Bounded Linear Operators on Banach Spaces
$begingroup$
let $1<p_1<p<p_2<infty$. Let $Y$ be a Banach space and let $T$ be a bounded linear operator from $L^{p_1}$ to $Y$ and from $L^{p_2}$ to $Y$. Show that $T$ is then also a bounded linear operator from $L^p$ to Y.
I'm trying to use the lemma which states that for any $uin L^p$ there exist $u_1in L^{p_1}$ and $u_2in L^{p_2}$ such that $u=u_1+u_2$.
So there exists $C_1$ such that $$||Tx||_yleq C_1||x||_{p_1}$$ for any $xin L^{p_1}$.
Also there exists $C_2$ such that $$||Tx||_yleq C_2||x||_{p_2}$$ for any $xin L^{p_2}$.
Hence if $sin L^p$ then we can write $s=u_1+u_2$ such that $u_1in L^{p_1}$ and $u_2in L^{p_2}$. Then $$||Ts||_yleq||Tu_1||_y+||Tu_2||_yleq C_1||u_1||_{p_1}+C_2||u_2||_{p_2}leq C[||u_1||_{p_1}+||u_2||_{p_2}],$$ for some constant $C$.
I don't know if it's possible to have inequality like $$||u_1||_{p_1}+||u_2||_{p_2}leq C^* ||u_1+u_2||_p ??$$
real-analysis banach-spaces
asked Jan 18 at 19:37
S_AlexS_Alex
1679
$endgroup$
add a comment |
$begingroup$
let $1<p_1<p<p_2<infty$. Let $Y$ be a Banach space and let $T$ be a bounded linear operator from $L^{p_1}$ to $Y$ and from $L^{p_2}$ to $Y$. Show that $T$ is then also a bounded linear operator from $L^p$ to Y.
I'm trying to use the lemma which states that for any $uin L^p$ there exist $u_1in L^{p_1}$ and $u_2in L^{p_2}$ such that $u=u_1+u_2$.
So there exists $C_1$ such that $$||Tx||_yleq C_1||x||_{p_1}$$ for any $xin L^{p_1}$.
Also there exists $C_2$ such that $$||Tx||_yleq C_2||x||_{p_2}$$ for any $xin L^{p_2}$.
Hence if $sin L^p$ then we can write $s=u_1+u_2$ such that $u_1in L^{p_1}$ and $u_2in L^{p_2}$. Then $$||Ts||_yleq||Tu_1||_y+||Tu_2||_yleq C_1||u_1||_{p_1}+C_2||u_2||_{p_2}leq C[||u_1||_{p_1}+||u_2||_{p_2}],$$ for some constant $C$.
I don't know if it's possible to have inequality like $$||u_1||_{p_1}+||u_2||_{p_2}leq C^* ||u_1+u_2||_p ??$$
real-analysis banach-spaces
asked Jan 18 at 19:37
S_AlexS_Alex
1679
$endgroup$
$begingroup$
As is, no. However, you should be able to construct a pair $(u_1,u_2)$ such that $|u_1|_{p_1} leq C’_1|x|_p$ and same for $u_2$. Do not forget to show that the value of $T$ will not depend on the decomposition chosen!
$endgroup$
– Mindlack
Jan 18 at 20:24
add a comment |
$begingroup$
let $1<p_1<p<p_2<infty$. Let $Y$ be a Banach space and let $T$ be a bounded linear operator from $L^{p_1}$ to $Y$ and from $L^{p_2}$ to $Y$. Show that $T$ is then also a bounded linear operator from $L^p$ to Y.
I'm trying to use the lemma which states that for any $uin L^p$ there exist $u_1in L^{p_1}$ and $u_2in L^{p_2}$ such that $u=u_1+u_2$.
So there exists $C_1$ such that $$||Tx||_yleq C_1||x||_{p_1}$$ for any $xin L^{p_1}$.
Also there exists $C_2$ such that $$||Tx||_yleq C_2||x||_{p_2}$$ for any $xin L^{p_2}$.
Hence if $sin L^p$ then we can write $s=u_1+u_2$ such that $u_1in L^{p_1}$ and $u_2in L^{p_2}$. Then $$||Ts||_yleq||Tu_1||_y+||Tu_2||_yleq C_1||u_1||_{p_1}+C_2||u_2||_{p_2}leq C[||u_1||_{p_1}+||u_2||_{p_2}],$$ for some constant $C$.
I don't know if it's possible to have inequality like $$||u_1||_{p_1}+||u_2||_{p_2}leq C^* ||u_1+u_2||_p ??$$
real-analysis banach-spaces
asked Jan 18 at 19:37
S_AlexS_Alex
1679
$endgroup$
let $1<p_1<p<p_2<infty$. Let $Y$ be a Banach space and let $T$ be a bounded linear operator from $L^{p_1}$ to $Y$ and from $L^{p_2}$ to $Y$. Show that $T$ is then also a bounded linear operator from $L^p$ to Y.
I'm trying to use the lemma which states that for any $uin L^p$ there exist $u_1in L^{p_1}$ and $u_2in L^{p_2}$ such that $u=u_1+u_2$.
So there exists $C_1$ such that $$||Tx||_yleq C_1||x||_{p_1}$$ for any $xin L^{p_1}$.
Also there exists $C_2$ such that $$||Tx||_yleq C_2||x||_{p_2}$$ for any $xin L^{p_2}$.
Hence if $sin L^p$ then we can write $s=u_1+u_2$ such that $u_1in L^{p_1}$ and $u_2in L^{p_2}$. Then $$||Ts||_yleq||Tu_1||_y+||Tu_2||_yleq C_1||u_1||_{p_1}+C_2||u_2||_{p_2}leq C[||u_1||_{p_1}+||u_2||_{p_2}],$$ for some constant $C$.
I don't know if it's possible to have inequality like $$||u_1||_{p_1}+||u_2||_{p_2}leq C^* ||u_1+u_2||_p ??$$
real-analysis banach-spaces
real-analysis banach-spaces
asked Jan 18 at 19:37
S_AlexS_Alex
1679
asked Jan 18 at 19:37
S_AlexS_Alex
1679
asked Jan 18 at 19:37
S_AlexS_Alex
1679
asked Jan 18 at 19:37
S_AlexS_Alex
1679
asked Jan 18 at 19:37
S_AlexS_Alex
1679
1679
$begingroup$
As is, no. However, you should be able to construct a pair $(u_1,u_2)$ such that $|u_1|_{p_1} leq C’_1|x|_p$ and same for $u_2$. Do not forget to show that the value of $T$ will not depend on the decomposition chosen!
$endgroup$
– Mindlack
Jan 18 at 20:24
add a comment |
$begingroup$
As is, no. However, you should be able to construct a pair $(u_1,u_2)$ such that $|u_1|_{p_1} leq C’_1|x|_p$ and same for $u_2$. Do not forget to show that the value of $T$ will not depend on the decomposition chosen!
$endgroup$
– Mindlack
Jan 18 at 20:24
$begingroup$
As is, no. However, you should be able to construct a pair $(u_1,u_2)$ such that $|u_1|_{p_1} leq C’_1|x|_p$ and same for $u_2$. Do not forget to show that the value of $T$ will not depend on the decomposition chosen!
$endgroup$
– Mindlack
Jan 18 at 20:24
$begingroup$
As is, no. However, you should be able to construct a pair $(u_1,u_2)$ such that $|u_1|_{p_1} leq C’_1|x|_p$ and same for $u_2$. Do not forget to show that the value of $T$ will not depend on the decomposition chosen!
$endgroup$
– Mindlack
Jan 18 at 20:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Assume $fin L^p$ such that $|f|_p = 1$ is given. What we want to show is the existence of $C>0$ (which does not depend on $f$) such that $$
|Tf|_Yle C.
$$ Now, we can decompose
$$
f=f1_{{|f|le 1}}+f1_{{|f|>1}}=f_1+f_2.
$$ Note that $int |f_1|^{p_1}leint |f|^p= 1$ and $int |f_2|^{p_2}leint |f|^p= 1$. Hence, $|f_1|_{p_1}le$ and $|f_2|_{p_2}le 1$ holds. Thus, we have
$$
|Tf|_Yle |Tf_1|_Y+|Tf_2|_Yle C_1+C_2
$$ and the desired conclusion holds for $C=C_1+C_2$.
Note: In fact, Riesz-Thorin method (using complex method) provides a better bound
$$
|T|_{p}le |T|_{p_1}^alpha|T|_{p_2}^{1-alpha}
$$ where $frac{1}{p}=frac{alpha}{p_1}+frac{1-alpha}{p_2}.$
answered Jan 18 at 21:58
SongSong
13.1k632
$endgroup$
$begingroup$
Oh, never mind. I misread the problem and was thinking of something more like Riesz-Thorin, where $T$ is bounded from $L^{p_1} to L^{p_1}$ and from $L^{p_2} to L^{p_2}$. Didn't realize that the codomain was the same space everywhere.
$endgroup$
– Nate Eldredge
Jan 18 at 22:03
$begingroup$
@NateEldredge Never mind! I think your concern is legitimate :)
$endgroup$
– Song
Jan 18 at 22:04
add a comment |
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$begingroup$
Assume $fin L^p$ such that $|f|_p = 1$ is given. What we want to show is the existence of $C>0$ (which does not depend on $f$) such that $$
|Tf|_Yle C.
$$ Now, we can decompose
$$
f=f1_{{|f|le 1}}+f1_{{|f|>1}}=f_1+f_2.
$$ Note that $int |f_1|^{p_1}leint |f|^p= 1$ and $int |f_2|^{p_2}leint |f|^p= 1$. Hence, $|f_1|_{p_1}le$ and $|f_2|_{p_2}le 1$ holds. Thus, we have
$$
|Tf|_Yle |Tf_1|_Y+|Tf_2|_Yle C_1+C_2
$$ and the desired conclusion holds for $C=C_1+C_2$.
Note: In fact, Riesz-Thorin method (using complex method) provides a better bound
$$
|T|_{p}le |T|_{p_1}^alpha|T|_{p_2}^{1-alpha}
$$ where $frac{1}{p}=frac{alpha}{p_1}+frac{1-alpha}{p_2}.$
answered Jan 18 at 21:58
SongSong
13.1k632
$endgroup$
$begingroup$
Oh, never mind. I misread the problem and was thinking of something more like Riesz-Thorin, where $T$ is bounded from $L^{p_1} to L^{p_1}$ and from $L^{p_2} to L^{p_2}$. Didn't realize that the codomain was the same space everywhere.
$endgroup$
– Nate Eldredge
Jan 18 at 22:03
$begingroup$
@NateEldredge Never mind! I think your concern is legitimate :)
$endgroup$
– Song
Jan 18 at 22:04
add a comment |
$begingroup$
Assume $fin L^p$ such that $|f|_p = 1$ is given. What we want to show is the existence of $C>0$ (which does not depend on $f$) such that $$
|Tf|_Yle C.
$$ Now, we can decompose
$$
f=f1_{{|f|le 1}}+f1_{{|f|>1}}=f_1+f_2.
$$ Note that $int |f_1|^{p_1}leint |f|^p= 1$ and $int |f_2|^{p_2}leint |f|^p= 1$. Hence, $|f_1|_{p_1}le$ and $|f_2|_{p_2}le 1$ holds. Thus, we have
$$
|Tf|_Yle |Tf_1|_Y+|Tf_2|_Yle C_1+C_2
$$ and the desired conclusion holds for $C=C_1+C_2$.
Note: In fact, Riesz-Thorin method (using complex method) provides a better bound
$$
|T|_{p}le |T|_{p_1}^alpha|T|_{p_2}^{1-alpha}
$$ where $frac{1}{p}=frac{alpha}{p_1}+frac{1-alpha}{p_2}.$
answered Jan 18 at 21:58
SongSong
13.1k632
$endgroup$
$begingroup$
Oh, never mind. I misread the problem and was thinking of something more like Riesz-Thorin, where $T$ is bounded from $L^{p_1} to L^{p_1}$ and from $L^{p_2} to L^{p_2}$. Didn't realize that the codomain was the same space everywhere.
$endgroup$
– Nate Eldredge
Jan 18 at 22:03
$begingroup$
@NateEldredge Never mind! I think your concern is legitimate :)
$endgroup$
– Song
Jan 18 at 22:04
add a comment |
$begingroup$
Assume $fin L^p$ such that $|f|_p = 1$ is given. What we want to show is the existence of $C>0$ (which does not depend on $f$) such that $$
|Tf|_Yle C.
$$ Now, we can decompose
$$
f=f1_{{|f|le 1}}+f1_{{|f|>1}}=f_1+f_2.
$$ Note that $int |f_1|^{p_1}leint |f|^p= 1$ and $int |f_2|^{p_2}leint |f|^p= 1$. Hence, $|f_1|_{p_1}le$ and $|f_2|_{p_2}le 1$ holds. Thus, we have
$$
|Tf|_Yle |Tf_1|_Y+|Tf_2|_Yle C_1+C_2
$$ and the desired conclusion holds for $C=C_1+C_2$.
Note: In fact, Riesz-Thorin method (using complex method) provides a better bound
$$
|T|_{p}le |T|_{p_1}^alpha|T|_{p_2}^{1-alpha}
$$ where $frac{1}{p}=frac{alpha}{p_1}+frac{1-alpha}{p_2}.$
answered Jan 18 at 21:58
SongSong
13.1k632
$endgroup$
Assume $fin L^p$ such that $|f|_p = 1$ is given. What we want to show is the existence of $C>0$ (which does not depend on $f$) such that $$
|Tf|_Yle C.
$$ Now, we can decompose
$$
f=f1_{{|f|le 1}}+f1_{{|f|>1}}=f_1+f_2.
$$ Note that $int |f_1|^{p_1}leint |f|^p= 1$ and $int |f_2|^{p_2}leint |f|^p= 1$. Hence, $|f_1|_{p_1}le$ and $|f_2|_{p_2}le 1$ holds. Thus, we have
$$
|Tf|_Yle |Tf_1|_Y+|Tf_2|_Yle C_1+C_2
$$ and the desired conclusion holds for $C=C_1+C_2$.
Note: In fact, Riesz-Thorin method (using complex method) provides a better bound
$$
|T|_{p}le |T|_{p_1}^alpha|T|_{p_2}^{1-alpha}
$$ where $frac{1}{p}=frac{alpha}{p_1}+frac{1-alpha}{p_2}.$
answered Jan 18 at 21:58
SongSong
13.1k632
edited Jan 18 at 22:08
answered Jan 18 at 21:58
SongSong
13.1k632
answered Jan 18 at 21:58
SongSong
13.1k632
answered Jan 18 at 21:58
SongSong
13.1k632
13.1k632
$begingroup$
Oh, never mind. I misread the problem and was thinking of something more like Riesz-Thorin, where $T$ is bounded from $L^{p_1} to L^{p_1}$ and from $L^{p_2} to L^{p_2}$. Didn't realize that the codomain was the same space everywhere.
$endgroup$
– Nate Eldredge
Jan 18 at 22:03
$begingroup$
@NateEldredge Never mind! I think your concern is legitimate :)
$endgroup$
– Song
Jan 18 at 22:04
add a comment |
$begingroup$
Oh, never mind. I misread the problem and was thinking of something more like Riesz-Thorin, where $T$ is bounded from $L^{p_1} to L^{p_1}$ and from $L^{p_2} to L^{p_2}$. Didn't realize that the codomain was the same space everywhere.
$endgroup$
– Nate Eldredge
Jan 18 at 22:03
$begingroup$
@NateEldredge Never mind! I think your concern is legitimate :)
$endgroup$
– Song
Jan 18 at 22:04
$begingroup$
Oh, never mind. I misread the problem and was thinking of something more like Riesz-Thorin, where $T$ is bounded from $L^{p_1} to L^{p_1}$ and from $L^{p_2} to L^{p_2}$. Didn't realize that the codomain was the same space everywhere.
$endgroup$
– Nate Eldredge
Jan 18 at 22:03
$begingroup$
Oh, never mind. I misread the problem and was thinking of something more like Riesz-Thorin, where $T$ is bounded from $L^{p_1} to L^{p_1}$ and from $L^{p_2} to L^{p_2}$. Didn't realize that the codomain was the same space everywhere.
$endgroup$
– Nate Eldredge
Jan 18 at 22:03
$begingroup$
@NateEldredge Never mind! I think your concern is legitimate :)
$endgroup$
– Song
Jan 18 at 22:04
$begingroup$
@NateEldredge Never mind! I think your concern is legitimate :)
$endgroup$
– Song
Jan 18 at 22:04
add a comment |
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$begingroup$
As is, no. However, you should be able to construct a pair $(u_1,u_2)$ such that $|u_1|_{p_1} leq C’_1|x|_p$ and same for $u_2$. Do not forget to show that the value of $T$ will not depend on the decomposition chosen!
$endgroup$
– Mindlack
Jan 18 at 20:24