Mixing
Dual space of $L^p(Omega,mathcal{A},mu,mathbb{R}^d)$.
$begingroup$
I want to show that for $pin(1,+infty)$ the dual space of $L^p(Omega,mathcal{A},mu,mathbb{R}^d)$ is isometrically isomorphic to $L^q(Omega,mathcal{A},mu,mathbb{R}^d)$, where $frac{1}{p}+frac{1}{q}=1$.
We have already shown the analogous result for $mathbb{R}$ instead of $mathbb{R}^n$ and therefore considered the bounded linear functional
begin{align*}
phi_g:fmapstoint_Omega fg,dmu
end{align*}
for every $gin L^q(Omega,mathcal{A},mu,mathbb{R})$, where $fin L^p(Omega,mathcal{A},mu,mathbb{R})$. We then proved that the mapping $gmapstophi_g$ is an isometric isomorphism.
After several attempts, I have still not even figured out how to start, as I need to find an analogous functional $phi_g$. I can not use the previous one with $fg$ considered to be a component-wise product, since the result would not be a scalar anymore, right? I have also thought about
begin{align*}
phi_g:fmapstoint_Omega langle f(x),g(x)rangle,dmu(x),
end{align*}
where $langle.,.rangle$ is the standard scalar product in $mathbb{R}^d$. Is that an option?
Thanks in advance!
Edit: On $L^p(Omega,mathcal{A},mu,mathbb{R}^d)$ we define the norm
begin{align*}
|f|:=left(int_Omega|f(x)|_p^p,dmu(x)right)^{frac{1}{p}},
end{align*}
where $|.|_p$ is the $p$-norm on $mathbb{R}^d$ and obtain a Banach space (I've already shown that).
functional-analysis measure-theory lp-spaces dual-spaces
asked Jan 6 at 11:56
Lemma 5Lemma 5
63
$endgroup$
add a comment |
$begingroup$
I want to show that for $pin(1,+infty)$ the dual space of $L^p(Omega,mathcal{A},mu,mathbb{R}^d)$ is isometrically isomorphic to $L^q(Omega,mathcal{A},mu,mathbb{R}^d)$, where $frac{1}{p}+frac{1}{q}=1$.
We have already shown the analogous result for $mathbb{R}$ instead of $mathbb{R}^n$ and therefore considered the bounded linear functional
begin{align*}
phi_g:fmapstoint_Omega fg,dmu
end{align*}
for every $gin L^q(Omega,mathcal{A},mu,mathbb{R})$, where $fin L^p(Omega,mathcal{A},mu,mathbb{R})$. We then proved that the mapping $gmapstophi_g$ is an isometric isomorphism.
After several attempts, I have still not even figured out how to start, as I need to find an analogous functional $phi_g$. I can not use the previous one with $fg$ considered to be a component-wise product, since the result would not be a scalar anymore, right? I have also thought about
begin{align*}
phi_g:fmapstoint_Omega langle f(x),g(x)rangle,dmu(x),
end{align*}
where $langle.,.rangle$ is the standard scalar product in $mathbb{R}^d$. Is that an option?
Thanks in advance!
Edit: On $L^p(Omega,mathcal{A},mu,mathbb{R}^d)$ we define the norm
begin{align*}
|f|:=left(int_Omega|f(x)|_p^p,dmu(x)right)^{frac{1}{p}},
end{align*}
where $|.|_p$ is the $p$-norm on $mathbb{R}^d$ and obtain a Banach space (I've already shown that).
functional-analysis measure-theory lp-spaces dual-spaces
asked Jan 6 at 11:56
Lemma 5Lemma 5
63
$endgroup$
$begingroup$
Yes. That is the correct thing to do.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 11:59
$begingroup$
Good to hear. I would then continue showing that $|phi_g|leq|g|$, where $|phi_g|$ is the operator norm of $phi_g$. Here is my attempt: begin{align*} left|intlangle f,grangle,dmuright| &leq sum_{i=1}^dint |f_ig_i|,dmu \ &leq sum_{i=1}^dleft(int|f_i|^pright)^{frac{1}{p}}left(int|g_i|^qright)^{frac{1}{q}} \ &stackrel{(?)}{leq} left(intsum_{i=1}^d|f_i|^pright)^{frac{1}{p}}left(intsum_{i=1}^d|g_i|^qright)^{frac{1}{q}}. end{align*} As tagged, I am not sure how to reason the last inequality or if it is even correct.
$endgroup$
– Lemma 5
Jan 6 at 13:16
add a comment |
$begingroup$
I want to show that for $pin(1,+infty)$ the dual space of $L^p(Omega,mathcal{A},mu,mathbb{R}^d)$ is isometrically isomorphic to $L^q(Omega,mathcal{A},mu,mathbb{R}^d)$, where $frac{1}{p}+frac{1}{q}=1$.
We have already shown the analogous result for $mathbb{R}$ instead of $mathbb{R}^n$ and therefore considered the bounded linear functional
begin{align*}
phi_g:fmapstoint_Omega fg,dmu
end{align*}
for every $gin L^q(Omega,mathcal{A},mu,mathbb{R})$, where $fin L^p(Omega,mathcal{A},mu,mathbb{R})$. We then proved that the mapping $gmapstophi_g$ is an isometric isomorphism.
After several attempts, I have still not even figured out how to start, as I need to find an analogous functional $phi_g$. I can not use the previous one with $fg$ considered to be a component-wise product, since the result would not be a scalar anymore, right? I have also thought about
begin{align*}
phi_g:fmapstoint_Omega langle f(x),g(x)rangle,dmu(x),
end{align*}
where $langle.,.rangle$ is the standard scalar product in $mathbb{R}^d$. Is that an option?
Thanks in advance!
Edit: On $L^p(Omega,mathcal{A},mu,mathbb{R}^d)$ we define the norm
begin{align*}
|f|:=left(int_Omega|f(x)|_p^p,dmu(x)right)^{frac{1}{p}},
end{align*}
where $|.|_p$ is the $p$-norm on $mathbb{R}^d$ and obtain a Banach space (I've already shown that).
functional-analysis measure-theory lp-spaces dual-spaces
asked Jan 6 at 11:56
Lemma 5Lemma 5
63
$endgroup$
I want to show that for $pin(1,+infty)$ the dual space of $L^p(Omega,mathcal{A},mu,mathbb{R}^d)$ is isometrically isomorphic to $L^q(Omega,mathcal{A},mu,mathbb{R}^d)$, where $frac{1}{p}+frac{1}{q}=1$.
We have already shown the analogous result for $mathbb{R}$ instead of $mathbb{R}^n$ and therefore considered the bounded linear functional
begin{align*}
phi_g:fmapstoint_Omega fg,dmu
end{align*}
for every $gin L^q(Omega,mathcal{A},mu,mathbb{R})$, where $fin L^p(Omega,mathcal{A},mu,mathbb{R})$. We then proved that the mapping $gmapstophi_g$ is an isometric isomorphism.
After several attempts, I have still not even figured out how to start, as I need to find an analogous functional $phi_g$. I can not use the previous one with $fg$ considered to be a component-wise product, since the result would not be a scalar anymore, right? I have also thought about
begin{align*}
phi_g:fmapstoint_Omega langle f(x),g(x)rangle,dmu(x),
end{align*}
where $langle.,.rangle$ is the standard scalar product in $mathbb{R}^d$. Is that an option?
Thanks in advance!
Edit: On $L^p(Omega,mathcal{A},mu,mathbb{R}^d)$ we define the norm
begin{align*}
|f|:=left(int_Omega|f(x)|_p^p,dmu(x)right)^{frac{1}{p}},
end{align*}
where $|.|_p$ is the $p$-norm on $mathbb{R}^d$ and obtain a Banach space (I've already shown that).
functional-analysis measure-theory lp-spaces dual-spaces
functional-analysis measure-theory lp-spaces dual-spaces
asked Jan 6 at 11:56
Lemma 5Lemma 5
63
asked Jan 6 at 11:56
Lemma 5Lemma 5
63
edited Jan 6 at 12:54
Lemma 5
asked Jan 6 at 11:56
Lemma 5Lemma 5
63
asked Jan 6 at 11:56
Lemma 5Lemma 5
63
asked Jan 6 at 11:56
Lemma 5Lemma 5
63
63
$begingroup$
Yes. That is the correct thing to do.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 11:59
$begingroup$
Good to hear. I would then continue showing that $|phi_g|leq|g|$, where $|phi_g|$ is the operator norm of $phi_g$. Here is my attempt: begin{align*} left|intlangle f,grangle,dmuright| &leq sum_{i=1}^dint |f_ig_i|,dmu \ &leq sum_{i=1}^dleft(int|f_i|^pright)^{frac{1}{p}}left(int|g_i|^qright)^{frac{1}{q}} \ &stackrel{(?)}{leq} left(intsum_{i=1}^d|f_i|^pright)^{frac{1}{p}}left(intsum_{i=1}^d|g_i|^qright)^{frac{1}{q}}. end{align*} As tagged, I am not sure how to reason the last inequality or if it is even correct.
$endgroup$
– Lemma 5
Jan 6 at 13:16
add a comment |
$begingroup$
Yes. That is the correct thing to do.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 11:59
$begingroup$
Good to hear. I would then continue showing that $|phi_g|leq|g|$, where $|phi_g|$ is the operator norm of $phi_g$. Here is my attempt: begin{align*} left|intlangle f,grangle,dmuright| &leq sum_{i=1}^dint |f_ig_i|,dmu \ &leq sum_{i=1}^dleft(int|f_i|^pright)^{frac{1}{p}}left(int|g_i|^qright)^{frac{1}{q}} \ &stackrel{(?)}{leq} left(intsum_{i=1}^d|f_i|^pright)^{frac{1}{p}}left(intsum_{i=1}^d|g_i|^qright)^{frac{1}{q}}. end{align*} As tagged, I am not sure how to reason the last inequality or if it is even correct.
$endgroup$
– Lemma 5
Jan 6 at 13:16
$begingroup$
Yes. That is the correct thing to do.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 11:59
$begingroup$
Yes. That is the correct thing to do.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 11:59
$begingroup$
Good to hear. I would then continue showing that $|phi_g|leq|g|$, where $|phi_g|$ is the operator norm of $phi_g$. Here is my attempt: begin{align*} left|intlangle f,grangle,dmuright| &leq sum_{i=1}^dint |f_ig_i|,dmu \ &leq sum_{i=1}^dleft(int|f_i|^pright)^{frac{1}{p}}left(int|g_i|^qright)^{frac{1}{q}} \ &stackrel{(?)}{leq} left(intsum_{i=1}^d|f_i|^pright)^{frac{1}{p}}left(intsum_{i=1}^d|g_i|^qright)^{frac{1}{q}}. end{align*} As tagged, I am not sure how to reason the last inequality or if it is even correct.
$endgroup$
– Lemma 5
Jan 6 at 13:16
$begingroup$
Good to hear. I would then continue showing that $|phi_g|leq|g|$, where $|phi_g|$ is the operator norm of $phi_g$. Here is my attempt: begin{align*} left|intlangle f,grangle,dmuright| &leq sum_{i=1}^dint |f_ig_i|,dmu \ &leq sum_{i=1}^dleft(int|f_i|^pright)^{frac{1}{p}}left(int|g_i|^qright)^{frac{1}{q}} \ &stackrel{(?)}{leq} left(intsum_{i=1}^d|f_i|^pright)^{frac{1}{p}}left(intsum_{i=1}^d|g_i|^qright)^{frac{1}{q}}. end{align*} As tagged, I am not sure how to reason the last inequality or if it is even correct.
$endgroup$
– Lemma 5
Jan 6 at 13:16
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have this version of Hölder's Inequality (when using it with the discrete measure):
$$
left|sum_j a_jb_jright|leqleft(sum_j|a_j|^pright)^{1/p}left(sum_j|b_j|^qright)^{1/q}.
$$
Then
$$
sum_{i=1}^dleft(int|f_i|^pright)^{frac{1}{p}}left(int|g_i|^qright)^{frac{1}{q}}
leq left(sum_i int|f_i|^pright)^{1/p}left(sum_i int|g_i|^qright)^{1/q}
=left(intsum_i |f_i|^pright)^{1/p}left(intsum_i |g_i|^qright)^{1/q}
$$
answered Jan 6 at 20:55
Martin ArgeramiMartin Argerami
126k1182180
$endgroup$
$begingroup$
Oh I somehow completely forgot Hölder, although having used the same inequality in the line above :D Thanks a lot!
$endgroup$
– Lemma 5
Jan 6 at 22:20
add a comment |
$begingroup$
Another way to see this is to use the following result: for Banach spaces $X_1,dots,X_n,$ consider the direct sum,
$$ X = bigoplus_{i=1}^n X_i, $$
equipped with the $p$-norm,
$$ lVert (x_1,dots,x_n) rVert_{p} = left(lVert x_irVert_{X_i}^pright)^{1/p}. $$
Then we can show that,
$$ X^* cong bigoplus_{i=1}^n X_i^*, $$
equipped with the $p^*$-norm $lVert cdot rVert_{p^*}$ defined similarly. Here the duality pairing is what you expect:
$$ leftlangle (x_1,dots,x_n),(f_1,dots,f_n) rightrangle = sum_{i=1}^n langle x_i, f_i rangle. $$
Now we can use the fact that,
$$ L^p(Omega,mathcal{A},mu,mathbb R^d) cong oplus_{i=1}^n L^p(Omega,mathcal{A},mu), $$
equipped with the $p$-norm as above. Hence the duality result shows that,
$$ L^p(Omega,mathcal{A},mu,mathbb R^d)^* cong oplus_{i=1}^n L^p(Omega,mathcal{A},mu)^* cong oplus_{i=1}^n L^{p^*}(Omega,mathcal{A},mu) cong L^{p^*}(Omega,mathcal{A},mu,mathbb R^d), $$
as required (the sums are equipped with the $p^*$ norms here). Note this gives the same pairing as the one you described, and both approaches (direct vs using the above result) boil down to the same argument.
answered Jan 7 at 16:31
ktoiktoi
2,3961616
$endgroup$
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
$begingroup$
You have this version of Hölder's Inequality (when using it with the discrete measure):
$$
left|sum_j a_jb_jright|leqleft(sum_j|a_j|^pright)^{1/p}left(sum_j|b_j|^qright)^{1/q}.
$$
Then
$$
sum_{i=1}^dleft(int|f_i|^pright)^{frac{1}{p}}left(int|g_i|^qright)^{frac{1}{q}}
leq left(sum_i int|f_i|^pright)^{1/p}left(sum_i int|g_i|^qright)^{1/q}
=left(intsum_i |f_i|^pright)^{1/p}left(intsum_i |g_i|^qright)^{1/q}
$$
answered Jan 6 at 20:55
Martin ArgeramiMartin Argerami
126k1182180
$endgroup$
$begingroup$
Oh I somehow completely forgot Hölder, although having used the same inequality in the line above :D Thanks a lot!
$endgroup$
– Lemma 5
Jan 6 at 22:20
add a comment |
$begingroup$
You have this version of Hölder's Inequality (when using it with the discrete measure):
$$
left|sum_j a_jb_jright|leqleft(sum_j|a_j|^pright)^{1/p}left(sum_j|b_j|^qright)^{1/q}.
$$
Then
$$
sum_{i=1}^dleft(int|f_i|^pright)^{frac{1}{p}}left(int|g_i|^qright)^{frac{1}{q}}
leq left(sum_i int|f_i|^pright)^{1/p}left(sum_i int|g_i|^qright)^{1/q}
=left(intsum_i |f_i|^pright)^{1/p}left(intsum_i |g_i|^qright)^{1/q}
$$
answered Jan 6 at 20:55
Martin ArgeramiMartin Argerami
126k1182180
$endgroup$
$begingroup$
Oh I somehow completely forgot Hölder, although having used the same inequality in the line above :D Thanks a lot!
$endgroup$
– Lemma 5
Jan 6 at 22:20
add a comment |
$begingroup$
You have this version of Hölder's Inequality (when using it with the discrete measure):
$$
left|sum_j a_jb_jright|leqleft(sum_j|a_j|^pright)^{1/p}left(sum_j|b_j|^qright)^{1/q}.
$$
Then
$$
sum_{i=1}^dleft(int|f_i|^pright)^{frac{1}{p}}left(int|g_i|^qright)^{frac{1}{q}}
leq left(sum_i int|f_i|^pright)^{1/p}left(sum_i int|g_i|^qright)^{1/q}
=left(intsum_i |f_i|^pright)^{1/p}left(intsum_i |g_i|^qright)^{1/q}
$$
answered Jan 6 at 20:55
Martin ArgeramiMartin Argerami
126k1182180
$endgroup$
You have this version of Hölder's Inequality (when using it with the discrete measure):
$$
left|sum_j a_jb_jright|leqleft(sum_j|a_j|^pright)^{1/p}left(sum_j|b_j|^qright)^{1/q}.
$$
Then
$$
sum_{i=1}^dleft(int|f_i|^pright)^{frac{1}{p}}left(int|g_i|^qright)^{frac{1}{q}}
leq left(sum_i int|f_i|^pright)^{1/p}left(sum_i int|g_i|^qright)^{1/q}
=left(intsum_i |f_i|^pright)^{1/p}left(intsum_i |g_i|^qright)^{1/q}
$$
answered Jan 6 at 20:55
Martin ArgeramiMartin Argerami
126k1182180
answered Jan 6 at 20:55
Martin ArgeramiMartin Argerami
126k1182180
answered Jan 6 at 20:55
Martin ArgeramiMartin Argerami
126k1182180
answered Jan 6 at 20:55
Martin ArgeramiMartin Argerami
126k1182180
126k1182180
$begingroup$
Oh I somehow completely forgot Hölder, although having used the same inequality in the line above :D Thanks a lot!
$endgroup$
– Lemma 5
Jan 6 at 22:20
add a comment |
$begingroup$
Oh I somehow completely forgot Hölder, although having used the same inequality in the line above :D Thanks a lot!
$endgroup$
– Lemma 5
Jan 6 at 22:20
$begingroup$
Oh I somehow completely forgot Hölder, although having used the same inequality in the line above :D Thanks a lot!
$endgroup$
– Lemma 5
Jan 6 at 22:20
$begingroup$
Oh I somehow completely forgot Hölder, although having used the same inequality in the line above :D Thanks a lot!
$endgroup$
– Lemma 5
Jan 6 at 22:20
add a comment |
$begingroup$
Another way to see this is to use the following result: for Banach spaces $X_1,dots,X_n,$ consider the direct sum,
$$ X = bigoplus_{i=1}^n X_i, $$
equipped with the $p$-norm,
$$ lVert (x_1,dots,x_n) rVert_{p} = left(lVert x_irVert_{X_i}^pright)^{1/p}. $$
Then we can show that,
$$ X^* cong bigoplus_{i=1}^n X_i^*, $$
equipped with the $p^*$-norm $lVert cdot rVert_{p^*}$ defined similarly. Here the duality pairing is what you expect:
$$ leftlangle (x_1,dots,x_n),(f_1,dots,f_n) rightrangle = sum_{i=1}^n langle x_i, f_i rangle. $$
Now we can use the fact that,
$$ L^p(Omega,mathcal{A},mu,mathbb R^d) cong oplus_{i=1}^n L^p(Omega,mathcal{A},mu), $$
equipped with the $p$-norm as above. Hence the duality result shows that,
$$ L^p(Omega,mathcal{A},mu,mathbb R^d)^* cong oplus_{i=1}^n L^p(Omega,mathcal{A},mu)^* cong oplus_{i=1}^n L^{p^*}(Omega,mathcal{A},mu) cong L^{p^*}(Omega,mathcal{A},mu,mathbb R^d), $$
as required (the sums are equipped with the $p^*$ norms here). Note this gives the same pairing as the one you described, and both approaches (direct vs using the above result) boil down to the same argument.
answered Jan 7 at 16:31
ktoiktoi
2,3961616
$endgroup$
add a comment |
$begingroup$
Another way to see this is to use the following result: for Banach spaces $X_1,dots,X_n,$ consider the direct sum,
$$ X = bigoplus_{i=1}^n X_i, $$
equipped with the $p$-norm,
$$ lVert (x_1,dots,x_n) rVert_{p} = left(lVert x_irVert_{X_i}^pright)^{1/p}. $$
Then we can show that,
$$ X^* cong bigoplus_{i=1}^n X_i^*, $$
equipped with the $p^*$-norm $lVert cdot rVert_{p^*}$ defined similarly. Here the duality pairing is what you expect:
$$ leftlangle (x_1,dots,x_n),(f_1,dots,f_n) rightrangle = sum_{i=1}^n langle x_i, f_i rangle. $$
Now we can use the fact that,
$$ L^p(Omega,mathcal{A},mu,mathbb R^d) cong oplus_{i=1}^n L^p(Omega,mathcal{A},mu), $$
equipped with the $p$-norm as above. Hence the duality result shows that,
$$ L^p(Omega,mathcal{A},mu,mathbb R^d)^* cong oplus_{i=1}^n L^p(Omega,mathcal{A},mu)^* cong oplus_{i=1}^n L^{p^*}(Omega,mathcal{A},mu) cong L^{p^*}(Omega,mathcal{A},mu,mathbb R^d), $$
as required (the sums are equipped with the $p^*$ norms here). Note this gives the same pairing as the one you described, and both approaches (direct vs using the above result) boil down to the same argument.
answered Jan 7 at 16:31
ktoiktoi
2,3961616
$endgroup$
add a comment |
$begingroup$
Another way to see this is to use the following result: for Banach spaces $X_1,dots,X_n,$ consider the direct sum,
$$ X = bigoplus_{i=1}^n X_i, $$
equipped with the $p$-norm,
$$ lVert (x_1,dots,x_n) rVert_{p} = left(lVert x_irVert_{X_i}^pright)^{1/p}. $$
Then we can show that,
$$ X^* cong bigoplus_{i=1}^n X_i^*, $$
equipped with the $p^*$-norm $lVert cdot rVert_{p^*}$ defined similarly. Here the duality pairing is what you expect:
$$ leftlangle (x_1,dots,x_n),(f_1,dots,f_n) rightrangle = sum_{i=1}^n langle x_i, f_i rangle. $$
Now we can use the fact that,
$$ L^p(Omega,mathcal{A},mu,mathbb R^d) cong oplus_{i=1}^n L^p(Omega,mathcal{A},mu), $$
equipped with the $p$-norm as above. Hence the duality result shows that,
$$ L^p(Omega,mathcal{A},mu,mathbb R^d)^* cong oplus_{i=1}^n L^p(Omega,mathcal{A},mu)^* cong oplus_{i=1}^n L^{p^*}(Omega,mathcal{A},mu) cong L^{p^*}(Omega,mathcal{A},mu,mathbb R^d), $$
as required (the sums are equipped with the $p^*$ norms here). Note this gives the same pairing as the one you described, and both approaches (direct vs using the above result) boil down to the same argument.
answered Jan 7 at 16:31
ktoiktoi
2,3961616
$endgroup$
Another way to see this is to use the following result: for Banach spaces $X_1,dots,X_n,$ consider the direct sum,
$$ X = bigoplus_{i=1}^n X_i, $$
equipped with the $p$-norm,
$$ lVert (x_1,dots,x_n) rVert_{p} = left(lVert x_irVert_{X_i}^pright)^{1/p}. $$
Then we can show that,
$$ X^* cong bigoplus_{i=1}^n X_i^*, $$
equipped with the $p^*$-norm $lVert cdot rVert_{p^*}$ defined similarly. Here the duality pairing is what you expect:
$$ leftlangle (x_1,dots,x_n),(f_1,dots,f_n) rightrangle = sum_{i=1}^n langle x_i, f_i rangle. $$
Now we can use the fact that,
$$ L^p(Omega,mathcal{A},mu,mathbb R^d) cong oplus_{i=1}^n L^p(Omega,mathcal{A},mu), $$
equipped with the $p$-norm as above. Hence the duality result shows that,
$$ L^p(Omega,mathcal{A},mu,mathbb R^d)^* cong oplus_{i=1}^n L^p(Omega,mathcal{A},mu)^* cong oplus_{i=1}^n L^{p^*}(Omega,mathcal{A},mu) cong L^{p^*}(Omega,mathcal{A},mu,mathbb R^d), $$
as required (the sums are equipped with the $p^*$ norms here). Note this gives the same pairing as the one you described, and both approaches (direct vs using the above result) boil down to the same argument.
answered Jan 7 at 16:31
ktoiktoi
2,3961616
answered Jan 7 at 16:31
ktoiktoi
2,3961616
answered Jan 7 at 16:31
ktoiktoi
2,3961616
answered Jan 7 at 16:31
ktoiktoi
2,3961616
2,3961616
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$begingroup$
Yes. That is the correct thing to do.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 11:59
$begingroup$
Good to hear. I would then continue showing that $|phi_g|leq|g|$, where $|phi_g|$ is the operator norm of $phi_g$. Here is my attempt: begin{align*} left|intlangle f,grangle,dmuright| &leq sum_{i=1}^dint |f_ig_i|,dmu \ &leq sum_{i=1}^dleft(int|f_i|^pright)^{frac{1}{p}}left(int|g_i|^qright)^{frac{1}{q}} \ &stackrel{(?)}{leq} left(intsum_{i=1}^d|f_i|^pright)^{frac{1}{p}}left(intsum_{i=1}^d|g_i|^qright)^{frac{1}{q}}. end{align*} As tagged, I am not sure how to reason the last inequality or if it is even correct.
$endgroup$
– Lemma 5
Jan 6 at 13:16