mu(X)<infty$?
    $endgroup$
    – Eric Auld
    Apr 22 '14 at 22:24




    $begingroup$
    Doesn't your proof assume $mu(X)<infty$?
    $endgroup$
    – Eric Auld
    Apr 22 '14 at 22:24




    6




    6




    $begingroup$
    Yes. However, I think we can get rid of this assumption: we only have to consider the case where the measure space is $sigma$-finite, hence $X=bigcup_n A_n$ where $A_nuparrow X$ and each $A_n$ is of finite measure. Then we pick $n$ such that $mu(A_ncap S_{delta})$ is a positive real number.
    $endgroup$
    – Davide Giraudo
    Apr 23 '14 at 8:17




    $begingroup$
    Yes. However, I think we can get rid of this assumption: we only have to consider the case where the measure space is $sigma$-finite, hence $X=bigcup_n A_n$ where $A_nuparrow X$ and each $A_n$ is of finite measure. Then we pick $n$ such that $mu(A_ncap S_{delta})$ is a positive real number.
    $endgroup$
    – Davide Giraudo
    Apr 23 '14 at 8:17











    22












    $begingroup$

    Let $f:Xto mathbb{R}$. Assume that $f$ is measurable, and that $|f|_p<infty$ for all large $p$. Suppose for convenience that $fgeq 0$. (If not, just work with $f^*:=|f|$.) We define
    $$
    |f|_{infty}:=sup {rin mathbb{R}: muleft( {x:|f(x)|geq r} right)>0}.
    $$



    I claim without proof that $|f|_p < infty$ for large $p$ implies that $|f|_infty < infty$.



    If $|f|_{infty}=0$, we can see that the proposition holds trivially. If $|f|_{infty}neq 0$, let $M:=|f|_{infty}$.



    Fix $epsilon$ such that $0< epsilon < M$. Define $D:={x:f(x)geq M-epsilon}$. Observe that $mu(D)>0$ by definition of $|f|_{infty}$. Also, $mu(D)<infty$ since $f$ is integrable for all large $p$. Now we can establish $liminf_{ptoinfty }|f|_pgeq M-epsilon$ by
    $$
    left( int_{X}f(x)^p dx right)^{1/p} geq left( int_D (M-epsilon)^pdx right)^{1/p} = (M-epsilon)mu(D)^{1/p} xrightarrow{ptoinfty}(M-epsilon)
    $$



    Now we show $limsup_{ptoinfty}|f|_{p} leq M+epsilon$. Let $tilde{f}(x) := dfrac{f(x)}{M+epsilon}$. Observe that $0leq tilde{f}(x)leq M/(M+epsilon)<1$, and that
    $$
    left( int_{X} f(x)^p dx right)^{1/p} = (M+epsilon)left( int_{X} tilde{f}(x)^p dx right)^{1/p}.
    $$



    Now it suffices to show that $int_X tilde{f}(x)^p dx$ is bounded above by $1$ as $pto infty$, since then we have



    $$
    left( int_{X} f(x)^p dx right)^{1/p} = (M+epsilon)left( int_{X} tilde{f}(x)^p dx right)^{1/p} leq M+epsilon.
    $$



    But observe that
    $$
    int_{X} f(x)^{a+b} dx = int_{X} f(x)^{a}f(x)^b dx
    $$
    $$
    leq int_{X} f(x)^{a} left(frac{M}{M+epsilon}right) ^b dx = left(frac{M}{M+epsilon}right)^b int_{X} f(x)^{a} dx.
    $$

    Therefore $int_{X} f(x)^{p} dx$ will eventually be less than one. This shows $displaystylelimsup_{ptoinfty}|f|_{p} leq M+epsilon$ and completes the proof.






    share|cite|improve this answer











    $endgroup$


















      22












      $begingroup$

      Let $f:Xto mathbb{R}$. Assume that $f$ is measurable, and that $|f|_p<infty$ for all large $p$. Suppose for convenience that $fgeq 0$. (If not, just work with $f^*:=|f|$.) We define
      $$
      |f|_{infty}:=sup {rin mathbb{R}: muleft( {x:|f(x)|geq r} right)>0}.
      $$



      I claim without proof that $|f|_p < infty$ for large $p$ implies that $|f|_infty < infty$.



      If $|f|_{infty}=0$, we can see that the proposition holds trivially. If $|f|_{infty}neq 0$, let $M:=|f|_{infty}$.



      Fix $epsilon$ such that $0< epsilon < M$. Define $D:={x:f(x)geq M-epsilon}$. Observe that $mu(D)>0$ by definition of $|f|_{infty}$. Also, $mu(D)<infty$ since $f$ is integrable for all large $p$. Now we can establish $liminf_{ptoinfty }|f|_pgeq M-epsilon$ by
      $$
      left( int_{X}f(x)^p dx right)^{1/p} geq left( int_D (M-epsilon)^pdx right)^{1/p} = (M-epsilon)mu(D)^{1/p} xrightarrow{ptoinfty}(M-epsilon)
      $$



      Now we show $limsup_{ptoinfty}|f|_{p} leq M+epsilon$. Let $tilde{f}(x) := dfrac{f(x)}{M+epsilon}$. Observe that $0leq tilde{f}(x)leq M/(M+epsilon)<1$, and that
      $$
      left( int_{X} f(x)^p dx right)^{1/p} = (M+epsilon)left( int_{X} tilde{f}(x)^p dx right)^{1/p}.
      $$



      Now it suffices to show that $int_X tilde{f}(x)^p dx$ is bounded above by $1$ as $pto infty$, since then we have



      $$
      left( int_{X} f(x)^p dx right)^{1/p} = (M+epsilon)left( int_{X} tilde{f}(x)^p dx right)^{1/p} leq M+epsilon.
      $$



      But observe that
      $$
      int_{X} f(x)^{a+b} dx = int_{X} f(x)^{a}f(x)^b dx
      $$
      $$
      leq int_{X} f(x)^{a} left(frac{M}{M+epsilon}right) ^b dx = left(frac{M}{M+epsilon}right)^b int_{X} f(x)^{a} dx.
      $$

      Therefore $int_{X} f(x)^{p} dx$ will eventually be less than one. This shows $displaystylelimsup_{ptoinfty}|f|_{p} leq M+epsilon$ and completes the proof.






      share|cite|improve this answer











      $endgroup$
















        22












        22








        22





        $begingroup$

        Let $f:Xto mathbb{R}$. Assume that $f$ is measurable, and that $|f|_p<infty$ for all large $p$. Suppose for convenience that $fgeq 0$. (If not, just work with $f^*:=|f|$.) We define
        $$
        |f|_{infty}:=sup {rin mathbb{R}: muleft( {x:|f(x)|geq r} right)>0}.
        $$



        I claim without proof that $|f|_p < infty$ for large $p$ implies that $|f|_infty < infty$.



        If $|f|_{infty}=0$, we can see that the proposition holds trivially. If $|f|_{infty}neq 0$, let $M:=|f|_{infty}$.



        Fix $epsilon$ such that $0< epsilon < M$. Define $D:={x:f(x)geq M-epsilon}$. Observe that $mu(D)>0$ by definition of $|f|_{infty}$. Also, $mu(D)<infty$ since $f$ is integrable for all large $p$. Now we can establish $liminf_{ptoinfty }|f|_pgeq M-epsilon$ by
        $$
        left( int_{X}f(x)^p dx right)^{1/p} geq left( int_D (M-epsilon)^pdx right)^{1/p} = (M-epsilon)mu(D)^{1/p} xrightarrow{ptoinfty}(M-epsilon)
        $$



        Now we show $limsup_{ptoinfty}|f|_{p} leq M+epsilon$. Let $tilde{f}(x) := dfrac{f(x)}{M+epsilon}$. Observe that $0leq tilde{f}(x)leq M/(M+epsilon)<1$, and that
        $$
        left( int_{X} f(x)^p dx right)^{1/p} = (M+epsilon)left( int_{X} tilde{f}(x)^p dx right)^{1/p}.
        $$



        Now it suffices to show that $int_X tilde{f}(x)^p dx$ is bounded above by $1$ as $pto infty$, since then we have



        $$
        left( int_{X} f(x)^p dx right)^{1/p} = (M+epsilon)left( int_{X} tilde{f}(x)^p dx right)^{1/p} leq M+epsilon.
        $$



        But observe that
        $$
        int_{X} f(x)^{a+b} dx = int_{X} f(x)^{a}f(x)^b dx
        $$
        $$
        leq int_{X} f(x)^{a} left(frac{M}{M+epsilon}right) ^b dx = left(frac{M}{M+epsilon}right)^b int_{X} f(x)^{a} dx.
        $$

        Therefore $int_{X} f(x)^{p} dx$ will eventually be less than one. This shows $displaystylelimsup_{ptoinfty}|f|_{p} leq M+epsilon$ and completes the proof.






        share|cite|improve this answer











        $endgroup$



        Let $f:Xto mathbb{R}$. Assume that $f$ is measurable, and that $|f|_p<infty$ for all large $p$. Suppose for convenience that $fgeq 0$. (If not, just work with $f^*:=|f|$.) We define
        $$
        |f|_{infty}:=sup {rin mathbb{R}: muleft( {x:|f(x)|geq r} right)>0}.
        $$



        I claim without proof that $|f|_p < infty$ for large $p$ implies that $|f|_infty < infty$.



        If $|f|_{infty}=0$, we can see that the proposition holds trivially. If $|f|_{infty}neq 0$, let $M:=|f|_{infty}$.



        Fix $epsilon$ such that $0< epsilon < M$. Define $D:={x:f(x)geq M-epsilon}$. Observe that $mu(D)>0$ by definition of $|f|_{infty}$. Also, $mu(D)<infty$ since $f$ is integrable for all large $p$. Now we can establish $liminf_{ptoinfty }|f|_pgeq M-epsilon$ by
        $$
        left( int_{X}f(x)^p dx right)^{1/p} geq left( int_D (M-epsilon)^pdx right)^{1/p} = (M-epsilon)mu(D)^{1/p} xrightarrow{ptoinfty}(M-epsilon)
        $$



        Now we show $limsup_{ptoinfty}|f|_{p} leq M+epsilon$. Let $tilde{f}(x) := dfrac{f(x)}{M+epsilon}$. Observe that $0leq tilde{f}(x)leq M/(M+epsilon)<1$, and that
        $$
        left( int_{X} f(x)^p dx right)^{1/p} = (M+epsilon)left( int_{X} tilde{f}(x)^p dx right)^{1/p}.
        $$



        Now it suffices to show that $int_X tilde{f}(x)^p dx$ is bounded above by $1$ as $pto infty$, since then we have



        $$
        left( int_{X} f(x)^p dx right)^{1/p} = (M+epsilon)left( int_{X} tilde{f}(x)^p dx right)^{1/p} leq M+epsilon.
        $$



        But observe that
        $$
        int_{X} f(x)^{a+b} dx = int_{X} f(x)^{a}f(x)^b dx
        $$
        $$
        leq int_{X} f(x)^{a} left(frac{M}{M+epsilon}right) ^b dx = left(frac{M}{M+epsilon}right)^b int_{X} f(x)^{a} dx.
        $$

        Therefore $int_{X} f(x)^{p} dx$ will eventually be less than one. This shows $displaystylelimsup_{ptoinfty}|f|_{p} leq M+epsilon$ and completes the proof.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 2 at 17:02









        Namaste

        1




        1










        answered Apr 22 '14 at 23:12









        Eric AuldEric Auld

        13.2k432112




        13.2k432112






























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