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Inequality of norms for $H^infty$ and $L^infty$ spaces (part of a theorem on the Rudins' book Real and Complex Analysis)
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In Rudin's book Real and Complex Analysis, in the eleventh chapter there is a theorem that states:
To every $fin H^infty$ corresponds a function $f^*in L^infty (T),$ defined almost everywhere by $$f^*(e^itheta)=lim_rto 1^-f(re^itheta).$$ The equality $leftVert frightVert_infty=leftVert f^*rightVert_infty$ holds.
In the proof of the theorem for the equality $leftVert frightVert_infty=leftVert f^*rightVert_infty$ there is stated that $leftVert frightVert_inftygeqleftVert f^*rightVert_infty$ is obvious. My question is how to prove that $leftVert frightVert_inftygeqleftVert f^*rightVert_infty$ holds?
In this question $H^infty$ is the space of all holomorphic functions in the open unit ball $U$ equipped with the supremum norm, while $L^infty$ is the space of all essentially bounded functions on $T=partial U$ normed by the essential supremum norm, relative to the Lebesgue measure, so $$leftVert f^*rightVert_infty=infleft alphavert mleft( left( f^* right)^-1 left( left[ alpha,infty right)right) right)=0 right$$ for $f^*:Tto left[0, infty right).$
My attempt: Let suppose that $leftVert frightVert_infty<leftVert f^*rightVert_infty.$ Then there exists an $alpha '=infleft alphavert mleft( left( f^* right)^-1 left( left[ alpha,infty right)right) right)=0 right$ s.t. $alpha'>Vert fVert_infty$, so $mleft( left( f^* right)^-1 left( left[ Vert fVert_infty,infty right)right)right)>0.$
How should I continue until a contradiction?
complex-analysis holomorphic-functions
edited Jul 25 at 14:16
Bob
1,517522
asked Jul 25 at 11:05
Emin
1,27621330
 |Â
show 1 more comment
up vote
0
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In Rudin's book Real and Complex Analysis, in the eleventh chapter there is a theorem that states:
To every $fin H^infty$ corresponds a function $f^*in L^infty (T),$ defined almost everywhere by $$f^*(e^itheta)=lim_rto 1^-f(re^itheta).$$ The equality $leftVert frightVert_infty=leftVert f^*rightVert_infty$ holds.
In the proof of the theorem for the equality $leftVert frightVert_infty=leftVert f^*rightVert_infty$ there is stated that $leftVert frightVert_inftygeqleftVert f^*rightVert_infty$ is obvious. My question is how to prove that $leftVert frightVert_inftygeqleftVert f^*rightVert_infty$ holds?
In this question $H^infty$ is the space of all holomorphic functions in the open unit ball $U$ equipped with the supremum norm, while $L^infty$ is the space of all essentially bounded functions on $T=partial U$ normed by the essential supremum norm, relative to the Lebesgue measure, so $$leftVert f^*rightVert_infty=infleft alphavert mleft( left( f^* right)^-1 left( left[ alpha,infty right)right) right)=0 right$$ for $f^*:Tto left[0, infty right).$
My attempt: Let suppose that $leftVert frightVert_infty<leftVert f^*rightVert_infty.$ Then there exists an $alpha '=infleft alphavert mleft( left( f^* right)^-1 left( left[ alpha,infty right)right) right)=0 right$ s.t. $alpha'>Vert fVert_infty$, so $mleft( left( f^* right)^-1 left( left[ Vert fVert_infty,infty right)right)right)>0.$
How should I continue until a contradiction?
complex-analysis holomorphic-functions
edited Jul 25 at 14:16
Bob
1,517522
asked Jul 25 at 11:05
Emin
1,27621330
Use the Poisson integral
– user578878
Jul 25 at 11:18
2
Just do it directly. Since $lvert f(re^it)rvert leqslant lVert frVert_infty$ for all $t$ and all $r in [0,1)$, it follows that if the radial limit exists at some point, we have $lvert f^ast(e^it)rvert leqslant lVert frVert_infty$ there. Since the set where the radial limit doesn't exist is a null set, we're done.
– Daniel Fischer♦
Jul 25 at 11:55
@DanielFischer how do you connect $vert f^*(e^it)vert$ with $Vert f^*Vert_infty$?
– Emin
Jul 25 at 12:31
1
Let $C$ be the set of points on the unit circle where the radial limit exists. For all $t$ with $e^it in C$ we know $lvert f^ast(e^it)rvert leqslant lVert frVert_infty$. The complement of $C$ has measure $0$, so $mbigl((f^ast)^-1([alpha,+infty))bigr) = 0$ for all $alpha > lVert frVert_infty$.
– Daniel Fischer♦
Jul 25 at 12:36
@DanielFischer Thanks mate!
– Emin
Jul 25 at 13:40
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In Rudin's book Real and Complex Analysis, in the eleventh chapter there is a theorem that states:
To every $fin H^infty$ corresponds a function $f^*in L^infty (T),$ defined almost everywhere by $$f^*(e^itheta)=lim_rto 1^-f(re^itheta).$$ The equality $leftVert frightVert_infty=leftVert f^*rightVert_infty$ holds.
In the proof of the theorem for the equality $leftVert frightVert_infty=leftVert f^*rightVert_infty$ there is stated that $leftVert frightVert_inftygeqleftVert f^*rightVert_infty$ is obvious. My question is how to prove that $leftVert frightVert_inftygeqleftVert f^*rightVert_infty$ holds?
In this question $H^infty$ is the space of all holomorphic functions in the open unit ball $U$ equipped with the supremum norm, while $L^infty$ is the space of all essentially bounded functions on $T=partial U$ normed by the essential supremum norm, relative to the Lebesgue measure, so $$leftVert f^*rightVert_infty=infleft alphavert mleft( left( f^* right)^-1 left( left[ alpha,infty right)right) right)=0 right$$ for $f^*:Tto left[0, infty right).$
My attempt: Let suppose that $leftVert frightVert_infty<leftVert f^*rightVert_infty.$ Then there exists an $alpha '=infleft alphavert mleft( left( f^* right)^-1 left( left[ alpha,infty right)right) right)=0 right$ s.t. $alpha'>Vert fVert_infty$, so $mleft( left( f^* right)^-1 left( left[ Vert fVert_infty,infty right)right)right)>0.$
How should I continue until a contradiction?
complex-analysis holomorphic-functions
edited Jul 25 at 14:16
Bob
1,517522
asked Jul 25 at 11:05
Emin
1,27621330
In Rudin's book Real and Complex Analysis, in the eleventh chapter there is a theorem that states:
To every $fin H^infty$ corresponds a function $f^*in L^infty (T),$ defined almost everywhere by $$f^*(e^itheta)=lim_rto 1^-f(re^itheta).$$ The equality $leftVert frightVert_infty=leftVert f^*rightVert_infty$ holds.
In the proof of the theorem for the equality $leftVert frightVert_infty=leftVert f^*rightVert_infty$ there is stated that $leftVert frightVert_inftygeqleftVert f^*rightVert_infty$ is obvious. My question is how to prove that $leftVert frightVert_inftygeqleftVert f^*rightVert_infty$ holds?
In this question $H^infty$ is the space of all holomorphic functions in the open unit ball $U$ equipped with the supremum norm, while $L^infty$ is the space of all essentially bounded functions on $T=partial U$ normed by the essential supremum norm, relative to the Lebesgue measure, so $$leftVert f^*rightVert_infty=infleft alphavert mleft( left( f^* right)^-1 left( left[ alpha,infty right)right) right)=0 right$$ for $f^*:Tto left[0, infty right).$
My attempt: Let suppose that $leftVert frightVert_infty<leftVert f^*rightVert_infty.$ Then there exists an $alpha '=infleft alphavert mleft( left( f^* right)^-1 left( left[ alpha,infty right)right) right)=0 right$ s.t. $alpha'>Vert fVert_infty$, so $mleft( left( f^* right)^-1 left( left[ Vert fVert_infty,infty right)right)right)>0.$
How should I continue until a contradiction?
complex-analysis holomorphic-functions
edited Jul 25 at 14:16
Bob
1,517522
asked Jul 25 at 11:05
Emin
1,27621330
edited Jul 25 at 14:16
Bob
1,517522
edited Jul 25 at 14:16
Bob
1,517522
edited Jul 25 at 14:16
Bob
1,517522
1,517522
asked Jul 25 at 11:05
Emin
1,27621330
asked Jul 25 at 11:05
Emin
1,27621330
asked Jul 25 at 11:05
Emin
1,27621330
1,27621330
Use the Poisson integral
– user578878
Jul 25 at 11:18
2
Just do it directly. Since $lvert f(re^it)rvert leqslant lVert frVert_infty$ for all $t$ and all $r in [0,1)$, it follows that if the radial limit exists at some point, we have $lvert f^ast(e^it)rvert leqslant lVert frVert_infty$ there. Since the set where the radial limit doesn't exist is a null set, we're done.
– Daniel Fischer♦
Jul 25 at 11:55
@DanielFischer how do you connect $vert f^*(e^it)vert$ with $Vert f^*Vert_infty$?
– Emin
Jul 25 at 12:31
1
Let $C$ be the set of points on the unit circle where the radial limit exists. For all $t$ with $e^it in C$ we know $lvert f^ast(e^it)rvert leqslant lVert frVert_infty$. The complement of $C$ has measure $0$, so $mbigl((f^ast)^-1([alpha,+infty))bigr) = 0$ for all $alpha > lVert frVert_infty$.
– Daniel Fischer♦
Jul 25 at 12:36
@DanielFischer Thanks mate!
– Emin
Jul 25 at 13:40
 |Â
show 1 more comment
Use the Poisson integral
– user578878
Jul 25 at 11:18
2
Just do it directly. Since $lvert f(re^it)rvert leqslant lVert frVert_infty$ for all $t$ and all $r in [0,1)$, it follows that if the radial limit exists at some point, we have $lvert f^ast(e^it)rvert leqslant lVert frVert_infty$ there. Since the set where the radial limit doesn't exist is a null set, we're done.
– Daniel Fischer♦
Jul 25 at 11:55
@DanielFischer how do you connect $vert f^*(e^it)vert$ with $Vert f^*Vert_infty$?
– Emin
Jul 25 at 12:31
1
Let $C$ be the set of points on the unit circle where the radial limit exists. For all $t$ with $e^it in C$ we know $lvert f^ast(e^it)rvert leqslant lVert frVert_infty$. The complement of $C$ has measure $0$, so $mbigl((f^ast)^-1([alpha,+infty))bigr) = 0$ for all $alpha > lVert frVert_infty$.
– Daniel Fischer♦
Jul 25 at 12:36
@DanielFischer Thanks mate!
– Emin
Jul 25 at 13:40
Use the Poisson integral
– user578878
Jul 25 at 11:18
Use the Poisson integral
– user578878
Jul 25 at 11:18
2
2
Just do it directly. Since $lvert f(re^it)rvert leqslant lVert frVert_infty$ for all $t$ and all $r in [0,1)$, it follows that if the radial limit exists at some point, we have $lvert f^ast(e^it)rvert leqslant lVert frVert_infty$ there. Since the set where the radial limit doesn't exist is a null set, we're done.
– Daniel Fischer♦
Jul 25 at 11:55
Just do it directly. Since $lvert f(re^it)rvert leqslant lVert frVert_infty$ for all $t$ and all $r in [0,1)$, it follows that if the radial limit exists at some point, we have $lvert f^ast(e^it)rvert leqslant lVert frVert_infty$ there. Since the set where the radial limit doesn't exist is a null set, we're done.
– Daniel Fischer♦
Jul 25 at 11:55
@DanielFischer how do you connect $vert f^*(e^it)vert$ with $Vert f^*Vert_infty$?
– Emin
Jul 25 at 12:31
@DanielFischer how do you connect $vert f^*(e^it)vert$ with $Vert f^*Vert_infty$?
– Emin
Jul 25 at 12:31
1
1
Let $C$ be the set of points on the unit circle where the radial limit exists. For all $t$ with $e^it in C$ we know $lvert f^ast(e^it)rvert leqslant lVert frVert_infty$. The complement of $C$ has measure $0$, so $mbigl((f^ast)^-1([alpha,+infty))bigr) = 0$ for all $alpha > lVert frVert_infty$.
– Daniel Fischer♦
Jul 25 at 12:36
Let $C$ be the set of points on the unit circle where the radial limit exists. For all $t$ with $e^it in C$ we know $lvert f^ast(e^it)rvert leqslant lVert frVert_infty$. The complement of $C$ has measure $0$, so $mbigl((f^ast)^-1([alpha,+infty))bigr) = 0$ for all $alpha > lVert frVert_infty$.
– Daniel Fischer♦
Jul 25 at 12:36
@DanielFischer Thanks mate!
– Emin
Jul 25 at 13:40
@DanielFischer Thanks mate!
– Emin
Jul 25 at 13:40
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Let $C subset T$ be the set where the radial limit
$$lim_r to 1^- f(re^itheta)$$
exists. By theorems 11.30 and 11.23, $m(Tsetminus C) = 0$, and $f$ is the Poisson integral of $f^ast$. On $Tsetminus C$, we can define $f^ast$ arbitrarily if we want an everywhere defined function $f^ast$.
Since $lvert f(re^itheta)rvert leqslant lVert frVert_infty$ for $r in [0,1)$, we have
$$lvert f^ast(e^itheta)rvert = lim_r to 1^-: lvert f(re^itheta)rvert leqslant lVert frVert_infty$$
for all $e^itheta in C$.
Let $alpha > lVert frVert_infty$. Then
$$bigl e^itheta : lvert f^ast(e^itheta)rvert geqslant alpha bigr subset Tsetminus C,,$$
so this is a null set. Hence
$$(lVert frVert_infty, +infty) subset bigl alpha in mathbbR : mbigl(bigl(lvert f^astrvertbigr)^-1([alpha, +infty))bigr) = 0bigr$$
and consequently
$$lVert f^astrVert_infty = inf: bigl alpha in mathbbR : mbigl(bigl(lvert f^astrvertbigr)^-1([alpha, +infty))bigr) = 0bigr leqslant inf: (lVert frVert_infty, +infty) = lVert frVert_infty,.$$
[Note that $f^ast$ is complex-valued, so we must consider $bigl(lvert f^astrvertbigr)^-1([alpha,+infty))$ and not $bigl(f^astbigr)^-1([alpha,+infty))$.]
answered Jul 25 at 19:51
Daniel Fischer♦
171k16154274
I think that the notation $left( f^* right)^-1 left( left[ alpha,infty right)right)=0$ is correct since the Lebesgue measure is defined on T.
– Emin
Jul 25 at 20:40
But we must look at the codomain. We need to know whether $bigl e^itheta : lvert f^ast(e^itheta)rvert geqslant alphabigr$ is a null set. The set $bigl(f^astbigr)^-1([alpha,+infty))$ is $bigl e^itheta : f^ast(e^itheta) in mathbbR land f^ast(e^itheta) geqslant alpha bigr$. But unless $f^ast$ is real-valued and takes only non-negative values, that set can have measure $0$ also for $alpha < lVert f^astrVert_infty$.
– Daniel Fischer♦
Jul 25 at 20:48
$f^*$ is given with $f^*:Tto left[0, infty right).$
– Emin
Jul 27 at 19:11
$H^infty$ is the space of bounded holomorphic functions on the unit disk, and for $f in H^infty$, $f^ast$ is defined as the radial limit function of $f$. E.g. for $f(z) = z$ we have $f^ast(e^itheta) = e^itheta$. Occasionally the image of $f^ast$ is contained in $[0,infty)$, but typically it isn't.
– Daniel Fischer♦
Jul 27 at 19:46
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $C subset T$ be the set where the radial limit
$$lim_r to 1^- f(re^itheta)$$
exists. By theorems 11.30 and 11.23, $m(Tsetminus C) = 0$, and $f$ is the Poisson integral of $f^ast$. On $Tsetminus C$, we can define $f^ast$ arbitrarily if we want an everywhere defined function $f^ast$.
Since $lvert f(re^itheta)rvert leqslant lVert frVert_infty$ for $r in [0,1)$, we have
$$lvert f^ast(e^itheta)rvert = lim_r to 1^-: lvert f(re^itheta)rvert leqslant lVert frVert_infty$$
for all $e^itheta in C$.
Let $alpha > lVert frVert_infty$. Then
$$bigl e^itheta : lvert f^ast(e^itheta)rvert geqslant alpha bigr subset Tsetminus C,,$$
so this is a null set. Hence
$$(lVert frVert_infty, +infty) subset bigl alpha in mathbbR : mbigl(bigl(lvert f^astrvertbigr)^-1([alpha, +infty))bigr) = 0bigr$$
and consequently
$$lVert f^astrVert_infty = inf: bigl alpha in mathbbR : mbigl(bigl(lvert f^astrvertbigr)^-1([alpha, +infty))bigr) = 0bigr leqslant inf: (lVert frVert_infty, +infty) = lVert frVert_infty,.$$
[Note that $f^ast$ is complex-valued, so we must consider $bigl(lvert f^astrvertbigr)^-1([alpha,+infty))$ and not $bigl(f^astbigr)^-1([alpha,+infty))$.]
answered Jul 25 at 19:51
Daniel Fischer♦
171k16154274
I think that the notation $left( f^* right)^-1 left( left[ alpha,infty right)right)=0$ is correct since the Lebesgue measure is defined on T.
– Emin
Jul 25 at 20:40
But we must look at the codomain. We need to know whether $bigl e^itheta : lvert f^ast(e^itheta)rvert geqslant alphabigr$ is a null set. The set $bigl(f^astbigr)^-1([alpha,+infty))$ is $bigl e^itheta : f^ast(e^itheta) in mathbbR land f^ast(e^itheta) geqslant alpha bigr$. But unless $f^ast$ is real-valued and takes only non-negative values, that set can have measure $0$ also for $alpha < lVert f^astrVert_infty$.
– Daniel Fischer♦
Jul 25 at 20:48
$f^*$ is given with $f^*:Tto left[0, infty right).$
– Emin
Jul 27 at 19:11
$H^infty$ is the space of bounded holomorphic functions on the unit disk, and for $f in H^infty$, $f^ast$ is defined as the radial limit function of $f$. E.g. for $f(z) = z$ we have $f^ast(e^itheta) = e^itheta$. Occasionally the image of $f^ast$ is contained in $[0,infty)$, but typically it isn't.
– Daniel Fischer♦
Jul 27 at 19:46
add a comment |Â
up vote
1
down vote
accepted
Let $C subset T$ be the set where the radial limit
$$lim_r to 1^- f(re^itheta)$$
exists. By theorems 11.30 and 11.23, $m(Tsetminus C) = 0$, and $f$ is the Poisson integral of $f^ast$. On $Tsetminus C$, we can define $f^ast$ arbitrarily if we want an everywhere defined function $f^ast$.
Since $lvert f(re^itheta)rvert leqslant lVert frVert_infty$ for $r in [0,1)$, we have
$$lvert f^ast(e^itheta)rvert = lim_r to 1^-: lvert f(re^itheta)rvert leqslant lVert frVert_infty$$
for all $e^itheta in C$.
Let $alpha > lVert frVert_infty$. Then
$$bigl e^itheta : lvert f^ast(e^itheta)rvert geqslant alpha bigr subset Tsetminus C,,$$
so this is a null set. Hence
$$(lVert frVert_infty, +infty) subset bigl alpha in mathbbR : mbigl(bigl(lvert f^astrvertbigr)^-1([alpha, +infty))bigr) = 0bigr$$
and consequently
$$lVert f^astrVert_infty = inf: bigl alpha in mathbbR : mbigl(bigl(lvert f^astrvertbigr)^-1([alpha, +infty))bigr) = 0bigr leqslant inf: (lVert frVert_infty, +infty) = lVert frVert_infty,.$$
[Note that $f^ast$ is complex-valued, so we must consider $bigl(lvert f^astrvertbigr)^-1([alpha,+infty))$ and not $bigl(f^astbigr)^-1([alpha,+infty))$.]
answered Jul 25 at 19:51
Daniel Fischer♦
171k16154274
I think that the notation $left( f^* right)^-1 left( left[ alpha,infty right)right)=0$ is correct since the Lebesgue measure is defined on T.
– Emin
Jul 25 at 20:40
But we must look at the codomain. We need to know whether $bigl e^itheta : lvert f^ast(e^itheta)rvert geqslant alphabigr$ is a null set. The set $bigl(f^astbigr)^-1([alpha,+infty))$ is $bigl e^itheta : f^ast(e^itheta) in mathbbR land f^ast(e^itheta) geqslant alpha bigr$. But unless $f^ast$ is real-valued and takes only non-negative values, that set can have measure $0$ also for $alpha < lVert f^astrVert_infty$.
– Daniel Fischer♦
Jul 25 at 20:48
$f^*$ is given with $f^*:Tto left[0, infty right).$
– Emin
Jul 27 at 19:11
$H^infty$ is the space of bounded holomorphic functions on the unit disk, and for $f in H^infty$, $f^ast$ is defined as the radial limit function of $f$. E.g. for $f(z) = z$ we have $f^ast(e^itheta) = e^itheta$. Occasionally the image of $f^ast$ is contained in $[0,infty)$, but typically it isn't.
– Daniel Fischer♦
Jul 27 at 19:46
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $C subset T$ be the set where the radial limit
$$lim_r to 1^- f(re^itheta)$$
exists. By theorems 11.30 and 11.23, $m(Tsetminus C) = 0$, and $f$ is the Poisson integral of $f^ast$. On $Tsetminus C$, we can define $f^ast$ arbitrarily if we want an everywhere defined function $f^ast$.
Since $lvert f(re^itheta)rvert leqslant lVert frVert_infty$ for $r in [0,1)$, we have
$$lvert f^ast(e^itheta)rvert = lim_r to 1^-: lvert f(re^itheta)rvert leqslant lVert frVert_infty$$
for all $e^itheta in C$.
Let $alpha > lVert frVert_infty$. Then
$$bigl e^itheta : lvert f^ast(e^itheta)rvert geqslant alpha bigr subset Tsetminus C,,$$
so this is a null set. Hence
$$(lVert frVert_infty, +infty) subset bigl alpha in mathbbR : mbigl(bigl(lvert f^astrvertbigr)^-1([alpha, +infty))bigr) = 0bigr$$
and consequently
$$lVert f^astrVert_infty = inf: bigl alpha in mathbbR : mbigl(bigl(lvert f^astrvertbigr)^-1([alpha, +infty))bigr) = 0bigr leqslant inf: (lVert frVert_infty, +infty) = lVert frVert_infty,.$$
[Note that $f^ast$ is complex-valued, so we must consider $bigl(lvert f^astrvertbigr)^-1([alpha,+infty))$ and not $bigl(f^astbigr)^-1([alpha,+infty))$.]
answered Jul 25 at 19:51
Daniel Fischer♦
171k16154274
Let $C subset T$ be the set where the radial limit
$$lim_r to 1^- f(re^itheta)$$
exists. By theorems 11.30 and 11.23, $m(Tsetminus C) = 0$, and $f$ is the Poisson integral of $f^ast$. On $Tsetminus C$, we can define $f^ast$ arbitrarily if we want an everywhere defined function $f^ast$.
Since $lvert f(re^itheta)rvert leqslant lVert frVert_infty$ for $r in [0,1)$, we have
$$lvert f^ast(e^itheta)rvert = lim_r to 1^-: lvert f(re^itheta)rvert leqslant lVert frVert_infty$$
for all $e^itheta in C$.
Let $alpha > lVert frVert_infty$. Then
$$bigl e^itheta : lvert f^ast(e^itheta)rvert geqslant alpha bigr subset Tsetminus C,,$$
so this is a null set. Hence
$$(lVert frVert_infty, +infty) subset bigl alpha in mathbbR : mbigl(bigl(lvert f^astrvertbigr)^-1([alpha, +infty))bigr) = 0bigr$$
and consequently
$$lVert f^astrVert_infty = inf: bigl alpha in mathbbR : mbigl(bigl(lvert f^astrvertbigr)^-1([alpha, +infty))bigr) = 0bigr leqslant inf: (lVert frVert_infty, +infty) = lVert frVert_infty,.$$
[Note that $f^ast$ is complex-valued, so we must consider $bigl(lvert f^astrvertbigr)^-1([alpha,+infty))$ and not $bigl(f^astbigr)^-1([alpha,+infty))$.]
answered Jul 25 at 19:51
Daniel Fischer♦
171k16154274
answered Jul 25 at 19:51
Daniel Fischer♦
171k16154274
answered Jul 25 at 19:51
Daniel Fischer♦
171k16154274
answered Jul 25 at 19:51
Daniel Fischer♦
171k16154274
171k16154274
I think that the notation $left( f^* right)^-1 left( left[ alpha,infty right)right)=0$ is correct since the Lebesgue measure is defined on T.
– Emin
Jul 25 at 20:40
But we must look at the codomain. We need to know whether $bigl e^itheta : lvert f^ast(e^itheta)rvert geqslant alphabigr$ is a null set. The set $bigl(f^astbigr)^-1([alpha,+infty))$ is $bigl e^itheta : f^ast(e^itheta) in mathbbR land f^ast(e^itheta) geqslant alpha bigr$. But unless $f^ast$ is real-valued and takes only non-negative values, that set can have measure $0$ also for $alpha < lVert f^astrVert_infty$.
– Daniel Fischer♦
Jul 25 at 20:48
$f^*$ is given with $f^*:Tto left[0, infty right).$
– Emin
Jul 27 at 19:11
$H^infty$ is the space of bounded holomorphic functions on the unit disk, and for $f in H^infty$, $f^ast$ is defined as the radial limit function of $f$. E.g. for $f(z) = z$ we have $f^ast(e^itheta) = e^itheta$. Occasionally the image of $f^ast$ is contained in $[0,infty)$, but typically it isn't.
– Daniel Fischer♦
Jul 27 at 19:46
add a comment |Â
I think that the notation $left( f^* right)^-1 left( left[ alpha,infty right)right)=0$ is correct since the Lebesgue measure is defined on T.
– Emin
Jul 25 at 20:40
But we must look at the codomain. We need to know whether $bigl e^itheta : lvert f^ast(e^itheta)rvert geqslant alphabigr$ is a null set. The set $bigl(f^astbigr)^-1([alpha,+infty))$ is $bigl e^itheta : f^ast(e^itheta) in mathbbR land f^ast(e^itheta) geqslant alpha bigr$. But unless $f^ast$ is real-valued and takes only non-negative values, that set can have measure $0$ also for $alpha < lVert f^astrVert_infty$.
– Daniel Fischer♦
Jul 25 at 20:48
$f^*$ is given with $f^*:Tto left[0, infty right).$
– Emin
Jul 27 at 19:11
$H^infty$ is the space of bounded holomorphic functions on the unit disk, and for $f in H^infty$, $f^ast$ is defined as the radial limit function of $f$. E.g. for $f(z) = z$ we have $f^ast(e^itheta) = e^itheta$. Occasionally the image of $f^ast$ is contained in $[0,infty)$, but typically it isn't.
– Daniel Fischer♦
Jul 27 at 19:46
I think that the notation $left( f^* right)^-1 left( left[ alpha,infty right)right)=0$ is correct since the Lebesgue measure is defined on T.
– Emin
Jul 25 at 20:40
I think that the notation $left( f^* right)^-1 left( left[ alpha,infty right)right)=0$ is correct since the Lebesgue measure is defined on T.
– Emin
Jul 25 at 20:40
But we must look at the codomain. We need to know whether $bigl e^itheta : lvert f^ast(e^itheta)rvert geqslant alphabigr$ is a null set. The set $bigl(f^astbigr)^-1([alpha,+infty))$ is $bigl e^itheta : f^ast(e^itheta) in mathbbR land f^ast(e^itheta) geqslant alpha bigr$. But unless $f^ast$ is real-valued and takes only non-negative values, that set can have measure $0$ also for $alpha < lVert f^astrVert_infty$.
– Daniel Fischer♦
Jul 25 at 20:48
But we must look at the codomain. We need to know whether $bigl e^itheta : lvert f^ast(e^itheta)rvert geqslant alphabigr$ is a null set. The set $bigl(f^astbigr)^-1([alpha,+infty))$ is $bigl e^itheta : f^ast(e^itheta) in mathbbR land f^ast(e^itheta) geqslant alpha bigr$. But unless $f^ast$ is real-valued and takes only non-negative values, that set can have measure $0$ also for $alpha < lVert f^astrVert_infty$.
– Daniel Fischer♦
Jul 25 at 20:48
$f^*$ is given with $f^*:Tto left[0, infty right).$
– Emin
Jul 27 at 19:11
$f^*$ is given with $f^*:Tto left[0, infty right).$
– Emin
Jul 27 at 19:11
$H^infty$ is the space of bounded holomorphic functions on the unit disk, and for $f in H^infty$, $f^ast$ is defined as the radial limit function of $f$. E.g. for $f(z) = z$ we have $f^ast(e^itheta) = e^itheta$. Occasionally the image of $f^ast$ is contained in $[0,infty)$, but typically it isn't.
– Daniel Fischer♦
Jul 27 at 19:46
$H^infty$ is the space of bounded holomorphic functions on the unit disk, and for $f in H^infty$, $f^ast$ is defined as the radial limit function of $f$. E.g. for $f(z) = z$ we have $f^ast(e^itheta) = e^itheta$. Occasionally the image of $f^ast$ is contained in $[0,infty)$, but typically it isn't.
– Daniel Fischer♦
Jul 27 at 19:46
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Use the Poisson integral
– user578878
Jul 25 at 11:18
2
Just do it directly. Since $lvert f(re^it)rvert leqslant lVert frVert_infty$ for all $t$ and all $r in [0,1)$, it follows that if the radial limit exists at some point, we have $lvert f^ast(e^it)rvert leqslant lVert frVert_infty$ there. Since the set where the radial limit doesn't exist is a null set, we're done.
– Daniel Fischer♦
Jul 25 at 11:55
@DanielFischer how do you connect $vert f^*(e^it)vert$ with $Vert f^*Vert_infty$?
– Emin
Jul 25 at 12:31
1
Let $C$ be the set of points on the unit circle where the radial limit exists. For all $t$ with $e^it in C$ we know $lvert f^ast(e^it)rvert leqslant lVert frVert_infty$. The complement of $C$ has measure $0$, so $mbigl((f^ast)^-1([alpha,+infty))bigr) = 0$ for all $alpha > lVert frVert_infty$.
– Daniel Fischer♦
Jul 25 at 12:36
@DanielFischer Thanks mate!
– Emin
Jul 25 at 13:40