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Rudin functional analysis theorem 4.13, (a) and (b)

Lemma 10.8 Rudin functional analysis, proof.Theorem 1.24 Rudin, Functional analysis (first half)Theorem 1.14 (b) Rudin functional analysis, few clarifications.Rudin's functional analysis theorem 3.12Theorem 3.18, Rudin's functional analysisRudin Functional Analysis Theorem 3.25Rudin functional analysis, theorem 3.27Rudin functional analysis theorem 3.28, application of Reisz representation theorem.Rudin functional analysis theorem 3.28, $P$ is weak*-compact in $C(Q)^*$.Rudin's functional analysis, theorem 4.1.(completeness of $mathcal{B}(X,Y)$)

0

$begingroup$

Let $U$ and $V$ be the open unit balls in the Banach spaces $X$ and $Y$, respectively. If $T in mathcal{B}(X,Y)$ and $delta > 0$, the the implications
$$
(a)to(b)to(c)to(d)
$$
hold among the following statements

(a) $lVert T^* y^* rVert geq delta lVert y^* rVert$

(b) $delta V subset overline{T(U)}$

There're (c) and (d) as well, but I focus only on the first two.

Proof : Assume (a), and pick $y_0 notin overline{T(U)}$. Since $overline{T(U)}$ is convex, closed and balanced, theorem 3.7 shows that there's a $y^*$ such that $|<y,y^*>|leq 1$ for every $y in overline{T(U)}$, but $|<y_0,y^*>| > 1$. If $x in U$, it follows that
$$
left| leftlangle x,T^* y^* rightrangle right| = left|leftlangle Tx, y^* rightrangle right| leq 1
$$
Thus $lVert T^* y^* rVert leq 1$, and now (a) gives
$$
delta < delta left| leftlangle y_0,y^* rightrangle right| leq delta lVert y_0 rVert lVert y^* rVert leq lVert y_0 rVert lVert T^* y^* rVert leq lVert y_0 rVert.
$$
It follows $y in overline{T(U)}$ if $lVert y rVert leq delta$

First question, why the very last statement? namely why:

It follows $y in overline{T(U)}$ if $lVert y rVert leq delta$
?

For (b) part I'll just write the bit I don't get, not the full proof.

Next, assume (b). Take $delta = 1$, without loss of generality. Then $overline{V} subset overline{T(U)}$. To every $y in Y$ and every $epsilon > 0$ corresponds an $x in X$ with $lVert x rVert leq lVert y rVert$ and $lVert y - Tx rVert < epsilon$.

Basically I don't understand the full sentence after the period, does it follow from continuity of $T$ somehow?

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

share|cite|improve this question

asked Mar 11 at 10:44

user8469759user8469759

1,5581618

$endgroup$

add a comment | 

0

$begingroup$

Let $U$ and $V$ be the open unit balls in the Banach spaces $X$ and $Y$, respectively. If $T in mathcal{B}(X,Y)$ and $delta > 0$, the the implications
$$
(a)to(b)to(c)to(d)
$$
hold among the following statements

(a) $lVert T^* y^* rVert geq delta lVert y^* rVert$

(b) $delta V subset overline{T(U)}$

There're (c) and (d) as well, but I focus only on the first two.

Proof : Assume (a), and pick $y_0 notin overline{T(U)}$. Since $overline{T(U)}$ is convex, closed and balanced, theorem 3.7 shows that there's a $y^*$ such that $|<y,y^*>|leq 1$ for every $y in overline{T(U)}$, but $|<y_0,y^*>| > 1$. If $x in U$, it follows that
$$
left| leftlangle x,T^* y^* rightrangle right| = left|leftlangle Tx, y^* rightrangle right| leq 1
$$
Thus $lVert T^* y^* rVert leq 1$, and now (a) gives
$$
delta < delta left| leftlangle y_0,y^* rightrangle right| leq delta lVert y_0 rVert lVert y^* rVert leq lVert y_0 rVert lVert T^* y^* rVert leq lVert y_0 rVert.
$$
It follows $y in overline{T(U)}$ if $lVert y rVert leq delta$

First question, why the very last statement? namely why:

It follows $y in overline{T(U)}$ if $lVert y rVert leq delta$
?

For (b) part I'll just write the bit I don't get, not the full proof.

Next, assume (b). Take $delta = 1$, without loss of generality. Then $overline{V} subset overline{T(U)}$. To every $y in Y$ and every $epsilon > 0$ corresponds an $x in X$ with $lVert x rVert leq lVert y rVert$ and $lVert y - Tx rVert < epsilon$.

Basically I don't understand the full sentence after the period, does it follow from continuity of $T$ somehow?

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

share|cite|improve this question

asked Mar 11 at 10:44

user8469759user8469759

1,5581618

$endgroup$

add a comment | 

$begingroup$

Let $U$ and $V$ be the open unit balls in the Banach spaces $X$ and $Y$, respectively. If $T in mathcal{B}(X,Y)$ and $delta > 0$, the the implications
$$
(a)to(b)to(c)to(d)
$$
hold among the following statements

(a) $lVert T^* y^* rVert geq delta lVert y^* rVert$

(b) $delta V subset overline{T(U)}$

There're (c) and (d) as well, but I focus only on the first two.

Proof : Assume (a), and pick $y_0 notin overline{T(U)}$. Since $overline{T(U)}$ is convex, closed and balanced, theorem 3.7 shows that there's a $y^*$ such that $|<y,y^*>|leq 1$ for every $y in overline{T(U)}$, but $|<y_0,y^*>| > 1$. If $x in U$, it follows that
$$
left| leftlangle x,T^* y^* rightrangle right| = left|leftlangle Tx, y^* rightrangle right| leq 1
$$
Thus $lVert T^* y^* rVert leq 1$, and now (a) gives
$$
delta < delta left| leftlangle y_0,y^* rightrangle right| leq delta lVert y_0 rVert lVert y^* rVert leq lVert y_0 rVert lVert T^* y^* rVert leq lVert y_0 rVert.
$$
It follows $y in overline{T(U)}$ if $lVert y rVert leq delta$

First question, why the very last statement? namely why:

It follows $y in overline{T(U)}$ if $lVert y rVert leq delta$
?

For (b) part I'll just write the bit I don't get, not the full proof.

Next, assume (b). Take $delta = 1$, without loss of generality. Then $overline{V} subset overline{T(U)}$. To every $y in Y$ and every $epsilon > 0$ corresponds an $x in X$ with $lVert x rVert leq lVert y rVert$ and $lVert y - Tx rVert < epsilon$.

Basically I don't understand the full sentence after the period, does it follow from continuity of $T$ somehow?

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

share|cite|improve this question

asked Mar 11 at 10:44

user8469759user8469759

1,5581618

$endgroup$

share|cite|improve this question

asked Mar 11 at 10:44

user8469759user8469759

1,5581618

share|cite|improve this question

asked Mar 11 at 10:44

user8469759user8469759

1,5581618

asked Mar 11 at 10:44

user8469759user8469759

1,5581618

asked Mar 11 at 10:44

user8469759user8469759

1,5581618

1 Answer
1

active

oldest

votes

2

$begingroup$

In a) implies b) it is shown that $y_0 notin overline {T(U)}$ implies $delta <|y_0|$. The contrapositive of this statement is $|y| leq delta$ implies $y in overline {T(U)}$.

In b) implies a) if $y neq 0$ then $frac y {|y|} in overset {-} V$. Hence
$frac y {|y|} in overline {T(U)}$. Hence for every $epsilon >0$ there exists $x_1 in U$ such that $|frac y {|y|} -Tx_1| <epsilon /|y|$. Now take $x=|y| x_1$. Then $|y-Tx| <epsilon $ and $|x| leq |y|$.

share|cite|improve this answer

answered Mar 11 at 11:57

Kavi Rama MurthyKavi Rama Murthy

67.6k53067

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very subtle for a) implies b)..., shouldn't it be $lVert y_0 rVert leq delta$ implies $y_0 in overline{T(U)}$? As far as I know if $A Rightarrow B$ the contrapositive should be $overline{B} Rightarrow overline{A}$
  $endgroup$
  – user8469759
  Mar 11 at 13:16
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In these statements $y_0$ and $Y$ are 'dummy' variables. The meaning of the implication does not change if you change $y_0$ to any other symbol.
  $endgroup$
  – Kavi Rama Murthy
  Mar 11 at 23:16
  

  
  
  
  

add a comment | 

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2

$begingroup$

In a) implies b) it is shown that $y_0 notin overline {T(U)}$ implies $delta <|y_0|$. The contrapositive of this statement is $|y| leq delta$ implies $y in overline {T(U)}$.

In b) implies a) if $y neq 0$ then $frac y {|y|} in overset {-} V$. Hence
$frac y {|y|} in overline {T(U)}$. Hence for every $epsilon >0$ there exists $x_1 in U$ such that $|frac y {|y|} -Tx_1| <epsilon /|y|$. Now take $x=|y| x_1$. Then $|y-Tx| <epsilon $ and $|x| leq |y|$.

share|cite|improve this answer

answered Mar 11 at 11:57

Kavi Rama MurthyKavi Rama Murthy

67.6k53067

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very subtle for a) implies b)..., shouldn't it be $lVert y_0 rVert leq delta$ implies $y_0 in overline{T(U)}$? As far as I know if $A Rightarrow B$ the contrapositive should be $overline{B} Rightarrow overline{A}$
  $endgroup$
  – user8469759
  Mar 11 at 13:16
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In these statements $y_0$ and $Y$ are 'dummy' variables. The meaning of the implication does not change if you change $y_0$ to any other symbol.
  $endgroup$
  – Kavi Rama Murthy
  Mar 11 at 23:16
  

  
  
  
  

add a comment | 

2

$begingroup$

In a) implies b) it is shown that $y_0 notin overline {T(U)}$ implies $delta <|y_0|$. The contrapositive of this statement is $|y| leq delta$ implies $y in overline {T(U)}$.

In b) implies a) if $y neq 0$ then $frac y {|y|} in overset {-} V$. Hence
$frac y {|y|} in overline {T(U)}$. Hence for every $epsilon >0$ there exists $x_1 in U$ such that $|frac y {|y|} -Tx_1| <epsilon /|y|$. Now take $x=|y| x_1$. Then $|y-Tx| <epsilon $ and $|x| leq |y|$.

share|cite|improve this answer

answered Mar 11 at 11:57

Kavi Rama MurthyKavi Rama Murthy

67.6k53067

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very subtle for a) implies b)..., shouldn't it be $lVert y_0 rVert leq delta$ implies $y_0 in overline{T(U)}$? As far as I know if $A Rightarrow B$ the contrapositive should be $overline{B} Rightarrow overline{A}$
  $endgroup$
  – user8469759
  Mar 11 at 13:16
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In these statements $y_0$ and $Y$ are 'dummy' variables. The meaning of the implication does not change if you change $y_0$ to any other symbol.
  $endgroup$
  – Kavi Rama Murthy
  Mar 11 at 23:16
  

  
  
  
  

add a comment | 

$begingroup$

In a) implies b) it is shown that $y_0 notin overline {T(U)}$ implies $delta <|y_0|$. The contrapositive of this statement is $|y| leq delta$ implies $y in overline {T(U)}$.

In b) implies a) if $y neq 0$ then $frac y {|y|} in overset {-} V$. Hence
$frac y {|y|} in overline {T(U)}$. Hence for every $epsilon >0$ there exists $x_1 in U$ such that $|frac y {|y|} -Tx_1| <epsilon /|y|$. Now take $x=|y| x_1$. Then $|y-Tx| <epsilon $ and $|x| leq |y|$.

share|cite|improve this answer

answered Mar 11 at 11:57

Kavi Rama MurthyKavi Rama Murthy

67.6k53067

$endgroup$

share|cite|improve this answer

answered Mar 11 at 11:57

Kavi Rama MurthyKavi Rama Murthy

67.6k53067

answered Mar 11 at 11:57

Kavi Rama MurthyKavi Rama Murthy

67.6k53067

answered Mar 11 at 11:57

Kavi Rama MurthyKavi Rama Murthy

67.6k53067