Limit of $L^p$ norm
$begingroup$
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
edited Jan 2 at 16:54
klirk
1
asked Nov 22 '12 at 19:51
ParakeeParakee
1,04321022
$endgroup$
add a comment |
$begingroup$
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
edited Jan 2 at 16:54
klirk
1
asked Nov 22 '12 at 19:51
ParakeeParakee
1,04321022
$endgroup$
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
add a comment |
$begingroup$
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
edited Jan 2 at 16:54
klirk
1
asked Nov 22 '12 at 19:51
ParakeeParakee
1,04321022
$endgroup$
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
edited Jan 2 at 16:54
klirk
1
asked Nov 22 '12 at 19:51
ParakeeParakee
1,04321022
edited Jan 2 at 16:54
klirk
1
asked Nov 22 '12 at 19:51
ParakeeParakee
1,04321022
edited Jan 2 at 16:54
klirk
1
edited Jan 2 at 16:54
klirk
1
edited Jan 2 at 16:54
klirk
1
1
asked Nov 22 '12 at 19:51
ParakeeParakee
1,04321022
asked Nov 22 '12 at 19:51
ParakeeParakee
1,04321022
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
edited Apr 22 '13 at 18:08
Community♦
1
answered Nov 22 '12 at 20:21
Davide GiraudoDavide Giraudo
127k16154268
$endgroup$
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
|
show 13 more comments
$begingroup$
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.
edited Jan 2 at 17:02
Namaste
1
answered Apr 22 '14 at 23:12
Eric AuldEric Auld
13.2k432112
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
edited Apr 22 '13 at 18:08
Community♦
1
answered Nov 22 '12 at 20:21
Davide GiraudoDavide Giraudo
127k16154268
$endgroup$
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
|
show 13 more comments
$begingroup$
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.
edited Apr 22 '13 at 18:08
Community♦
1
answered Nov 22 '12 at 20:21
Davide GiraudoDavide Giraudo
127k16154268
$endgroup$
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
|
show 13 more comments
$begingroup$
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.
edited Apr 22 '13 at 18:08
Community♦
1
answered Nov 22 '12 at 20:21
Davide GiraudoDavide Giraudo
127k16154268
$endgroup$
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.
edited Apr 22 '13 at 18:08
Community♦
1
answered Nov 22 '12 at 20:21
Davide GiraudoDavide Giraudo
127k16154268
edited Apr 22 '13 at 18:08
Community♦
1
edited Apr 22 '13 at 18:08
Community♦
1
edited Apr 22 '13 at 18:08
Community♦
1
1
answered Nov 22 '12 at 20:21
Davide GiraudoDavide Giraudo
127k16154268
answered Nov 22 '12 at 20:21
Davide GiraudoDavide Giraudo
127k16154268
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
|
show 13 more comments
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
|
show 13 more comments
$begingroup$
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.
edited Jan 2 at 17:02
Namaste
1
answered Apr 22 '14 at 23:12
Eric AuldEric Auld
13.2k432112
$endgroup$
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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.
edited Jan 2 at 17:02
Namaste
1
answered Apr 22 '14 at 23:12
Eric AuldEric Auld
13.2k432112
$endgroup$
add a comment |
$begingroup$
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.
edited Jan 2 at 17:02
Namaste
1
answered Apr 22 '14 at 23:12
Eric AuldEric Auld
13.2k432112
$endgroup$
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.
edited Jan 2 at 17:02
Namaste
1
answered Apr 22 '14 at 23:12
Eric AuldEric Auld
13.2k432112
edited Jan 2 at 17:02
Namaste
1
edited Jan 2 at 17:02
Namaste
1
edited Jan 2 at 17:02
Namaste
1
1
answered Apr 22 '14 at 23:12
Eric AuldEric Auld
13.2k432112
answered Apr 22 '14 at 23:12
Eric AuldEric Auld
13.2k432112
answered Apr 22 '14 at 23:12
Eric AuldEric Auld
13.2k432112
13.2k432112
add a comment |
add a comment |
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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.
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– Ryan
Nov 22 '12 at 20:10
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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.
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– Parakee
Nov 22 '12 at 20:19