Properties of Norm spaces
$begingroup$
Suppose $m(E)<infty$ and $finmathcal{L}^{infty}(E)$. The goal of this problem is to show
begin{align*}
lim_{pto infty}|f|_p=|f|_{infty}.
end{align*}
First, prove that
begin{align*}
lim_{ptoinfty}|f|_pleq |f|_{infty}.
end{align*}Proof:
Assume $m(E)<infty$ and $finmathcal{L}^{infty}(E)$. By definition,
begin{align*}
|f|_p=left(int_E|f|^pdmright)^{1/p}leq left(int_E|f|_{infty}^pright)^{1/p}=|f|_{infty}cdot m(E)^{1/p}<infty.
end{align*}
Letting $pto infty$, we have that $m(E)^{1/p}to 1$. Therefore,
begin{align*}
lim_{ptoinfty}|f|_pleq |f|_{infty}.
end{align*}
I'm not sure if I am missing anything here or need to justify being able to bring the limit inside and applying to $m(E)$.
Next, I need to prove that
begin{align*}
lim_{ptoinfty}|f|_pgeq |f|_{infty}-epsilon
end{align*}
for any $epsilon>0$. Hint: Look at the set
begin{align*}
F={xin E:|f|>|f|_{infty}-epsilon}.
end{align*}
I'm stuck on this part and would appreciate any help, thanks.
functional-analysis
asked Dec 4 '18 at 19:18
TNTTNT
596
$endgroup$
add a comment |
$begingroup$
Suppose $m(E)<infty$ and $finmathcal{L}^{infty}(E)$. The goal of this problem is to show
begin{align*}
lim_{pto infty}|f|_p=|f|_{infty}.
end{align*}
First, prove that
begin{align*}
lim_{ptoinfty}|f|_pleq |f|_{infty}.
end{align*}Proof:
Assume $m(E)<infty$ and $finmathcal{L}^{infty}(E)$. By definition,
begin{align*}
|f|_p=left(int_E|f|^pdmright)^{1/p}leq left(int_E|f|_{infty}^pright)^{1/p}=|f|_{infty}cdot m(E)^{1/p}<infty.
end{align*}
Letting $pto infty$, we have that $m(E)^{1/p}to 1$. Therefore,
begin{align*}
lim_{ptoinfty}|f|_pleq |f|_{infty}.
end{align*}
I'm not sure if I am missing anything here or need to justify being able to bring the limit inside and applying to $m(E)$.
Next, I need to prove that
begin{align*}
lim_{ptoinfty}|f|_pgeq |f|_{infty}-epsilon
end{align*}
for any $epsilon>0$. Hint: Look at the set
begin{align*}
F={xin E:|f|>|f|_{infty}-epsilon}.
end{align*}
I'm stuck on this part and would appreciate any help, thanks.
functional-analysis
asked Dec 4 '18 at 19:18
TNTTNT
596
$endgroup$
add a comment |
$begingroup$
Suppose $m(E)<infty$ and $finmathcal{L}^{infty}(E)$. The goal of this problem is to show
begin{align*}
lim_{pto infty}|f|_p=|f|_{infty}.
end{align*}
First, prove that
begin{align*}
lim_{ptoinfty}|f|_pleq |f|_{infty}.
end{align*}Proof:
Assume $m(E)<infty$ and $finmathcal{L}^{infty}(E)$. By definition,
begin{align*}
|f|_p=left(int_E|f|^pdmright)^{1/p}leq left(int_E|f|_{infty}^pright)^{1/p}=|f|_{infty}cdot m(E)^{1/p}<infty.
end{align*}
Letting $pto infty$, we have that $m(E)^{1/p}to 1$. Therefore,
begin{align*}
lim_{ptoinfty}|f|_pleq |f|_{infty}.
end{align*}
I'm not sure if I am missing anything here or need to justify being able to bring the limit inside and applying to $m(E)$.
Next, I need to prove that
begin{align*}
lim_{ptoinfty}|f|_pgeq |f|_{infty}-epsilon
end{align*}
for any $epsilon>0$. Hint: Look at the set
begin{align*}
F={xin E:|f|>|f|_{infty}-epsilon}.
end{align*}
I'm stuck on this part and would appreciate any help, thanks.
functional-analysis
asked Dec 4 '18 at 19:18
TNTTNT
596
$endgroup$
Suppose $m(E)<infty$ and $finmathcal{L}^{infty}(E)$. The goal of this problem is to show
begin{align*}
lim_{pto infty}|f|_p=|f|_{infty}.
end{align*}
First, prove that
begin{align*}
lim_{ptoinfty}|f|_pleq |f|_{infty}.
end{align*}Proof:
Assume $m(E)<infty$ and $finmathcal{L}^{infty}(E)$. By definition,
begin{align*}
|f|_p=left(int_E|f|^pdmright)^{1/p}leq left(int_E|f|_{infty}^pright)^{1/p}=|f|_{infty}cdot m(E)^{1/p}<infty.
end{align*}
Letting $pto infty$, we have that $m(E)^{1/p}to 1$. Therefore,
begin{align*}
lim_{ptoinfty}|f|_pleq |f|_{infty}.
end{align*}
I'm not sure if I am missing anything here or need to justify being able to bring the limit inside and applying to $m(E)$.
Next, I need to prove that
begin{align*}
lim_{ptoinfty}|f|_pgeq |f|_{infty}-epsilon
end{align*}
for any $epsilon>0$. Hint: Look at the set
begin{align*}
F={xin E:|f|>|f|_{infty}-epsilon}.
end{align*}
I'm stuck on this part and would appreciate any help, thanks.
functional-analysis
functional-analysis
asked Dec 4 '18 at 19:18
TNTTNT
596
asked Dec 4 '18 at 19:18
TNTTNT
596
asked Dec 4 '18 at 19:18
TNTTNT
596
asked Dec 4 '18 at 19:18
TNTTNT
596
asked Dec 4 '18 at 19:18
TNTTNT
596
596
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have that
$||f||_p^p=int_F|f|^p+int_{F^c}|f|^p>int_F|f|^p >$
$>int_F (||f||_infty -epsilon)^p=(||f||_infty -epsilon)^pm(F) $
Then
$||f||_p> (||f||_infty -epsilon)(m(F))^frac{1}{p}$
So if you fixed $epsilon$ (and $ F=F_{epsilon}$ ) you have that for $pto infty$
$lim_p||f||_p>||f||_infty -epsilon$
for every $epsilon>0$ so
$lim_p ||f||_p>||f||_infty$
answered Dec 4 '18 at 19:27
Federico FalluccaFederico Fallucca
2,185210
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You have that
$||f||_p^p=int_F|f|^p+int_{F^c}|f|^p>int_F|f|^p >$
$>int_F (||f||_infty -epsilon)^p=(||f||_infty -epsilon)^pm(F) $
Then
$||f||_p> (||f||_infty -epsilon)(m(F))^frac{1}{p}$
So if you fixed $epsilon$ (and $ F=F_{epsilon}$ ) you have that for $pto infty$
$lim_p||f||_p>||f||_infty -epsilon$
for every $epsilon>0$ so
$lim_p ||f||_p>||f||_infty$
answered Dec 4 '18 at 19:27
Federico FalluccaFederico Fallucca
2,185210
$endgroup$
add a comment |
$begingroup$
You have that
$||f||_p^p=int_F|f|^p+int_{F^c}|f|^p>int_F|f|^p >$
$>int_F (||f||_infty -epsilon)^p=(||f||_infty -epsilon)^pm(F) $
Then
$||f||_p> (||f||_infty -epsilon)(m(F))^frac{1}{p}$
So if you fixed $epsilon$ (and $ F=F_{epsilon}$ ) you have that for $pto infty$
$lim_p||f||_p>||f||_infty -epsilon$
for every $epsilon>0$ so
$lim_p ||f||_p>||f||_infty$
answered Dec 4 '18 at 19:27
Federico FalluccaFederico Fallucca
2,185210
$endgroup$
add a comment |
$begingroup$
You have that
$||f||_p^p=int_F|f|^p+int_{F^c}|f|^p>int_F|f|^p >$
$>int_F (||f||_infty -epsilon)^p=(||f||_infty -epsilon)^pm(F) $
Then
$||f||_p> (||f||_infty -epsilon)(m(F))^frac{1}{p}$
So if you fixed $epsilon$ (and $ F=F_{epsilon}$ ) you have that for $pto infty$
$lim_p||f||_p>||f||_infty -epsilon$
for every $epsilon>0$ so
$lim_p ||f||_p>||f||_infty$
answered Dec 4 '18 at 19:27
Federico FalluccaFederico Fallucca
2,185210
$endgroup$
You have that
$||f||_p^p=int_F|f|^p+int_{F^c}|f|^p>int_F|f|^p >$
$>int_F (||f||_infty -epsilon)^p=(||f||_infty -epsilon)^pm(F) $
Then
$||f||_p> (||f||_infty -epsilon)(m(F))^frac{1}{p}$
So if you fixed $epsilon$ (and $ F=F_{epsilon}$ ) you have that for $pto infty$
$lim_p||f||_p>||f||_infty -epsilon$
for every $epsilon>0$ so
$lim_p ||f||_p>||f||_infty$
answered Dec 4 '18 at 19:27
Federico FalluccaFederico Fallucca
2,185210
edited Dec 4 '18 at 19:38
answered Dec 4 '18 at 19:27
Federico FalluccaFederico Fallucca
2,185210
answered Dec 4 '18 at 19:27
Federico FalluccaFederico Fallucca
2,185210
answered Dec 4 '18 at 19:27
Federico FalluccaFederico Fallucca
2,185210
2,185210
add a comment |
add a comment |
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