Why $sup_hint |m(x)|^2|h(x)|dx=sup__L^2leq 1int |Th(x)|^2dx$?A specific linear operator between Banach spacesDefining a bounded operator on $l^p$$phi : L^1(A) to mathbbC$ where $f mapsto int gf$ for $g in L^infty(A)$ for $A subseteq mathbbR$ compact isometric isomorphismBounded linear operator, why do we only consider points in the unit circleWhy is the linear functional $F:C[a,b] rightarrow mathbbR$ such that $F(f)=int f dmu$ continuous?Computing the $L^2$ norm of a linear operator over a Hilbert spaceFor $f$ that is integrable on any set of finite measure $|f|_p=sup__p'=1left|int fgright|$.Help showing a linear functional is boundedWhy an operator $A:D(A)to F$ is called “unbounded”?Showing the follow linear operator is bounded
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Why $sup_int |m(x)|^2|h(x)|dx=sup_hint |Th(x)|^2dx$?
A specific linear operator between Banach spacesDefining a bounded operator on $l^p$$phi : L^1(A) to mathbbC$ where $f mapsto int gf$ for $g in L^infty(A)$ for $A subseteq mathbbR$ compact isometric isomorphismBounded linear operator, why do we only consider points in the unit circleWhy is the linear functional $F:C[a,b] rightarrow mathbbR$ such that $F(f)=int f dmu$ continuous?Computing the $L^2$ norm of a linear operator over a Hilbert spaceFor $f$ that is integrable on any set of finite measure $|f|_p=sup__p'=1left|int fgright|$.Help showing a linear functional is boundedWhy an operator $A:D(A)to F$ is called “unbounded”?Showing the follow linear operator is bounded
$begingroup$
Let $T:L^2to L^2$ a bounded linear operator defined by $Tg(x)=n(x)g(x)$ where $min L^infty (mathbb R^n)$.
I don't see why
$$sup_int |m(x)|^2|h(x)|dx=sup_int |Ttilde h(x)|^2dxleq |T|_L^2to L^2.$$
The first inequality looks to come from nowhere. For the second, $$int|Ttilde h|^2=|Th|_L^2^2$$ and thus $$sup_hint |Th|^2leqsup_h |Th|_L^2^2,$$
but it's not $sup__L^2leq 2|Th|_L^2$... any idea ?
functional-analysis
asked Mar 28 at 20:07
user657324user657324
3509
New contributor
user657324 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let $T:L^2to L^2$ a bounded linear operator defined by $Tg(x)=n(x)g(x)$ where $min L^infty (mathbb R^n)$.
I don't see why
$$sup_int |m(x)|^2|h(x)|dx=sup_int |Ttilde h(x)|^2dxleq |T|_L^2to L^2.$$
The first inequality looks to come from nowhere. For the second, $$int|Ttilde h|^2=|Th|_L^2^2$$ and thus $$sup_hint |Th|^2leqsup_h |Th|_L^2^2,$$
but it's not $sup__L^2leq 2|Th|_L^2$... any idea ?
functional-analysis
asked Mar 28 at 20:07
user657324user657324
3509
New contributor
user657324 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
For the second, the thing to notice is that the norm in question is an operator norm $|T|_L^2 to L^2$.
$endgroup$
– Gary Moon
Mar 28 at 21:39
$begingroup$
@GaryMoon: And so ?
$endgroup$
– user657324
Mar 28 at 21:42
$begingroup$
One typically defines $|T|_textop = sup_ |Tv|$. That definition should explain the second inequality.
$endgroup$
– Gary Moon
Mar 28 at 21:48
add a comment |
$begingroup$
Let $T:L^2to L^2$ a bounded linear operator defined by $Tg(x)=n(x)g(x)$ where $min L^infty (mathbb R^n)$.
I don't see why
$$sup_int |m(x)|^2|h(x)|dx=sup_int |Ttilde h(x)|^2dxleq |T|_L^2to L^2.$$
The first inequality looks to come from nowhere. For the second, $$int|Ttilde h|^2=|Th|_L^2^2$$ and thus $$sup_hint |Th|^2leqsup_h |Th|_L^2^2,$$
but it's not $sup__L^2leq 2|Th|_L^2$... any idea ?
functional-analysis
asked Mar 28 at 20:07
user657324user657324
3509
New contributor
user657324 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let $T:L^2to L^2$ a bounded linear operator defined by $Tg(x)=n(x)g(x)$ where $min L^infty (mathbb R^n)$.
I don't see why
$$sup_int |m(x)|^2|h(x)|dx=sup_int |Ttilde h(x)|^2dxleq |T|_L^2to L^2.$$
The first inequality looks to come from nowhere. For the second, $$int|Ttilde h|^2=|Th|_L^2^2$$ and thus $$sup_hint |Th|^2leqsup_h |Th|_L^2^2,$$
but it's not $sup__L^2leq 2|Th|_L^2$... any idea ?
functional-analysis
functional-analysis
asked Mar 28 at 20:07
user657324user657324
3509
New contributor
user657324 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 20:07
user657324user657324
3509
New contributor
user657324 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 20:07
user657324user657324
3509
New contributor
user657324 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 20:07
user657324user657324
3509
asked Mar 28 at 20:07
user657324user657324
3509
3509
New contributor
user657324 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user657324 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user657324 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
For the second, the thing to notice is that the norm in question is an operator norm $|T|_L^2 to L^2$.
$endgroup$
– Gary Moon
Mar 28 at 21:39
$begingroup$
@GaryMoon: And so ?
$endgroup$
– user657324
Mar 28 at 21:42
$begingroup$
One typically defines $|T|_textop = sup_ |Tv|$. That definition should explain the second inequality.
$endgroup$
– Gary Moon
Mar 28 at 21:48
add a comment |
$begingroup$
For the second, the thing to notice is that the norm in question is an operator norm $|T|_L^2 to L^2$.
$endgroup$
– Gary Moon
Mar 28 at 21:39
$begingroup$
@GaryMoon: And so ?
$endgroup$
– user657324
Mar 28 at 21:42
$begingroup$
One typically defines $|T|_textop = sup_ |Tv|$. That definition should explain the second inequality.
$endgroup$
– Gary Moon
Mar 28 at 21:48
$begingroup$
For the second, the thing to notice is that the norm in question is an operator norm $|T|_L^2 to L^2$.
$endgroup$
– Gary Moon
Mar 28 at 21:39
$begingroup$
For the second, the thing to notice is that the norm in question is an operator norm $|T|_L^2 to L^2$.
$endgroup$
– Gary Moon
Mar 28 at 21:39
$begingroup$
@GaryMoon: And so ?
$endgroup$
– user657324
Mar 28 at 21:42
$begingroup$
@GaryMoon: And so ?
$endgroup$
– user657324
Mar 28 at 21:42
$begingroup$
One typically defines $|T|_textop = sup_ |Tv|$. That definition should explain the second inequality.
$endgroup$
– Gary Moon
Mar 28 at 21:48
$begingroup$
One typically defines $|T|_textop = sup_ |Tv|$. That definition should explain the second inequality.
$endgroup$
– Gary Moon
Mar 28 at 21:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For the second inequality, note that the norm in question is $|T|_L^2to L^2$, where we can define
$$|T|_L^2to L^2 = sup_tildeh |Ttildeh|_L^2.$$
For the first inequality, I imagine it's quite a simple move, perhaps something like taking $tildeh = h^frac12$.
answered Mar 28 at 21:59
Gary MoonGary Moon
92127
$endgroup$
$begingroup$
but we have $sup |Th|_L^2^2$ not $sup |Th|_L^2$. How can you pass from the first to the second one ? in other words why $sup |Th|_L^2^2leq sup |Th|_L^2$ ?
$endgroup$
– user657324
Mar 28 at 22:00
$begingroup$
The restriction on the $L^2$ norm of $tildeh$ (plus the structure of $T$) should allow for the given comparison.
$endgroup$
– Gary Moon
Mar 28 at 22:07
add a comment |
Your Answer
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1 Answer
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oldest
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
For the second inequality, note that the norm in question is $|T|_L^2to L^2$, where we can define
$$|T|_L^2to L^2 = sup_tildeh |Ttildeh|_L^2.$$
For the first inequality, I imagine it's quite a simple move, perhaps something like taking $tildeh = h^frac12$.
answered Mar 28 at 21:59
Gary MoonGary Moon
92127
$endgroup$
$begingroup$
but we have $sup |Th|_L^2^2$ not $sup |Th|_L^2$. How can you pass from the first to the second one ? in other words why $sup |Th|_L^2^2leq sup |Th|_L^2$ ?
$endgroup$
– user657324
Mar 28 at 22:00
$begingroup$
The restriction on the $L^2$ norm of $tildeh$ (plus the structure of $T$) should allow for the given comparison.
$endgroup$
– Gary Moon
Mar 28 at 22:07
add a comment |
$begingroup$
For the second inequality, note that the norm in question is $|T|_L^2to L^2$, where we can define
$$|T|_L^2to L^2 = sup_tildeh |Ttildeh|_L^2.$$
For the first inequality, I imagine it's quite a simple move, perhaps something like taking $tildeh = h^frac12$.
answered Mar 28 at 21:59
Gary MoonGary Moon
92127
$endgroup$
$begingroup$
but we have $sup |Th|_L^2^2$ not $sup |Th|_L^2$. How can you pass from the first to the second one ? in other words why $sup |Th|_L^2^2leq sup |Th|_L^2$ ?
$endgroup$
– user657324
Mar 28 at 22:00
$begingroup$
The restriction on the $L^2$ norm of $tildeh$ (plus the structure of $T$) should allow for the given comparison.
$endgroup$
– Gary Moon
Mar 28 at 22:07
add a comment |
$begingroup$
For the second inequality, note that the norm in question is $|T|_L^2to L^2$, where we can define
$$|T|_L^2to L^2 = sup_tildeh |Ttildeh|_L^2.$$
For the first inequality, I imagine it's quite a simple move, perhaps something like taking $tildeh = h^frac12$.
answered Mar 28 at 21:59
Gary MoonGary Moon
92127
$endgroup$
For the second inequality, note that the norm in question is $|T|_L^2to L^2$, where we can define
$$|T|_L^2to L^2 = sup_tildeh |Ttildeh|_L^2.$$
For the first inequality, I imagine it's quite a simple move, perhaps something like taking $tildeh = h^frac12$.
answered Mar 28 at 21:59
Gary MoonGary Moon
92127
edited Mar 28 at 22:05
answered Mar 28 at 21:59
Gary MoonGary Moon
92127
answered Mar 28 at 21:59
Gary MoonGary Moon
92127
answered Mar 28 at 21:59
Gary MoonGary Moon
92127
92127
$begingroup$
but we have $sup |Th|_L^2^2$ not $sup |Th|_L^2$. How can you pass from the first to the second one ? in other words why $sup |Th|_L^2^2leq sup |Th|_L^2$ ?
$endgroup$
– user657324
Mar 28 at 22:00
$begingroup$
The restriction on the $L^2$ norm of $tildeh$ (plus the structure of $T$) should allow for the given comparison.
$endgroup$
– Gary Moon
Mar 28 at 22:07
add a comment |
$begingroup$
but we have $sup |Th|_L^2^2$ not $sup |Th|_L^2$. How can you pass from the first to the second one ? in other words why $sup |Th|_L^2^2leq sup |Th|_L^2$ ?
$endgroup$
– user657324
Mar 28 at 22:00
$begingroup$
The restriction on the $L^2$ norm of $tildeh$ (plus the structure of $T$) should allow for the given comparison.
$endgroup$
– Gary Moon
Mar 28 at 22:07
$begingroup$
but we have $sup |Th|_L^2^2$ not $sup |Th|_L^2$. How can you pass from the first to the second one ? in other words why $sup |Th|_L^2^2leq sup |Th|_L^2$ ?
$endgroup$
– user657324
Mar 28 at 22:00
$begingroup$
but we have $sup |Th|_L^2^2$ not $sup |Th|_L^2$. How can you pass from the first to the second one ? in other words why $sup |Th|_L^2^2leq sup |Th|_L^2$ ?
$endgroup$
– user657324
Mar 28 at 22:00
$begingroup$
The restriction on the $L^2$ norm of $tildeh$ (plus the structure of $T$) should allow for the given comparison.
$endgroup$
– Gary Moon
Mar 28 at 22:07
$begingroup$
The restriction on the $L^2$ norm of $tildeh$ (plus the structure of $T$) should allow for the given comparison.
$endgroup$
– Gary Moon
Mar 28 at 22:07
add a comment |
user657324 is a new contributor. Be nice, and check out our Code of Conduct.
user657324 is a new contributor. Be nice, and check out our Code of Conduct.
user657324 is a new contributor. Be nice, and check out our Code of Conduct.
user657324 is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
For the second, the thing to notice is that the norm in question is an operator norm $|T|_L^2 to L^2$.
$endgroup$
– Gary Moon
Mar 28 at 21:39
$begingroup$
@GaryMoon: And so ?
$endgroup$
– user657324
Mar 28 at 21:42
$begingroup$
One typically defines $|T|_textop = sup_ |Tv|$. That definition should explain the second inequality.
$endgroup$
– Gary Moon
Mar 28 at 21:48