Mixing
$ p geq q geq 1$, then $L^p_{[a,b]} subset L^q_{[a,b]}$
$ p geq q geq 1$, then $L^p_{[a,b]} subset L^q_{[a,b]}$
I want to prove it with Holder inequality for integrals. However I am not sure how to proceed. This is what I did:
Let $f in L^p_{[a,b]}$.
Then I can write:
$$ int_a^b vert f(x) vert^p dx < infty $$
$$ int_a^b vert f(x) vert^p dx = int_a^b vert f(x) vert^{p-q+q} dx $$
$$ int_a^b vert f(x) vert^{p+q} vert f(x) vert^{-q}dx $$
But from here I do not know how to proceed. I am not even sure this is the right way to approach the problem.
EDIT: As suggested in the comments I tried to prove that $frac{1}{p} = frac{1}{q} + frac{1}{r}$ and Holder inequality gives: $Vert fg Vert _p leq Vert f Vert_q Vert g Vert_r$
By holder inequality and the equality that I have written before I can write:
$$ Vert fg Vert_{frac{1}{p}} leq Vert f Vert_{frac{1}{q}} Vert g Vert_{frac{1}{p}} $$
Then the inequality follows. However I am struggling to see how I can write $Vert f Vert_q$ less than something that converges.
functional-analysis
asked Nov 20 '18 at 21:21
qcc101
458113
add a comment |
$ p geq q geq 1$, then $L^p_{[a,b]} subset L^q_{[a,b]}$
I want to prove it with Holder inequality for integrals. However I am not sure how to proceed. This is what I did:
Let $f in L^p_{[a,b]}$.
Then I can write:
$$ int_a^b vert f(x) vert^p dx < infty $$
$$ int_a^b vert f(x) vert^p dx = int_a^b vert f(x) vert^{p-q+q} dx $$
$$ int_a^b vert f(x) vert^{p+q} vert f(x) vert^{-q}dx $$
But from here I do not know how to proceed. I am not even sure this is the right way to approach the problem.
EDIT: As suggested in the comments I tried to prove that $frac{1}{p} = frac{1}{q} + frac{1}{r}$ and Holder inequality gives: $Vert fg Vert _p leq Vert f Vert_q Vert g Vert_r$
By holder inequality and the equality that I have written before I can write:
$$ Vert fg Vert_{frac{1}{p}} leq Vert f Vert_{frac{1}{q}} Vert g Vert_{frac{1}{p}} $$
Then the inequality follows. However I am struggling to see how I can write $Vert f Vert_q$ less than something that converges.
functional-analysis
asked Nov 20 '18 at 21:21
qcc101
458113
Essentially this: math.stackexchange.com/q/66029/587192
– user587192
Nov 20 '18 at 21:41
I saw that, but I want to prove it using strictly things from functional analysis.
– qcc101
Nov 21 '18 at 6:53
1
@qcc101 Does Holder inequality counts as functional analysis?
– Jacky Chong
Nov 21 '18 at 6:55
Yes, but the link by @user587192 has a flavor of measure theory and I want something strictly analytical if that makes sense.
– qcc101
Nov 21 '18 at 6:56
Do you mean you are not comfortable with abstract measure theory? Because you won't be able to avoid measure theory completely - that's just how the $L^p$ spaces are defined. And the arguments given the answers are exactly the same as for the linked question.
– MaoWao
Nov 21 '18 at 13:22
add a comment |
$ p geq q geq 1$, then $L^p_{[a,b]} subset L^q_{[a,b]}$
I want to prove it with Holder inequality for integrals. However I am not sure how to proceed. This is what I did:
Let $f in L^p_{[a,b]}$.
Then I can write:
$$ int_a^b vert f(x) vert^p dx < infty $$
$$ int_a^b vert f(x) vert^p dx = int_a^b vert f(x) vert^{p-q+q} dx $$
$$ int_a^b vert f(x) vert^{p+q} vert f(x) vert^{-q}dx $$
But from here I do not know how to proceed. I am not even sure this is the right way to approach the problem.
EDIT: As suggested in the comments I tried to prove that $frac{1}{p} = frac{1}{q} + frac{1}{r}$ and Holder inequality gives: $Vert fg Vert _p leq Vert f Vert_q Vert g Vert_r$
By holder inequality and the equality that I have written before I can write:
$$ Vert fg Vert_{frac{1}{p}} leq Vert f Vert_{frac{1}{q}} Vert g Vert_{frac{1}{p}} $$
Then the inequality follows. However I am struggling to see how I can write $Vert f Vert_q$ less than something that converges.
functional-analysis
asked Nov 20 '18 at 21:21
qcc101
458113
$ p geq q geq 1$, then $L^p_{[a,b]} subset L^q_{[a,b]}$
I want to prove it with Holder inequality for integrals. However I am not sure how to proceed. This is what I did:
Let $f in L^p_{[a,b]}$.
Then I can write:
$$ int_a^b vert f(x) vert^p dx < infty $$
$$ int_a^b vert f(x) vert^p dx = int_a^b vert f(x) vert^{p-q+q} dx $$
$$ int_a^b vert f(x) vert^{p+q} vert f(x) vert^{-q}dx $$
But from here I do not know how to proceed. I am not even sure this is the right way to approach the problem.
EDIT: As suggested in the comments I tried to prove that $frac{1}{p} = frac{1}{q} + frac{1}{r}$ and Holder inequality gives: $Vert fg Vert _p leq Vert f Vert_q Vert g Vert_r$
By holder inequality and the equality that I have written before I can write:
$$ Vert fg Vert_{frac{1}{p}} leq Vert f Vert_{frac{1}{q}} Vert g Vert_{frac{1}{p}} $$
Then the inequality follows. However I am struggling to see how I can write $Vert f Vert_q$ less than something that converges.
functional-analysis
functional-analysis
asked Nov 20 '18 at 21:21
qcc101
458113
asked Nov 20 '18 at 21:21
qcc101
458113
edited Nov 21 '18 at 7:04
asked Nov 20 '18 at 21:21
qcc101
458113
asked Nov 20 '18 at 21:21
qcc101
458113
asked Nov 20 '18 at 21:21
qcc101
458113
458113
Essentially this: math.stackexchange.com/q/66029/587192
– user587192
Nov 20 '18 at 21:41
I saw that, but I want to prove it using strictly things from functional analysis.
– qcc101
Nov 21 '18 at 6:53
1
@qcc101 Does Holder inequality counts as functional analysis?
– Jacky Chong
Nov 21 '18 at 6:55
Yes, but the link by @user587192 has a flavor of measure theory and I want something strictly analytical if that makes sense.
– qcc101
Nov 21 '18 at 6:56
Do you mean you are not comfortable with abstract measure theory? Because you won't be able to avoid measure theory completely - that's just how the $L^p$ spaces are defined. And the arguments given the answers are exactly the same as for the linked question.
– MaoWao
Nov 21 '18 at 13:22
add a comment |
Essentially this: math.stackexchange.com/q/66029/587192
– user587192
Nov 20 '18 at 21:41
I saw that, but I want to prove it using strictly things from functional analysis.
– qcc101
Nov 21 '18 at 6:53
1
@qcc101 Does Holder inequality counts as functional analysis?
– Jacky Chong
Nov 21 '18 at 6:55
Yes, but the link by @user587192 has a flavor of measure theory and I want something strictly analytical if that makes sense.
– qcc101
Nov 21 '18 at 6:56
Do you mean you are not comfortable with abstract measure theory? Because you won't be able to avoid measure theory completely - that's just how the $L^p$ spaces are defined. And the arguments given the answers are exactly the same as for the linked question.
– MaoWao
Nov 21 '18 at 13:22
Essentially this: math.stackexchange.com/q/66029/587192
– user587192
Nov 20 '18 at 21:41
Essentially this: math.stackexchange.com/q/66029/587192
– user587192
Nov 20 '18 at 21:41
I saw that, but I want to prove it using strictly things from functional analysis.
– qcc101
Nov 21 '18 at 6:53
I saw that, but I want to prove it using strictly things from functional analysis.
– qcc101
Nov 21 '18 at 6:53
1
1
@qcc101 Does Holder inequality counts as functional analysis?
– Jacky Chong
Nov 21 '18 at 6:55
@qcc101 Does Holder inequality counts as functional analysis?
– Jacky Chong
Nov 21 '18 at 6:55
Yes, but the link by @user587192 has a flavor of measure theory and I want something strictly analytical if that makes sense.
– qcc101
Nov 21 '18 at 6:56
Yes, but the link by @user587192 has a flavor of measure theory and I want something strictly analytical if that makes sense.
– qcc101
Nov 21 '18 at 6:56
Do you mean you are not comfortable with abstract measure theory? Because you won't be able to avoid measure theory completely - that's just how the $L^p$ spaces are defined. And the arguments given the answers are exactly the same as for the linked question.
– MaoWao
Nov 21 '18 at 13:22
Do you mean you are not comfortable with abstract measure theory? Because you won't be able to avoid measure theory completely - that's just how the $L^p$ spaces are defined. And the arguments given the answers are exactly the same as for the linked question.
– MaoWao
Nov 21 '18 at 13:22
add a comment |
2 Answers
2
active
oldest
votes
Using Holder's inequality one has
$$
|f|_q^q=int_a^b |f(x)|^qcdot 1 dxle
bigVert|f|^qbigVert_{p/q}|1|_{p/(p-q)}=Vert fVert_p^q(b-a)^{(p-q)/p}
$$
answered Nov 21 '18 at 13:08
user587192
1,782215
add a comment |
Hint: Prove the following: If $p, q, rgeq 1$ and
begin{align}
frac{1}{p}=frac{1}{q}+frac{1}{r}
end{align}
then
begin{align}
|fg|_p leq |f|_q|g|_r.
end{align}
Additional Hint: The proof of the above hint uses Holder inequality. Once you are done proving the hint set $g=ldots$. I will let you figure this out.
answered Nov 20 '18 at 21:29
Jacky Chong
17.8k21128
I do not see how from that inequality I have $Vert f Vert_q$ less than something that converges. Can you shed some light?
– qcc101
Nov 21 '18 at 6:52
At first I tried putting $g = g/f$ but I don't think it works
– qcc101
Nov 21 '18 at 9:16
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Using Holder's inequality one has
$$
|f|_q^q=int_a^b |f(x)|^qcdot 1 dxle
bigVert|f|^qbigVert_{p/q}|1|_{p/(p-q)}=Vert fVert_p^q(b-a)^{(p-q)/p}
$$
answered Nov 21 '18 at 13:08
user587192
1,782215
add a comment |
Using Holder's inequality one has
$$
|f|_q^q=int_a^b |f(x)|^qcdot 1 dxle
bigVert|f|^qbigVert_{p/q}|1|_{p/(p-q)}=Vert fVert_p^q(b-a)^{(p-q)/p}
$$
answered Nov 21 '18 at 13:08
user587192
1,782215
add a comment |
Using Holder's inequality one has
$$
|f|_q^q=int_a^b |f(x)|^qcdot 1 dxle
bigVert|f|^qbigVert_{p/q}|1|_{p/(p-q)}=Vert fVert_p^q(b-a)^{(p-q)/p}
$$
answered Nov 21 '18 at 13:08
user587192
1,782215
Using Holder's inequality one has
$$
|f|_q^q=int_a^b |f(x)|^qcdot 1 dxle
bigVert|f|^qbigVert_{p/q}|1|_{p/(p-q)}=Vert fVert_p^q(b-a)^{(p-q)/p}
$$
answered Nov 21 '18 at 13:08
user587192
1,782215
answered Nov 21 '18 at 13:08
user587192
1,782215
answered Nov 21 '18 at 13:08
user587192
1,782215
answered Nov 21 '18 at 13:08
user587192
1,782215
1,782215
add a comment |
add a comment |
Hint: Prove the following: If $p, q, rgeq 1$ and
begin{align}
frac{1}{p}=frac{1}{q}+frac{1}{r}
end{align}
then
begin{align}
|fg|_p leq |f|_q|g|_r.
end{align}
Additional Hint: The proof of the above hint uses Holder inequality. Once you are done proving the hint set $g=ldots$. I will let you figure this out.
answered Nov 20 '18 at 21:29
Jacky Chong
17.8k21128
I do not see how from that inequality I have $Vert f Vert_q$ less than something that converges. Can you shed some light?
– qcc101
Nov 21 '18 at 6:52
At first I tried putting $g = g/f$ but I don't think it works
– qcc101
Nov 21 '18 at 9:16
add a comment |
Hint: Prove the following: If $p, q, rgeq 1$ and
begin{align}
frac{1}{p}=frac{1}{q}+frac{1}{r}
end{align}
then
begin{align}
|fg|_p leq |f|_q|g|_r.
end{align}
Additional Hint: The proof of the above hint uses Holder inequality. Once you are done proving the hint set $g=ldots$. I will let you figure this out.
answered Nov 20 '18 at 21:29
Jacky Chong
17.8k21128
I do not see how from that inequality I have $Vert f Vert_q$ less than something that converges. Can you shed some light?
– qcc101
Nov 21 '18 at 6:52
At first I tried putting $g = g/f$ but I don't think it works
– qcc101
Nov 21 '18 at 9:16
add a comment |
Hint: Prove the following: If $p, q, rgeq 1$ and
begin{align}
frac{1}{p}=frac{1}{q}+frac{1}{r}
end{align}
then
begin{align}
|fg|_p leq |f|_q|g|_r.
end{align}
Additional Hint: The proof of the above hint uses Holder inequality. Once you are done proving the hint set $g=ldots$. I will let you figure this out.
answered Nov 20 '18 at 21:29
Jacky Chong
17.8k21128
Hint: Prove the following: If $p, q, rgeq 1$ and
begin{align}
frac{1}{p}=frac{1}{q}+frac{1}{r}
end{align}
then
begin{align}
|fg|_p leq |f|_q|g|_r.
end{align}
Additional Hint: The proof of the above hint uses Holder inequality. Once you are done proving the hint set $g=ldots$. I will let you figure this out.
answered Nov 20 '18 at 21:29
Jacky Chong
17.8k21128
answered Nov 20 '18 at 21:29
Jacky Chong
17.8k21128
answered Nov 20 '18 at 21:29
Jacky Chong
17.8k21128
answered Nov 20 '18 at 21:29
Jacky Chong
17.8k21128
17.8k21128
I do not see how from that inequality I have $Vert f Vert_q$ less than something that converges. Can you shed some light?
– qcc101
Nov 21 '18 at 6:52
At first I tried putting $g = g/f$ but I don't think it works
– qcc101
Nov 21 '18 at 9:16
add a comment |
I do not see how from that inequality I have $Vert f Vert_q$ less than something that converges. Can you shed some light?
– qcc101
Nov 21 '18 at 6:52
At first I tried putting $g = g/f$ but I don't think it works
– qcc101
Nov 21 '18 at 9:16
I do not see how from that inequality I have $Vert f Vert_q$ less than something that converges. Can you shed some light?
– qcc101
Nov 21 '18 at 6:52
I do not see how from that inequality I have $Vert f Vert_q$ less than something that converges. Can you shed some light?
– qcc101
Nov 21 '18 at 6:52
At first I tried putting $g = g/f$ but I don't think it works
– qcc101
Nov 21 '18 at 9:16
At first I tried putting $g = g/f$ but I don't think it works
– qcc101
Nov 21 '18 at 9:16
add a comment |
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Essentially this: math.stackexchange.com/q/66029/587192
– user587192
Nov 20 '18 at 21:41
I saw that, but I want to prove it using strictly things from functional analysis.
– qcc101
Nov 21 '18 at 6:53
1
@qcc101 Does Holder inequality counts as functional analysis?
– Jacky Chong
Nov 21 '18 at 6:55
Yes, but the link by @user587192 has a flavor of measure theory and I want something strictly analytical if that makes sense.
– qcc101
Nov 21 '18 at 6:56
Do you mean you are not comfortable with abstract measure theory? Because you won't be able to avoid measure theory completely - that's just how the $L^p$ spaces are defined. And the arguments given the answers are exactly the same as for the linked question.
– MaoWao
Nov 21 '18 at 13:22