How find the norm $||f_x||$ of linear functional $f_x:C[a,b] to mathbb{R}$? [closed]
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Consider the Banach space $C[a,b]$ with sup norm. For any fixed $xin C[a,b]$ define $f_x(y)$ for any $y in C[a,b]$ as follows:
$$f_x(y)=int_a^b x(t)y(t) dt$$
Show that $f_x$ is a Bounded linear functional and compute $||f_x||$ .
functional-analysis
edited Nov 23 at 18:03
Yadati Kiran
1,638519
asked Nov 23 at 17:37
Victor Daniel Mendoza Rubio
22
closed as off-topic by Davide Giraudo, Lord_Farin, Leucippus, Shailesh, KReiser Nov 29 at 1:05
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, Lord_Farin, Leucippus, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-1
down vote
favorite
Consider the Banach space $C[a,b]$ with sup norm. For any fixed $xin C[a,b]$ define $f_x(y)$ for any $y in C[a,b]$ as follows:
$$f_x(y)=int_a^b x(t)y(t) dt$$
Show that $f_x$ is a Bounded linear functional and compute $||f_x||$ .
functional-analysis
edited Nov 23 at 18:03
Yadati Kiran
1,638519
asked Nov 23 at 17:37
Victor Daniel Mendoza Rubio
22
closed as off-topic by Davide Giraudo, Lord_Farin, Leucippus, Shailesh, KReiser Nov 29 at 1:05
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, Lord_Farin, Leucippus, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
Do you know the Hahn-Banach Theorem? Or the Riesz Representaton theorem for $C[a,b]^*$
– DisintegratingByParts
Nov 24 at 22:54
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Consider the Banach space $C[a,b]$ with sup norm. For any fixed $xin C[a,b]$ define $f_x(y)$ for any $y in C[a,b]$ as follows:
$$f_x(y)=int_a^b x(t)y(t) dt$$
Show that $f_x$ is a Bounded linear functional and compute $||f_x||$ .
functional-analysis
edited Nov 23 at 18:03
Yadati Kiran
1,638519
asked Nov 23 at 17:37
Victor Daniel Mendoza Rubio
22
Consider the Banach space $C[a,b]$ with sup norm. For any fixed $xin C[a,b]$ define $f_x(y)$ for any $y in C[a,b]$ as follows:
$$f_x(y)=int_a^b x(t)y(t) dt$$
Show that $f_x$ is a Bounded linear functional and compute $||f_x||$ .
functional-analysis
functional-analysis
edited Nov 23 at 18:03
Yadati Kiran
1,638519
asked Nov 23 at 17:37
Victor Daniel Mendoza Rubio
22
edited Nov 23 at 18:03
Yadati Kiran
1,638519
asked Nov 23 at 17:37
Victor Daniel Mendoza Rubio
22
edited Nov 23 at 18:03
Yadati Kiran
1,638519
edited Nov 23 at 18:03
Yadati Kiran
1,638519
edited Nov 23 at 18:03
Yadati Kiran
1,638519
1,638519
asked Nov 23 at 17:37
Victor Daniel Mendoza Rubio
22
asked Nov 23 at 17:37
Victor Daniel Mendoza Rubio
22
asked Nov 23 at 17:37
Victor Daniel Mendoza Rubio
22
22
closed as off-topic by Davide Giraudo, Lord_Farin, Leucippus, Shailesh, KReiser Nov 29 at 1:05
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, Lord_Farin, Leucippus, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Davide Giraudo, Lord_Farin, Leucippus, Shailesh, KReiser Nov 29 at 1:05
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, Lord_Farin, Leucippus, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
Do you know the Hahn-Banach Theorem? Or the Riesz Representaton theorem for $C[a,b]^*$
– DisintegratingByParts
Nov 24 at 22:54
add a comment |
Do you know the Hahn-Banach Theorem? Or the Riesz Representaton theorem for $C[a,b]^*$
– DisintegratingByParts
Nov 24 at 22:54
Do you know the Hahn-Banach Theorem? Or the Riesz Representaton theorem for $C[a,b]^*$
– DisintegratingByParts
Nov 24 at 22:54
Do you know the Hahn-Banach Theorem? Or the Riesz Representaton theorem for $C[a,b]^*$
– DisintegratingByParts
Nov 24 at 22:54
add a comment |
1 Answer
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1
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You have
$$
|f_x(y)| = left|int_a^b x(t)y(t),dt right| leq int_a^b |x(t)y(t)|,dt
leq |x|_1|y|_infty ,
$$
so $|f_x| leq |x|_1$. Now try proving that they are indeed equal.
Hint: you would like to take $y(t)=mathop{mathrm{sign}}(x(t))$.
Some more hints: call $Z={t:x(t)=0}$ and for every $ninmathbb N$ define $y_n(t)=mathop{mathrm{sign}}(x(t))min(n d_Z(t),1)$. Then:
$|y_n|_inftyleq1$,
$y_n$ is $n$-Lipschitz, hence continuous,
$y_n(t)tomathop{mathrm{sign}}(x(t))$ pointwise,
$f_x(y_n) = int_a^b x(t)y_n(t),dt to int_a^b x(t)mathop{mathrm{sign}}(x(t)),dt = int_a^b|x(t)|,dt = |x|_1$.
Edit: since a picture is worth a thousand words...
The blue line is $x(t)$, the orange line is $y_5(t)$:
answered Nov 23 at 17:51
Federico
4,184512
1
No, of course not. Hence, you would like to take...
– Federico
Nov 23 at 17:58
1
Call $Z={t:x(t)=0}$ and consider $y_n(t)=mathrm{sign}(x(t)) (n d_Z(t) wedge 1)$.
– Federico
Nov 23 at 18:01
1
These functions are $n$-Lip, they converge pointwise to $mathrm{sign}(x)$ and they are dominated (by $1$).
– Federico
Nov 23 at 18:02
1
Why complicated? All the required checks listed above are immediate. Obvious by the definition of $y_n$. What is that you don't understand?
– Federico
Nov 23 at 18:08
1
No but I'm taking it easy. I just asked what you find difficult to prove. It seems to me that the verification is immediate
– Federico
Nov 23 at 18:13
|
show 10 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You have
$$
|f_x(y)| = left|int_a^b x(t)y(t),dt right| leq int_a^b |x(t)y(t)|,dt
leq |x|_1|y|_infty ,
$$
so $|f_x| leq |x|_1$. Now try proving that they are indeed equal.
Hint: you would like to take $y(t)=mathop{mathrm{sign}}(x(t))$.
Some more hints: call $Z={t:x(t)=0}$ and for every $ninmathbb N$ define $y_n(t)=mathop{mathrm{sign}}(x(t))min(n d_Z(t),1)$. Then:
$|y_n|_inftyleq1$,
$y_n$ is $n$-Lipschitz, hence continuous,
$y_n(t)tomathop{mathrm{sign}}(x(t))$ pointwise,
$f_x(y_n) = int_a^b x(t)y_n(t),dt to int_a^b x(t)mathop{mathrm{sign}}(x(t)),dt = int_a^b|x(t)|,dt = |x|_1$.
Edit: since a picture is worth a thousand words...
The blue line is $x(t)$, the orange line is $y_5(t)$:
answered Nov 23 at 17:51
Federico
4,184512
1
No, of course not. Hence, you would like to take...
– Federico
Nov 23 at 17:58
1
Call $Z={t:x(t)=0}$ and consider $y_n(t)=mathrm{sign}(x(t)) (n d_Z(t) wedge 1)$.
– Federico
Nov 23 at 18:01
1
These functions are $n$-Lip, they converge pointwise to $mathrm{sign}(x)$ and they are dominated (by $1$).
– Federico
Nov 23 at 18:02
1
Why complicated? All the required checks listed above are immediate. Obvious by the definition of $y_n$. What is that you don't understand?
– Federico
Nov 23 at 18:08
1
No but I'm taking it easy. I just asked what you find difficult to prove. It seems to me that the verification is immediate
– Federico
Nov 23 at 18:13
|
show 10 more comments
up vote
1
down vote
You have
$$
|f_x(y)| = left|int_a^b x(t)y(t),dt right| leq int_a^b |x(t)y(t)|,dt
leq |x|_1|y|_infty ,
$$
so $|f_x| leq |x|_1$. Now try proving that they are indeed equal.
Hint: you would like to take $y(t)=mathop{mathrm{sign}}(x(t))$.
Some more hints: call $Z={t:x(t)=0}$ and for every $ninmathbb N$ define $y_n(t)=mathop{mathrm{sign}}(x(t))min(n d_Z(t),1)$. Then:
$|y_n|_inftyleq1$,
$y_n$ is $n$-Lipschitz, hence continuous,
$y_n(t)tomathop{mathrm{sign}}(x(t))$ pointwise,
$f_x(y_n) = int_a^b x(t)y_n(t),dt to int_a^b x(t)mathop{mathrm{sign}}(x(t)),dt = int_a^b|x(t)|,dt = |x|_1$.
Edit: since a picture is worth a thousand words...
The blue line is $x(t)$, the orange line is $y_5(t)$:
answered Nov 23 at 17:51
Federico
4,184512
1
No, of course not. Hence, you would like to take...
– Federico
Nov 23 at 17:58
1
Call $Z={t:x(t)=0}$ and consider $y_n(t)=mathrm{sign}(x(t)) (n d_Z(t) wedge 1)$.
– Federico
Nov 23 at 18:01
1
These functions are $n$-Lip, they converge pointwise to $mathrm{sign}(x)$ and they are dominated (by $1$).
– Federico
Nov 23 at 18:02
1
Why complicated? All the required checks listed above are immediate. Obvious by the definition of $y_n$. What is that you don't understand?
– Federico
Nov 23 at 18:08
1
No but I'm taking it easy. I just asked what you find difficult to prove. It seems to me that the verification is immediate
– Federico
Nov 23 at 18:13
|
show 10 more comments
up vote
1
down vote
up vote
1
down vote
You have
$$
|f_x(y)| = left|int_a^b x(t)y(t),dt right| leq int_a^b |x(t)y(t)|,dt
leq |x|_1|y|_infty ,
$$
so $|f_x| leq |x|_1$. Now try proving that they are indeed equal.
Hint: you would like to take $y(t)=mathop{mathrm{sign}}(x(t))$.
Some more hints: call $Z={t:x(t)=0}$ and for every $ninmathbb N$ define $y_n(t)=mathop{mathrm{sign}}(x(t))min(n d_Z(t),1)$. Then:
$|y_n|_inftyleq1$,
$y_n$ is $n$-Lipschitz, hence continuous,
$y_n(t)tomathop{mathrm{sign}}(x(t))$ pointwise,
$f_x(y_n) = int_a^b x(t)y_n(t),dt to int_a^b x(t)mathop{mathrm{sign}}(x(t)),dt = int_a^b|x(t)|,dt = |x|_1$.
Edit: since a picture is worth a thousand words...
The blue line is $x(t)$, the orange line is $y_5(t)$:
answered Nov 23 at 17:51
Federico
4,184512
You have
$$
|f_x(y)| = left|int_a^b x(t)y(t),dt right| leq int_a^b |x(t)y(t)|,dt
leq |x|_1|y|_infty ,
$$
so $|f_x| leq |x|_1$. Now try proving that they are indeed equal.
Hint: you would like to take $y(t)=mathop{mathrm{sign}}(x(t))$.
Some more hints: call $Z={t:x(t)=0}$ and for every $ninmathbb N$ define $y_n(t)=mathop{mathrm{sign}}(x(t))min(n d_Z(t),1)$. Then:
$|y_n|_inftyleq1$,
$y_n$ is $n$-Lipschitz, hence continuous,
$y_n(t)tomathop{mathrm{sign}}(x(t))$ pointwise,
$f_x(y_n) = int_a^b x(t)y_n(t),dt to int_a^b x(t)mathop{mathrm{sign}}(x(t)),dt = int_a^b|x(t)|,dt = |x|_1$.
Edit: since a picture is worth a thousand words...
The blue line is $x(t)$, the orange line is $y_5(t)$:
answered Nov 23 at 17:51
Federico
4,184512
edited Nov 28 at 16:38
answered Nov 23 at 17:51
Federico
4,184512
answered Nov 23 at 17:51
Federico
4,184512
answered Nov 23 at 17:51
Federico
4,184512
4,184512
1
No, of course not. Hence, you would like to take...
– Federico
Nov 23 at 17:58
1
Call $Z={t:x(t)=0}$ and consider $y_n(t)=mathrm{sign}(x(t)) (n d_Z(t) wedge 1)$.
– Federico
Nov 23 at 18:01
1
These functions are $n$-Lip, they converge pointwise to $mathrm{sign}(x)$ and they are dominated (by $1$).
– Federico
Nov 23 at 18:02
1
Why complicated? All the required checks listed above are immediate. Obvious by the definition of $y_n$. What is that you don't understand?
– Federico
Nov 23 at 18:08
1
No but I'm taking it easy. I just asked what you find difficult to prove. It seems to me that the verification is immediate
– Federico
Nov 23 at 18:13
|
show 10 more comments
1
No, of course not. Hence, you would like to take...
– Federico
Nov 23 at 17:58
1
Call $Z={t:x(t)=0}$ and consider $y_n(t)=mathrm{sign}(x(t)) (n d_Z(t) wedge 1)$.
– Federico
Nov 23 at 18:01
1
These functions are $n$-Lip, they converge pointwise to $mathrm{sign}(x)$ and they are dominated (by $1$).
– Federico
Nov 23 at 18:02
1
Why complicated? All the required checks listed above are immediate. Obvious by the definition of $y_n$. What is that you don't understand?
– Federico
Nov 23 at 18:08
1
No but I'm taking it easy. I just asked what you find difficult to prove. It seems to me that the verification is immediate
– Federico
Nov 23 at 18:13
1
1
No, of course not. Hence, you would like to take...
– Federico
Nov 23 at 17:58
No, of course not. Hence, you would like to take...
– Federico
Nov 23 at 17:58
1
1
Call $Z={t:x(t)=0}$ and consider $y_n(t)=mathrm{sign}(x(t)) (n d_Z(t) wedge 1)$.
– Federico
Nov 23 at 18:01
Call $Z={t:x(t)=0}$ and consider $y_n(t)=mathrm{sign}(x(t)) (n d_Z(t) wedge 1)$.
– Federico
Nov 23 at 18:01
1
1
These functions are $n$-Lip, they converge pointwise to $mathrm{sign}(x)$ and they are dominated (by $1$).
– Federico
Nov 23 at 18:02
These functions are $n$-Lip, they converge pointwise to $mathrm{sign}(x)$ and they are dominated (by $1$).
– Federico
Nov 23 at 18:02
1
1
Why complicated? All the required checks listed above are immediate. Obvious by the definition of $y_n$. What is that you don't understand?
– Federico
Nov 23 at 18:08
Why complicated? All the required checks listed above are immediate. Obvious by the definition of $y_n$. What is that you don't understand?
– Federico
Nov 23 at 18:08
1
1
No but I'm taking it easy. I just asked what you find difficult to prove. It seems to me that the verification is immediate
– Federico
Nov 23 at 18:13
No but I'm taking it easy. I just asked what you find difficult to prove. It seems to me that the verification is immediate
– Federico
Nov 23 at 18:13
|
show 10 more comments
Do you know the Hahn-Banach Theorem? Or the Riesz Representaton theorem for $C[a,b]^*$
– DisintegratingByParts
Nov 24 at 22:54