Mixing
Use Open Mapping Theorem to prove reverse norm inequality
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I am working through problems in Conway studying for an exam, and I have become stumped on one. The problem is in the section on Open Mapping Theorem (OMT) and Closed Graph Theorem (CGT).
III.13.6: Let $X$ be compact and suppose that $mathscrX$ is a Banach subspace of $C(X)$. If $E$ is a closed [hence compact] subset of $X$ such that for every $g in C(E)$ there is an $fin mathscrX$ with $f_E = g$, show that there is a constant $c > 0$ such that for each $g in C(E)$ there is an $f in mathscrX$ with $f_E$ and $maxf(x) leq c maxg(x)$.
My current solution: Of course the final statement is equivalent to stating that $|f|_L^infty(C(X)) leq c |g|_L^infty(C(E))$. I note that $A : mathscrX to C(E)$ via $Af = f|_E$ is a bounded linear operator (in fact $|A| leq 1$). Since we have the precondition that for every $g in C(E)$ we can find an $f in mathscrX$ such that $Af = g$, then $A$ is a surjective map. So the OMT applies. Note that $|Af| = |f|_E|_L^infty(C(E))$, so we really want to show that $1/c|Af| geq |f|$. In the previous problem in the book, we in fact have that this holds if and only if $textran(A)$ is closed and $textker(A) = 0$ (we needed the OMT, in fact the inverse mapping theorem, to prove the backwards direction). Now the first condition most certainly holds because a subset of a Banach space is Banach if and only if it is closed and we have surjectivity, but the second I suspect is not necessarily true.
My question: is there a way I can adjust $A$ so as to make it injective, am I missing something, or should I be going about this in a different manner? Thanks.
functional-analysis
edited Jul 18 at 23:34
Bernard
110k635103
asked Jul 18 at 23:06
mathishard.butweloveit
1069
 |Â
show 5 more comments
up vote
0
down vote
favorite
I am working through problems in Conway studying for an exam, and I have become stumped on one. The problem is in the section on Open Mapping Theorem (OMT) and Closed Graph Theorem (CGT).
III.13.6: Let $X$ be compact and suppose that $mathscrX$ is a Banach subspace of $C(X)$. If $E$ is a closed [hence compact] subset of $X$ such that for every $g in C(E)$ there is an $fin mathscrX$ with $f_E = g$, show that there is a constant $c > 0$ such that for each $g in C(E)$ there is an $f in mathscrX$ with $f_E$ and $maxf(x) leq c maxg(x)$.
My current solution: Of course the final statement is equivalent to stating that $|f|_L^infty(C(X)) leq c |g|_L^infty(C(E))$. I note that $A : mathscrX to C(E)$ via $Af = f|_E$ is a bounded linear operator (in fact $|A| leq 1$). Since we have the precondition that for every $g in C(E)$ we can find an $f in mathscrX$ such that $Af = g$, then $A$ is a surjective map. So the OMT applies. Note that $|Af| = |f|_E|_L^infty(C(E))$, so we really want to show that $1/c|Af| geq |f|$. In the previous problem in the book, we in fact have that this holds if and only if $textran(A)$ is closed and $textker(A) = 0$ (we needed the OMT, in fact the inverse mapping theorem, to prove the backwards direction). Now the first condition most certainly holds because a subset of a Banach space is Banach if and only if it is closed and we have surjectivity, but the second I suspect is not necessarily true.
My question: is there a way I can adjust $A$ so as to make it injective, am I missing something, or should I be going about this in a different manner? Thanks.
functional-analysis
edited Jul 18 at 23:34
Bernard
110k635103
asked Jul 18 at 23:06
mathishard.butweloveit
1069
1
I might be missing something but consider $X = [0,1]$, $E = left[0, frac12right]$ and $mathscrX = C[0,1]$. Every $f in Cleft[0, frac12right]$ can be extended to a function in $C[0,1]$ but $|cdot|_infty, [0,1]$ is not dominated by a constant times $|cdot|_infty, left[0,frac12right]$.
– mechanodroid
Jul 18 at 23:22
2
Look at $hatA:C(X)/Ker(A)to C(E)$. It is onto, continuous, and injective. The open mapping theorem gives you that the inverse is bounded. Now, the finiteness of the norm of the inverse gives you $inf_min Ker(A)|f+m|leq |hatA^-1||f|_E|$. So, you can take $c$ to be $> |hatA^-1|$, and since the infimum before is strictly smaller you can pick an $m$ such that the inequality holds.
– user574889
Jul 18 at 23:22
@mechanodroid "there is an $f in mathscrX$ ..."
– user574889
Jul 18 at 23:23
@cactus Sorry I don't get it.
– mechanodroid
Jul 18 at 23:24
@mechanodroid The conclusion is that there is a $c$ such that for every $g$ there is an extension $f$ such that the norm of that particular extension is dominated by the norm of $g$ times $c$.
– user574889
Jul 18 at 23:25
 |Â
show 5 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am working through problems in Conway studying for an exam, and I have become stumped on one. The problem is in the section on Open Mapping Theorem (OMT) and Closed Graph Theorem (CGT).
III.13.6: Let $X$ be compact and suppose that $mathscrX$ is a Banach subspace of $C(X)$. If $E$ is a closed [hence compact] subset of $X$ such that for every $g in C(E)$ there is an $fin mathscrX$ with $f_E = g$, show that there is a constant $c > 0$ such that for each $g in C(E)$ there is an $f in mathscrX$ with $f_E$ and $maxf(x) leq c maxg(x)$.
My current solution: Of course the final statement is equivalent to stating that $|f|_L^infty(C(X)) leq c |g|_L^infty(C(E))$. I note that $A : mathscrX to C(E)$ via $Af = f|_E$ is a bounded linear operator (in fact $|A| leq 1$). Since we have the precondition that for every $g in C(E)$ we can find an $f in mathscrX$ such that $Af = g$, then $A$ is a surjective map. So the OMT applies. Note that $|Af| = |f|_E|_L^infty(C(E))$, so we really want to show that $1/c|Af| geq |f|$. In the previous problem in the book, we in fact have that this holds if and only if $textran(A)$ is closed and $textker(A) = 0$ (we needed the OMT, in fact the inverse mapping theorem, to prove the backwards direction). Now the first condition most certainly holds because a subset of a Banach space is Banach if and only if it is closed and we have surjectivity, but the second I suspect is not necessarily true.
My question: is there a way I can adjust $A$ so as to make it injective, am I missing something, or should I be going about this in a different manner? Thanks.
functional-analysis
edited Jul 18 at 23:34
Bernard
110k635103
asked Jul 18 at 23:06
mathishard.butweloveit
1069
I am working through problems in Conway studying for an exam, and I have become stumped on one. The problem is in the section on Open Mapping Theorem (OMT) and Closed Graph Theorem (CGT).
III.13.6: Let $X$ be compact and suppose that $mathscrX$ is a Banach subspace of $C(X)$. If $E$ is a closed [hence compact] subset of $X$ such that for every $g in C(E)$ there is an $fin mathscrX$ with $f_E = g$, show that there is a constant $c > 0$ such that for each $g in C(E)$ there is an $f in mathscrX$ with $f_E$ and $maxf(x) leq c maxg(x)$.
My current solution: Of course the final statement is equivalent to stating that $|f|_L^infty(C(X)) leq c |g|_L^infty(C(E))$. I note that $A : mathscrX to C(E)$ via $Af = f|_E$ is a bounded linear operator (in fact $|A| leq 1$). Since we have the precondition that for every $g in C(E)$ we can find an $f in mathscrX$ such that $Af = g$, then $A$ is a surjective map. So the OMT applies. Note that $|Af| = |f|_E|_L^infty(C(E))$, so we really want to show that $1/c|Af| geq |f|$. In the previous problem in the book, we in fact have that this holds if and only if $textran(A)$ is closed and $textker(A) = 0$ (we needed the OMT, in fact the inverse mapping theorem, to prove the backwards direction). Now the first condition most certainly holds because a subset of a Banach space is Banach if and only if it is closed and we have surjectivity, but the second I suspect is not necessarily true.
My question: is there a way I can adjust $A$ so as to make it injective, am I missing something, or should I be going about this in a different manner? Thanks.
functional-analysis
edited Jul 18 at 23:34
Bernard
110k635103
asked Jul 18 at 23:06
mathishard.butweloveit
1069
edited Jul 18 at 23:34
Bernard
110k635103
edited Jul 18 at 23:34
Bernard
110k635103
edited Jul 18 at 23:34
Bernard
110k635103
110k635103
asked Jul 18 at 23:06
mathishard.butweloveit
1069
asked Jul 18 at 23:06
mathishard.butweloveit
1069
asked Jul 18 at 23:06
mathishard.butweloveit
1069
1069
1
I might be missing something but consider $X = [0,1]$, $E = left[0, frac12right]$ and $mathscrX = C[0,1]$. Every $f in Cleft[0, frac12right]$ can be extended to a function in $C[0,1]$ but $|cdot|_infty, [0,1]$ is not dominated by a constant times $|cdot|_infty, left[0,frac12right]$.
– mechanodroid
Jul 18 at 23:22
2
Look at $hatA:C(X)/Ker(A)to C(E)$. It is onto, continuous, and injective. The open mapping theorem gives you that the inverse is bounded. Now, the finiteness of the norm of the inverse gives you $inf_min Ker(A)|f+m|leq |hatA^-1||f|_E|$. So, you can take $c$ to be $> |hatA^-1|$, and since the infimum before is strictly smaller you can pick an $m$ such that the inequality holds.
– user574889
Jul 18 at 23:22
@mechanodroid "there is an $f in mathscrX$ ..."
– user574889
Jul 18 at 23:23
@cactus Sorry I don't get it.
– mechanodroid
Jul 18 at 23:24
@mechanodroid The conclusion is that there is a $c$ such that for every $g$ there is an extension $f$ such that the norm of that particular extension is dominated by the norm of $g$ times $c$.
– user574889
Jul 18 at 23:25
 |Â
show 5 more comments
1
I might be missing something but consider $X = [0,1]$, $E = left[0, frac12right]$ and $mathscrX = C[0,1]$. Every $f in Cleft[0, frac12right]$ can be extended to a function in $C[0,1]$ but $|cdot|_infty, [0,1]$ is not dominated by a constant times $|cdot|_infty, left[0,frac12right]$.
– mechanodroid
Jul 18 at 23:22
2
Look at $hatA:C(X)/Ker(A)to C(E)$. It is onto, continuous, and injective. The open mapping theorem gives you that the inverse is bounded. Now, the finiteness of the norm of the inverse gives you $inf_min Ker(A)|f+m|leq |hatA^-1||f|_E|$. So, you can take $c$ to be $> |hatA^-1|$, and since the infimum before is strictly smaller you can pick an $m$ such that the inequality holds.
– user574889
Jul 18 at 23:22
@mechanodroid "there is an $f in mathscrX$ ..."
– user574889
Jul 18 at 23:23
@cactus Sorry I don't get it.
– mechanodroid
Jul 18 at 23:24
@mechanodroid The conclusion is that there is a $c$ such that for every $g$ there is an extension $f$ such that the norm of that particular extension is dominated by the norm of $g$ times $c$.
– user574889
Jul 18 at 23:25
1
1
I might be missing something but consider $X = [0,1]$, $E = left[0, frac12right]$ and $mathscrX = C[0,1]$. Every $f in Cleft[0, frac12right]$ can be extended to a function in $C[0,1]$ but $|cdot|_infty, [0,1]$ is not dominated by a constant times $|cdot|_infty, left[0,frac12right]$.
– mechanodroid
Jul 18 at 23:22
I might be missing something but consider $X = [0,1]$, $E = left[0, frac12right]$ and $mathscrX = C[0,1]$. Every $f in Cleft[0, frac12right]$ can be extended to a function in $C[0,1]$ but $|cdot|_infty, [0,1]$ is not dominated by a constant times $|cdot|_infty, left[0,frac12right]$.
– mechanodroid
Jul 18 at 23:22
2
2
Look at $hatA:C(X)/Ker(A)to C(E)$. It is onto, continuous, and injective. The open mapping theorem gives you that the inverse is bounded. Now, the finiteness of the norm of the inverse gives you $inf_min Ker(A)|f+m|leq |hatA^-1||f|_E|$. So, you can take $c$ to be $> |hatA^-1|$, and since the infimum before is strictly smaller you can pick an $m$ such that the inequality holds.
– user574889
Jul 18 at 23:22
Look at $hatA:C(X)/Ker(A)to C(E)$. It is onto, continuous, and injective. The open mapping theorem gives you that the inverse is bounded. Now, the finiteness of the norm of the inverse gives you $inf_min Ker(A)|f+m|leq |hatA^-1||f|_E|$. So, you can take $c$ to be $> |hatA^-1|$, and since the infimum before is strictly smaller you can pick an $m$ such that the inequality holds.
– user574889
Jul 18 at 23:22
@mechanodroid "there is an $f in mathscrX$ ..."
– user574889
Jul 18 at 23:23
@mechanodroid "there is an $f in mathscrX$ ..."
– user574889
Jul 18 at 23:23
@cactus Sorry I don't get it.
– mechanodroid
Jul 18 at 23:24
@cactus Sorry I don't get it.
– mechanodroid
Jul 18 at 23:24
@mechanodroid The conclusion is that there is a $c$ such that for every $g$ there is an extension $f$ such that the norm of that particular extension is dominated by the norm of $g$ times $c$.
– user574889
Jul 18 at 23:25
@mechanodroid The conclusion is that there is a $c$ such that for every $g$ there is an extension $f$ such that the norm of that particular extension is dominated by the norm of $g$ times $c$.
– user574889
Jul 18 at 23:25
 |Â
show 5 more comments
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2856104%2fuse-open-mapping-theorem-to-prove-reverse-norm-inequality%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
I might be missing something but consider $X = [0,1]$, $E = left[0, frac12right]$ and $mathscrX = C[0,1]$. Every $f in Cleft[0, frac12right]$ can be extended to a function in $C[0,1]$ but $|cdot|_infty, [0,1]$ is not dominated by a constant times $|cdot|_infty, left[0,frac12right]$.
– mechanodroid
Jul 18 at 23:22
2
Look at $hatA:C(X)/Ker(A)to C(E)$. It is onto, continuous, and injective. The open mapping theorem gives you that the inverse is bounded. Now, the finiteness of the norm of the inverse gives you $inf_min Ker(A)|f+m|leq |hatA^-1||f|_E|$. So, you can take $c$ to be $> |hatA^-1|$, and since the infimum before is strictly smaller you can pick an $m$ such that the inequality holds.
– user574889
Jul 18 at 23:22
@mechanodroid "there is an $f in mathscrX$ ..."
– user574889
Jul 18 at 23:23
@cactus Sorry I don't get it.
– mechanodroid
Jul 18 at 23:24
@mechanodroid The conclusion is that there is a $c$ such that for every $g$ there is an extension $f$ such that the norm of that particular extension is dominated by the norm of $g$ times $c$.
– user574889
Jul 18 at 23:25