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Rudin Functional analysis, theorem 4.12 corollary (b)

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0

$begingroup$

Suppose $X$ and $Y$ are Banach spaces, and $T in mathcalB(X,Y)$ Then
$$
mathcalN(T^*) = mathcalR(T)^perp ;;textand;; mathcalN(T) = ^perpmathcalR(T^*)
$$

The corollary in the title

b) $mathcalR(T)$ is dense in $Y$ iff $T^*$ is one-to-one

The proof is really short but there's a specific bit I don't get (fully at least)

$mathcalR(T)$ is dense in $Y$ iff $mathcalR(T)^perp = left 0right$

I've tried to prove this bit by exercise, and I think the most promosing was to show that

$$
^perp(mathcalR(T)^perp) = Y iff mathcalR(T) = left0right
$$

The reason of the equality on the left of the double implication is because of theorem 4.7, which states that the orthogonal of the orthogonal of a set is the norm closure of a set.

However I'm not entirely sure how to exploit this, can you help?

functional-analysis proof-explanation normed-spaces topological-vector-spaces duality-theorems

share|cite|improve this question

edited Mar 11 at 11:51

user8469759

asked Mar 11 at 10:24

user8469759user8469759

1,5561618

$endgroup$

add a comment | 

0

$begingroup$

Suppose $X$ and $Y$ are Banach spaces, and $T in mathcalB(X,Y)$ Then
$$
mathcalN(T^*) = mathcalR(T)^perp ;;textand;; mathcalN(T) = ^perpmathcalR(T^*)
$$

The corollary in the title

b) $mathcalR(T)$ is dense in $Y$ iff $T^*$ is one-to-one

The proof is really short but there's a specific bit I don't get (fully at least)

$mathcalR(T)$ is dense in $Y$ iff $mathcalR(T)^perp = left 0right$

I've tried to prove this bit by exercise, and I think the most promosing was to show that

$$
^perp(mathcalR(T)^perp) = Y iff mathcalR(T) = left0right
$$

The reason of the equality on the left of the double implication is because of theorem 4.7, which states that the orthogonal of the orthogonal of a set is the norm closure of a set.

However I'm not entirely sure how to exploit this, can you help?

functional-analysis proof-explanation normed-spaces topological-vector-spaces duality-theorems

share|cite|improve this question

edited Mar 11 at 11:51

user8469759

asked Mar 11 at 10:24

user8469759user8469759

1,5561618

$endgroup$

add a comment | 

$begingroup$

Suppose $X$ and $Y$ are Banach spaces, and $T in mathcalB(X,Y)$ Then
$$
mathcalN(T^*) = mathcalR(T)^perp ;;textand;; mathcalN(T) = ^perpmathcalR(T^*)
$$

The corollary in the title

b) $mathcalR(T)$ is dense in $Y$ iff $T^*$ is one-to-one

The proof is really short but there's a specific bit I don't get (fully at least)

$mathcalR(T)$ is dense in $Y$ iff $mathcalR(T)^perp = left 0right$

I've tried to prove this bit by exercise, and I think the most promosing was to show that

$$
^perp(mathcalR(T)^perp) = Y iff mathcalR(T) = left0right
$$

The reason of the equality on the left of the double implication is because of theorem 4.7, which states that the orthogonal of the orthogonal of a set is the norm closure of a set.

However I'm not entirely sure how to exploit this, can you help?

functional-analysis proof-explanation normed-spaces topological-vector-spaces duality-theorems

share|cite|improve this question

edited Mar 11 at 11:51

user8469759

asked Mar 11 at 10:24

user8469759user8469759

1,5561618

$endgroup$

share|cite|improve this question

edited Mar 11 at 11:51

user8469759

asked Mar 11 at 10:24

user8469759user8469759

1,5561618

share|cite|improve this question

edited Mar 11 at 11:51

user8469759

asked Mar 11 at 10:24

user8469759user8469759

1,5561618

asked Mar 11 at 10:24

user8469759user8469759

1,5561618

asked Mar 11 at 10:24

user8469759user8469759

1,5561618

1 Answer
1

active

oldest

votes

0

$begingroup$

Here is another approach:

Suppose $R(T)$ is dense. Any linear functional that vanishes on $R(T)$ must vanish on the entire space by continuity considerations. That is, $R(T)^perp$ consists only of the zero functional.

Conversely suppose $R(T)$ is not dense. Then $overlineR(T)$ is a proper closed subspace of $Y$ and there exists (per e.g. Hahn-Banach) a nontrivial linear function $phi$ that vanishes on $overlineR(T)$, meaning $R(T)^perp not= 0$.

share|cite|improve this answer

answered Mar 11 at 11:31

Umberto P.Umberto P.

39.9k13267

$endgroup$

add a comment | 

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0

$begingroup$

Here is another approach:

Suppose $R(T)$ is dense. Any linear functional that vanishes on $R(T)$ must vanish on the entire space by continuity considerations. That is, $R(T)^perp$ consists only of the zero functional.

Conversely suppose $R(T)$ is not dense. Then $overlineR(T)$ is a proper closed subspace of $Y$ and there exists (per e.g. Hahn-Banach) a nontrivial linear function $phi$ that vanishes on $overlineR(T)$, meaning $R(T)^perp not= 0$.

share|cite|improve this answer

answered Mar 11 at 11:31

Umberto P.Umberto P.

39.9k13267

$endgroup$

add a comment | 

0

$begingroup$

Here is another approach:

Suppose $R(T)$ is dense. Any linear functional that vanishes on $R(T)$ must vanish on the entire space by continuity considerations. That is, $R(T)^perp$ consists only of the zero functional.

Conversely suppose $R(T)$ is not dense. Then $overlineR(T)$ is a proper closed subspace of $Y$ and there exists (per e.g. Hahn-Banach) a nontrivial linear function $phi$ that vanishes on $overlineR(T)$, meaning $R(T)^perp not= 0$.

share|cite|improve this answer

answered Mar 11 at 11:31

Umberto P.Umberto P.

39.9k13267

$endgroup$

add a comment | 

$begingroup$

Here is another approach:

Suppose $R(T)$ is dense. Any linear functional that vanishes on $R(T)$ must vanish on the entire space by continuity considerations. That is, $R(T)^perp$ consists only of the zero functional.

Conversely suppose $R(T)$ is not dense. Then $overlineR(T)$ is a proper closed subspace of $Y$ and there exists (per e.g. Hahn-Banach) a nontrivial linear function $phi$ that vanishes on $overlineR(T)$, meaning $R(T)^perp not= 0$.

share|cite|improve this answer

answered Mar 11 at 11:31

Umberto P.Umberto P.

39.9k13267

$endgroup$

share|cite|improve this answer

answered Mar 11 at 11:31

Umberto P.Umberto P.

39.9k13267

answered Mar 11 at 11:31

Umberto P.Umberto P.

39.9k13267

answered Mar 11 at 11:31

Umberto P.Umberto P.

39.9k13267