norms of Hilbert space operators
$begingroup$
Let $A$ be a bounded linear operator on a complex Hilbert space $V$.
It is well known that
$$lVert ArVert=supleft{ frac{lVert AvrVert}{lVert vrVert};colon; vin Vtext{ with }vneq 0right}.$$
I want to understand why
$$begin{align*}
I &= inf{ c;colon; lVert AvrVertleq clVert vrVert text{ for all }vin V}\
&=supleft{ frac{lVert AvrVert}{lVert vrVert};colon; vin Vtext{ with }vneq 0right}.
end{align*}$$
Proof:
Now note that by definition of $|A|$ we have
$$ |Av| le |A| |v| quad forall v in V.$$
Then $I le |A|$.
On the other hand by definition of $sup$ we have
$$ I ge |Av_n| /|v_n| ge |A| - 1/n quad forall n.$$
Then $|A| = I$.
I don't understand why
$$I ge |Av_n| /|v_n|?$$
functional-analysis proof-explanation
asked Dec 14 '18 at 9:19
SchülerSchüler
1,5301421
$endgroup$
add a comment |
$begingroup$
Let $A$ be a bounded linear operator on a complex Hilbert space $V$.
It is well known that
$$lVert ArVert=supleft{ frac{lVert AvrVert}{lVert vrVert};colon; vin Vtext{ with }vneq 0right}.$$
I want to understand why
$$begin{align*}
I &= inf{ c;colon; lVert AvrVertleq clVert vrVert text{ for all }vin V}\
&=supleft{ frac{lVert AvrVert}{lVert vrVert};colon; vin Vtext{ with }vneq 0right}.
end{align*}$$
Proof:
Now note that by definition of $|A|$ we have
$$ |Av| le |A| |v| quad forall v in V.$$
Then $I le |A|$.
On the other hand by definition of $sup$ we have
$$ I ge |Av_n| /|v_n| ge |A| - 1/n quad forall n.$$
Then $|A| = I$.
I don't understand why
$$I ge |Av_n| /|v_n|?$$
functional-analysis proof-explanation
asked Dec 14 '18 at 9:19
SchülerSchüler
1,5301421
$endgroup$
add a comment |
$begingroup$
Let $A$ be a bounded linear operator on a complex Hilbert space $V$.
It is well known that
$$lVert ArVert=supleft{ frac{lVert AvrVert}{lVert vrVert};colon; vin Vtext{ with }vneq 0right}.$$
I want to understand why
$$begin{align*}
I &= inf{ c;colon; lVert AvrVertleq clVert vrVert text{ for all }vin V}\
&=supleft{ frac{lVert AvrVert}{lVert vrVert};colon; vin Vtext{ with }vneq 0right}.
end{align*}$$
Proof:
Now note that by definition of $|A|$ we have
$$ |Av| le |A| |v| quad forall v in V.$$
Then $I le |A|$.
On the other hand by definition of $sup$ we have
$$ I ge |Av_n| /|v_n| ge |A| - 1/n quad forall n.$$
Then $|A| = I$.
I don't understand why
$$I ge |Av_n| /|v_n|?$$
functional-analysis proof-explanation
asked Dec 14 '18 at 9:19
SchülerSchüler
1,5301421
$endgroup$
Let $A$ be a bounded linear operator on a complex Hilbert space $V$.
It is well known that
$$lVert ArVert=supleft{ frac{lVert AvrVert}{lVert vrVert};colon; vin Vtext{ with }vneq 0right}.$$
I want to understand why
$$begin{align*}
I &= inf{ c;colon; lVert AvrVertleq clVert vrVert text{ for all }vin V}\
&=supleft{ frac{lVert AvrVert}{lVert vrVert};colon; vin Vtext{ with }vneq 0right}.
end{align*}$$
Proof:
Now note that by definition of $|A|$ we have
$$ |Av| le |A| |v| quad forall v in V.$$
Then $I le |A|$.
On the other hand by definition of $sup$ we have
$$ I ge |Av_n| /|v_n| ge |A| - 1/n quad forall n.$$
Then $|A| = I$.
I don't understand why
$$I ge |Av_n| /|v_n|?$$
functional-analysis proof-explanation
functional-analysis proof-explanation
asked Dec 14 '18 at 9:19
SchülerSchüler
1,5301421
asked Dec 14 '18 at 9:19
SchülerSchüler
1,5301421
asked Dec 14 '18 at 9:19
SchülerSchüler
1,5301421
asked Dec 14 '18 at 9:19
SchülerSchüler
1,5301421
asked Dec 14 '18 at 9:19
SchülerSchüler
1,5301421
1,5301421
add a comment |
add a comment |
1 Answer
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$begingroup$
$I+frac 1 n> c_n$ for some $c_n$ with $|Av| leq c_n|v|$ for all $v$. Hence $I+frac 1 n> frac {|Av|} {|v|}$. Since this is true for all $v neq 0$ we get $I+frac 1 n>sup { frac {{|Av|}} {{|v|}}: v neq 0}$. Now let $n to infty$.
answered Dec 14 '18 at 9:25
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
add a comment |
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$begingroup$
$I+frac 1 n> c_n$ for some $c_n$ with $|Av| leq c_n|v|$ for all $v$. Hence $I+frac 1 n> frac {|Av|} {|v|}$. Since this is true for all $v neq 0$ we get $I+frac 1 n>sup { frac {{|Av|}} {{|v|}}: v neq 0}$. Now let $n to infty$.
answered Dec 14 '18 at 9:25
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
add a comment |
$begingroup$
$I+frac 1 n> c_n$ for some $c_n$ with $|Av| leq c_n|v|$ for all $v$. Hence $I+frac 1 n> frac {|Av|} {|v|}$. Since this is true for all $v neq 0$ we get $I+frac 1 n>sup { frac {{|Av|}} {{|v|}}: v neq 0}$. Now let $n to infty$.
answered Dec 14 '18 at 9:25
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
add a comment |
$begingroup$
$I+frac 1 n> c_n$ for some $c_n$ with $|Av| leq c_n|v|$ for all $v$. Hence $I+frac 1 n> frac {|Av|} {|v|}$. Since this is true for all $v neq 0$ we get $I+frac 1 n>sup { frac {{|Av|}} {{|v|}}: v neq 0}$. Now let $n to infty$.
answered Dec 14 '18 at 9:25
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
$I+frac 1 n> c_n$ for some $c_n$ with $|Av| leq c_n|v|$ for all $v$. Hence $I+frac 1 n> frac {|Av|} {|v|}$. Since this is true for all $v neq 0$ we get $I+frac 1 n>sup { frac {{|Av|}} {{|v|}}: v neq 0}$. Now let $n to infty$.
answered Dec 14 '18 at 9:25
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
answered Dec 14 '18 at 9:25
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
answered Dec 14 '18 at 9:25
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
answered Dec 14 '18 at 9:25
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
60.6k42161
add a comment |
add a comment |
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