Mixing
Boundedness of $f mapsto f(0)$ w.r.t. different norms [closed]
$begingroup$
Let $X = C[0, 1]$ be the space of all continuous real valued functions on $[0, 1]$. On $X$, we define two norms: For $f$ in $X,$
$$|f|_{infty}:= sup{|f(t)| : t in [0, 1]} quad text{and} quad |f|_1 := int_{0}^{1}|f(t)|, mathrm{d}t.$$
Let $X_1 := (X, | , cdot , |_1)$ and $X_2 := (X, | , cdot , |_{infty})$. Let $T : X rightarrow mathbb{R}$ be the linear map defined by
$T(f) := f(0)$. Which of the following statement(s) is (are) true?
$(a)$ $T$ is bounded on $X_1$ but not on $X_2$.
$(b)$ $T$ is bounded on $X_2$ but not on $X_1$.
$(c)$ $T$ is bounded on $X_1$ and $X_2$.
$(d)$ $T$ is neither bounded on $X_1$ nor on $X_2$.
I think option $(b)$ is true because $X_2$ is complete, and $X_1$ is not complete.
Is this the case?
Any hints/solution will be appreciated.
functional-analysis
edited Jan 7 at 18:06
SvanN
2,0421422
asked Jan 7 at 17:44
jasminejasmine
1,716417
$endgroup$
closed as off-topic by Saad, amWhy, jgon, Lord_Farin, Eevee Trainer Jan 10 at 2:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, amWhy, jgon, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $X = C[0, 1]$ be the space of all continuous real valued functions on $[0, 1]$. On $X$, we define two norms: For $f$ in $X,$
$$|f|_{infty}:= sup{|f(t)| : t in [0, 1]} quad text{and} quad |f|_1 := int_{0}^{1}|f(t)|, mathrm{d}t.$$
Let $X_1 := (X, | , cdot , |_1)$ and $X_2 := (X, | , cdot , |_{infty})$. Let $T : X rightarrow mathbb{R}$ be the linear map defined by
$T(f) := f(0)$. Which of the following statement(s) is (are) true?
$(a)$ $T$ is bounded on $X_1$ but not on $X_2$.
$(b)$ $T$ is bounded on $X_2$ but not on $X_1$.
$(c)$ $T$ is bounded on $X_1$ and $X_2$.
$(d)$ $T$ is neither bounded on $X_1$ nor on $X_2$.
I think option $(b)$ is true because $X_2$ is complete, and $X_1$ is not complete.
Is this the case?
Any hints/solution will be appreciated.
functional-analysis
edited Jan 7 at 18:06
SvanN
2,0421422
asked Jan 7 at 17:44
jasminejasmine
1,716417
$endgroup$
closed as off-topic by Saad, amWhy, jgon, Lord_Farin, Eevee Trainer Jan 10 at 2:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, amWhy, jgon, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
$||f||_1leq||f||_{infty}$.
$endgroup$
– Thomas Shelby
Jan 7 at 17:47
$begingroup$
@ThomasShelby that mean option b) is correct ?
$endgroup$
– jasmine
Jan 7 at 17:50
1
$begingroup$
Without further context, it means that the unit ball in $X_1$ contains the unit ball in $X_2$, so option (a) must be incorrect. Furthermore, $lvert T(f)rvertleq |f|_{infty}$, so all that is left is to determine whether $T$ is bounded on $X_1$.
$endgroup$
– Michael Lee
Jan 7 at 18:01
2
$begingroup$
a good way to get examples/counterexamples in functional analysis is trying to get "$delta$-functions", i. e. functions with a large peak but a "small" integral.
$endgroup$
– 0x539
Jan 7 at 18:05
add a comment |
$begingroup$
Let $X = C[0, 1]$ be the space of all continuous real valued functions on $[0, 1]$. On $X$, we define two norms: For $f$ in $X,$
$$|f|_{infty}:= sup{|f(t)| : t in [0, 1]} quad text{and} quad |f|_1 := int_{0}^{1}|f(t)|, mathrm{d}t.$$
Let $X_1 := (X, | , cdot , |_1)$ and $X_2 := (X, | , cdot , |_{infty})$. Let $T : X rightarrow mathbb{R}$ be the linear map defined by
$T(f) := f(0)$. Which of the following statement(s) is (are) true?
$(a)$ $T$ is bounded on $X_1$ but not on $X_2$.
$(b)$ $T$ is bounded on $X_2$ but not on $X_1$.
$(c)$ $T$ is bounded on $X_1$ and $X_2$.
$(d)$ $T$ is neither bounded on $X_1$ nor on $X_2$.
I think option $(b)$ is true because $X_2$ is complete, and $X_1$ is not complete.
Is this the case?
Any hints/solution will be appreciated.
functional-analysis
edited Jan 7 at 18:06
SvanN
2,0421422
asked Jan 7 at 17:44
jasminejasmine
1,716417
$endgroup$
Let $X = C[0, 1]$ be the space of all continuous real valued functions on $[0, 1]$. On $X$, we define two norms: For $f$ in $X,$
$$|f|_{infty}:= sup{|f(t)| : t in [0, 1]} quad text{and} quad |f|_1 := int_{0}^{1}|f(t)|, mathrm{d}t.$$
Let $X_1 := (X, | , cdot , |_1)$ and $X_2 := (X, | , cdot , |_{infty})$. Let $T : X rightarrow mathbb{R}$ be the linear map defined by
$T(f) := f(0)$. Which of the following statement(s) is (are) true?
$(a)$ $T$ is bounded on $X_1$ but not on $X_2$.
$(b)$ $T$ is bounded on $X_2$ but not on $X_1$.
$(c)$ $T$ is bounded on $X_1$ and $X_2$.
$(d)$ $T$ is neither bounded on $X_1$ nor on $X_2$.
I think option $(b)$ is true because $X_2$ is complete, and $X_1$ is not complete.
Is this the case?
Any hints/solution will be appreciated.
functional-analysis
functional-analysis
edited Jan 7 at 18:06
SvanN
2,0421422
asked Jan 7 at 17:44
jasminejasmine
1,716417
edited Jan 7 at 18:06
SvanN
2,0421422
asked Jan 7 at 17:44
jasminejasmine
1,716417
edited Jan 7 at 18:06
SvanN
2,0421422
edited Jan 7 at 18:06
SvanN
2,0421422
edited Jan 7 at 18:06
SvanN
2,0421422
2,0421422
asked Jan 7 at 17:44
jasminejasmine
1,716417
asked Jan 7 at 17:44
jasminejasmine
1,716417
asked Jan 7 at 17:44
jasminejasmine
1,716417
1,716417
closed as off-topic by Saad, amWhy, jgon, Lord_Farin, Eevee Trainer Jan 10 at 2:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, amWhy, jgon, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, amWhy, jgon, Lord_Farin, Eevee Trainer Jan 10 at 2:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, amWhy, jgon, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
$||f||_1leq||f||_{infty}$.
$endgroup$
– Thomas Shelby
Jan 7 at 17:47
$begingroup$
@ThomasShelby that mean option b) is correct ?
$endgroup$
– jasmine
Jan 7 at 17:50
1
$begingroup$
Without further context, it means that the unit ball in $X_1$ contains the unit ball in $X_2$, so option (a) must be incorrect. Furthermore, $lvert T(f)rvertleq |f|_{infty}$, so all that is left is to determine whether $T$ is bounded on $X_1$.
$endgroup$
– Michael Lee
Jan 7 at 18:01
2
$begingroup$
a good way to get examples/counterexamples in functional analysis is trying to get "$delta$-functions", i. e. functions with a large peak but a "small" integral.
$endgroup$
– 0x539
Jan 7 at 18:05
add a comment |
2
$begingroup$
$||f||_1leq||f||_{infty}$.
$endgroup$
– Thomas Shelby
Jan 7 at 17:47
$begingroup$
@ThomasShelby that mean option b) is correct ?
$endgroup$
– jasmine
Jan 7 at 17:50
1
$begingroup$
Without further context, it means that the unit ball in $X_1$ contains the unit ball in $X_2$, so option (a) must be incorrect. Furthermore, $lvert T(f)rvertleq |f|_{infty}$, so all that is left is to determine whether $T$ is bounded on $X_1$.
$endgroup$
– Michael Lee
Jan 7 at 18:01
2
$begingroup$
a good way to get examples/counterexamples in functional analysis is trying to get "$delta$-functions", i. e. functions with a large peak but a "small" integral.
$endgroup$
– 0x539
Jan 7 at 18:05
2
2
$begingroup$
$||f||_1leq||f||_{infty}$.
$endgroup$
– Thomas Shelby
Jan 7 at 17:47
$begingroup$
$||f||_1leq||f||_{infty}$.
$endgroup$
– Thomas Shelby
Jan 7 at 17:47
$begingroup$
@ThomasShelby that mean option b) is correct ?
$endgroup$
– jasmine
Jan 7 at 17:50
$begingroup$
@ThomasShelby that mean option b) is correct ?
$endgroup$
– jasmine
Jan 7 at 17:50
1
1
$begingroup$
Without further context, it means that the unit ball in $X_1$ contains the unit ball in $X_2$, so option (a) must be incorrect. Furthermore, $lvert T(f)rvertleq |f|_{infty}$, so all that is left is to determine whether $T$ is bounded on $X_1$.
$endgroup$
– Michael Lee
Jan 7 at 18:01
$begingroup$
Without further context, it means that the unit ball in $X_1$ contains the unit ball in $X_2$, so option (a) must be incorrect. Furthermore, $lvert T(f)rvertleq |f|_{infty}$, so all that is left is to determine whether $T$ is bounded on $X_1$.
$endgroup$
– Michael Lee
Jan 7 at 18:01
2
2
$begingroup$
a good way to get examples/counterexamples in functional analysis is trying to get "$delta$-functions", i. e. functions with a large peak but a "small" integral.
$endgroup$
– 0x539
Jan 7 at 18:05
$begingroup$
a good way to get examples/counterexamples in functional analysis is trying to get "$delta$-functions", i. e. functions with a large peak but a "small" integral.
$endgroup$
– 0x539
Jan 7 at 18:05
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I was wrong in my previous answer, because i mistook the definition of boundedness.
The answer is b:
1) T is bounded on $X_{1}$ because $$|T(f)|=|f(0)|lesup_{tin[0,1]}|f(t)|=||f||_{infty}$$
That means that $||T||_{op}le 1$ (in particular equals 1, in fact if you take a costant function you have the inverse inequality, too) on $X_{2}$.
2) T is not bounded on $X_{2}$.
Let $f_{n}(t)=begin{cases} 1-nt & text{if $tin[0,1/n]$} \
0 & text{otherwise}end{cases}$.
$||f_{n}||_{1}le 1/n$ and $T(f_{n})=f_{n}(0)=1$.
Let $c>0$, for $n>c$ $|T(f_{n})|=1>c/n>c||f_{n}||_{1}$, so $||T||_{op}=infty$, which means that T isn't bounded on $X_{1}$.
answered Jan 8 at 11:33
ecrinecrin
3827
$endgroup$
$begingroup$
thanks u @ecrin
$endgroup$
– jasmine
Jan 9 at 6:06
add a comment |
$begingroup$
The boundedness of a function depends on the topology of the codomain, in fact a function $fcolon Xto Y$ is said to be bounded if $f(X)subseteq Y$ is bounded, so the answer is the same for both $X_{1}$ and $X_{2}$.
In particular I think the answer is (d), in fact consider the function $f_{n}(t)=n$ $forall tin[0,1]$. $f_{n}in C([0,1],mathbb{R})$ $forall ninmathbb{N}$.
So $mathbb{N}subseteq T(X)$, and is not bounded, so neither $T(X)$.
answered Jan 7 at 23:00
ecrinecrin
3827
$endgroup$
1
$begingroup$
You are mistaken about the use of the term 'bounded'. In functional analysis, a map $T : V to W$ is called bounded when its operator norm exists (en.wikipedia.org/wiki/Operator_norm), not when its image is a bounded set.
$endgroup$
– SvanN
Jan 8 at 8:21
$begingroup$
You are right, thanks!
$endgroup$
– ecrin
Jan 8 at 8:33
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I was wrong in my previous answer, because i mistook the definition of boundedness.
The answer is b:
1) T is bounded on $X_{1}$ because $$|T(f)|=|f(0)|lesup_{tin[0,1]}|f(t)|=||f||_{infty}$$
That means that $||T||_{op}le 1$ (in particular equals 1, in fact if you take a costant function you have the inverse inequality, too) on $X_{2}$.
2) T is not bounded on $X_{2}$.
Let $f_{n}(t)=begin{cases} 1-nt & text{if $tin[0,1/n]$} \
0 & text{otherwise}end{cases}$.
$||f_{n}||_{1}le 1/n$ and $T(f_{n})=f_{n}(0)=1$.
Let $c>0$, for $n>c$ $|T(f_{n})|=1>c/n>c||f_{n}||_{1}$, so $||T||_{op}=infty$, which means that T isn't bounded on $X_{1}$.
answered Jan 8 at 11:33
ecrinecrin
3827
$endgroup$
$begingroup$
thanks u @ecrin
$endgroup$
– jasmine
Jan 9 at 6:06
add a comment |
$begingroup$
I was wrong in my previous answer, because i mistook the definition of boundedness.
The answer is b:
1) T is bounded on $X_{1}$ because $$|T(f)|=|f(0)|lesup_{tin[0,1]}|f(t)|=||f||_{infty}$$
That means that $||T||_{op}le 1$ (in particular equals 1, in fact if you take a costant function you have the inverse inequality, too) on $X_{2}$.
2) T is not bounded on $X_{2}$.
Let $f_{n}(t)=begin{cases} 1-nt & text{if $tin[0,1/n]$} \
0 & text{otherwise}end{cases}$.
$||f_{n}||_{1}le 1/n$ and $T(f_{n})=f_{n}(0)=1$.
Let $c>0$, for $n>c$ $|T(f_{n})|=1>c/n>c||f_{n}||_{1}$, so $||T||_{op}=infty$, which means that T isn't bounded on $X_{1}$.
answered Jan 8 at 11:33
ecrinecrin
3827
$endgroup$
$begingroup$
thanks u @ecrin
$endgroup$
– jasmine
Jan 9 at 6:06
add a comment |
$begingroup$
I was wrong in my previous answer, because i mistook the definition of boundedness.
The answer is b:
1) T is bounded on $X_{1}$ because $$|T(f)|=|f(0)|lesup_{tin[0,1]}|f(t)|=||f||_{infty}$$
That means that $||T||_{op}le 1$ (in particular equals 1, in fact if you take a costant function you have the inverse inequality, too) on $X_{2}$.
2) T is not bounded on $X_{2}$.
Let $f_{n}(t)=begin{cases} 1-nt & text{if $tin[0,1/n]$} \
0 & text{otherwise}end{cases}$.
$||f_{n}||_{1}le 1/n$ and $T(f_{n})=f_{n}(0)=1$.
Let $c>0$, for $n>c$ $|T(f_{n})|=1>c/n>c||f_{n}||_{1}$, so $||T||_{op}=infty$, which means that T isn't bounded on $X_{1}$.
answered Jan 8 at 11:33
ecrinecrin
3827
$endgroup$
I was wrong in my previous answer, because i mistook the definition of boundedness.
The answer is b:
1) T is bounded on $X_{1}$ because $$|T(f)|=|f(0)|lesup_{tin[0,1]}|f(t)|=||f||_{infty}$$
That means that $||T||_{op}le 1$ (in particular equals 1, in fact if you take a costant function you have the inverse inequality, too) on $X_{2}$.
2) T is not bounded on $X_{2}$.
Let $f_{n}(t)=begin{cases} 1-nt & text{if $tin[0,1/n]$} \
0 & text{otherwise}end{cases}$.
$||f_{n}||_{1}le 1/n$ and $T(f_{n})=f_{n}(0)=1$.
Let $c>0$, for $n>c$ $|T(f_{n})|=1>c/n>c||f_{n}||_{1}$, so $||T||_{op}=infty$, which means that T isn't bounded on $X_{1}$.
answered Jan 8 at 11:33
ecrinecrin
3827
answered Jan 8 at 11:33
ecrinecrin
3827
answered Jan 8 at 11:33
ecrinecrin
3827
answered Jan 8 at 11:33
ecrinecrin
3827
3827
$begingroup$
thanks u @ecrin
$endgroup$
– jasmine
Jan 9 at 6:06
add a comment |
$begingroup$
thanks u @ecrin
$endgroup$
– jasmine
Jan 9 at 6:06
$begingroup$
thanks u @ecrin
$endgroup$
– jasmine
Jan 9 at 6:06
$begingroup$
thanks u @ecrin
$endgroup$
– jasmine
Jan 9 at 6:06
add a comment |
$begingroup$
The boundedness of a function depends on the topology of the codomain, in fact a function $fcolon Xto Y$ is said to be bounded if $f(X)subseteq Y$ is bounded, so the answer is the same for both $X_{1}$ and $X_{2}$.
In particular I think the answer is (d), in fact consider the function $f_{n}(t)=n$ $forall tin[0,1]$. $f_{n}in C([0,1],mathbb{R})$ $forall ninmathbb{N}$.
So $mathbb{N}subseteq T(X)$, and is not bounded, so neither $T(X)$.
answered Jan 7 at 23:00
ecrinecrin
3827
$endgroup$
1
$begingroup$
You are mistaken about the use of the term 'bounded'. In functional analysis, a map $T : V to W$ is called bounded when its operator norm exists (en.wikipedia.org/wiki/Operator_norm), not when its image is a bounded set.
$endgroup$
– SvanN
Jan 8 at 8:21
$begingroup$
You are right, thanks!
$endgroup$
– ecrin
Jan 8 at 8:33
add a comment |
$begingroup$
The boundedness of a function depends on the topology of the codomain, in fact a function $fcolon Xto Y$ is said to be bounded if $f(X)subseteq Y$ is bounded, so the answer is the same for both $X_{1}$ and $X_{2}$.
In particular I think the answer is (d), in fact consider the function $f_{n}(t)=n$ $forall tin[0,1]$. $f_{n}in C([0,1],mathbb{R})$ $forall ninmathbb{N}$.
So $mathbb{N}subseteq T(X)$, and is not bounded, so neither $T(X)$.
answered Jan 7 at 23:00
ecrinecrin
3827
$endgroup$
1
$begingroup$
You are mistaken about the use of the term 'bounded'. In functional analysis, a map $T : V to W$ is called bounded when its operator norm exists (en.wikipedia.org/wiki/Operator_norm), not when its image is a bounded set.
$endgroup$
– SvanN
Jan 8 at 8:21
$begingroup$
You are right, thanks!
$endgroup$
– ecrin
Jan 8 at 8:33
add a comment |
$begingroup$
The boundedness of a function depends on the topology of the codomain, in fact a function $fcolon Xto Y$ is said to be bounded if $f(X)subseteq Y$ is bounded, so the answer is the same for both $X_{1}$ and $X_{2}$.
In particular I think the answer is (d), in fact consider the function $f_{n}(t)=n$ $forall tin[0,1]$. $f_{n}in C([0,1],mathbb{R})$ $forall ninmathbb{N}$.
So $mathbb{N}subseteq T(X)$, and is not bounded, so neither $T(X)$.
answered Jan 7 at 23:00
ecrinecrin
3827
$endgroup$
The boundedness of a function depends on the topology of the codomain, in fact a function $fcolon Xto Y$ is said to be bounded if $f(X)subseteq Y$ is bounded, so the answer is the same for both $X_{1}$ and $X_{2}$.
In particular I think the answer is (d), in fact consider the function $f_{n}(t)=n$ $forall tin[0,1]$. $f_{n}in C([0,1],mathbb{R})$ $forall ninmathbb{N}$.
So $mathbb{N}subseteq T(X)$, and is not bounded, so neither $T(X)$.
answered Jan 7 at 23:00
ecrinecrin
3827
answered Jan 7 at 23:00
ecrinecrin
3827
answered Jan 7 at 23:00
ecrinecrin
3827
answered Jan 7 at 23:00
ecrinecrin
3827
3827
1
$begingroup$
You are mistaken about the use of the term 'bounded'. In functional analysis, a map $T : V to W$ is called bounded when its operator norm exists (en.wikipedia.org/wiki/Operator_norm), not when its image is a bounded set.
$endgroup$
– SvanN
Jan 8 at 8:21
$begingroup$
You are right, thanks!
$endgroup$
– ecrin
Jan 8 at 8:33
add a comment |
1
$begingroup$
You are mistaken about the use of the term 'bounded'. In functional analysis, a map $T : V to W$ is called bounded when its operator norm exists (en.wikipedia.org/wiki/Operator_norm), not when its image is a bounded set.
$endgroup$
– SvanN
Jan 8 at 8:21
$begingroup$
You are right, thanks!
$endgroup$
– ecrin
Jan 8 at 8:33
1
1
$begingroup$
You are mistaken about the use of the term 'bounded'. In functional analysis, a map $T : V to W$ is called bounded when its operator norm exists (en.wikipedia.org/wiki/Operator_norm), not when its image is a bounded set.
$endgroup$
– SvanN
Jan 8 at 8:21
$begingroup$
You are mistaken about the use of the term 'bounded'. In functional analysis, a map $T : V to W$ is called bounded when its operator norm exists (en.wikipedia.org/wiki/Operator_norm), not when its image is a bounded set.
$endgroup$
– SvanN
Jan 8 at 8:21
$begingroup$
You are right, thanks!
$endgroup$
– ecrin
Jan 8 at 8:33
$begingroup$
You are right, thanks!
$endgroup$
– ecrin
Jan 8 at 8:33
add a comment |
2
$begingroup$
$||f||_1leq||f||_{infty}$.
$endgroup$
– Thomas Shelby
Jan 7 at 17:47
$begingroup$
@ThomasShelby that mean option b) is correct ?
$endgroup$
– jasmine
Jan 7 at 17:50
1
$begingroup$
Without further context, it means that the unit ball in $X_1$ contains the unit ball in $X_2$, so option (a) must be incorrect. Furthermore, $lvert T(f)rvertleq |f|_{infty}$, so all that is left is to determine whether $T$ is bounded on $X_1$.
$endgroup$
– Michael Lee
Jan 7 at 18:01
2
$begingroup$
a good way to get examples/counterexamples in functional analysis is trying to get "$delta$-functions", i. e. functions with a large peak but a "small" integral.
$endgroup$
– 0x539
Jan 7 at 18:05