What is the operator norm of this operator?
$begingroup$
Let $mathbb{X} = C^1([0,1], mathbb{R})$ and $|,f|_{mathbb{X}} = |,f|_{infty} + |,f'|_{infty}$. Let $A: mathbb{X} rightarrow mathbb{R}$ with $Af = f'left(frac{1}{2}right)$.
Find the operator norm of $A$
I am not too sure how to begin here. I have several definitions of the operator norm, the one I have been trying is that
$$|A|_{op} = infbig{c geq 0: |Av| leq c |v|big}$$
But I'm not even sure what the expression for $|Av|$ will be...
Any help is appreciated
operator-theory norm
edited Jan 8 at 5:43
Hanno
2,091425
asked Jan 27 '18 at 17:50
PhysicsMathsLovePhysicsMathsLove
1,189314
$endgroup$
add a comment |
$begingroup$
Let $mathbb{X} = C^1([0,1], mathbb{R})$ and $|,f|_{mathbb{X}} = |,f|_{infty} + |,f'|_{infty}$. Let $A: mathbb{X} rightarrow mathbb{R}$ with $Af = f'left(frac{1}{2}right)$.
Find the operator norm of $A$
I am not too sure how to begin here. I have several definitions of the operator norm, the one I have been trying is that
$$|A|_{op} = infbig{c geq 0: |Av| leq c |v|big}$$
But I'm not even sure what the expression for $|Av|$ will be...
Any help is appreciated
operator-theory norm
edited Jan 8 at 5:43
Hanno
2,091425
asked Jan 27 '18 at 17:50
PhysicsMathsLovePhysicsMathsLove
1,189314
$endgroup$
add a comment |
$begingroup$
Let $mathbb{X} = C^1([0,1], mathbb{R})$ and $|,f|_{mathbb{X}} = |,f|_{infty} + |,f'|_{infty}$. Let $A: mathbb{X} rightarrow mathbb{R}$ with $Af = f'left(frac{1}{2}right)$.
Find the operator norm of $A$
I am not too sure how to begin here. I have several definitions of the operator norm, the one I have been trying is that
$$|A|_{op} = infbig{c geq 0: |Av| leq c |v|big}$$
But I'm not even sure what the expression for $|Av|$ will be...
Any help is appreciated
operator-theory norm
edited Jan 8 at 5:43
Hanno
2,091425
asked Jan 27 '18 at 17:50
PhysicsMathsLovePhysicsMathsLove
1,189314
$endgroup$
Let $mathbb{X} = C^1([0,1], mathbb{R})$ and $|,f|_{mathbb{X}} = |,f|_{infty} + |,f'|_{infty}$. Let $A: mathbb{X} rightarrow mathbb{R}$ with $Af = f'left(frac{1}{2}right)$.
Find the operator norm of $A$
I am not too sure how to begin here. I have several definitions of the operator norm, the one I have been trying is that
$$|A|_{op} = infbig{c geq 0: |Av| leq c |v|big}$$
But I'm not even sure what the expression for $|Av|$ will be...
Any help is appreciated
operator-theory norm
operator-theory norm
edited Jan 8 at 5:43
Hanno
2,091425
asked Jan 27 '18 at 17:50
PhysicsMathsLovePhysicsMathsLove
1,189314
edited Jan 8 at 5:43
Hanno
2,091425
asked Jan 27 '18 at 17:50
PhysicsMathsLovePhysicsMathsLove
1,189314
edited Jan 8 at 5:43
Hanno
2,091425
edited Jan 8 at 5:43
Hanno
2,091425
edited Jan 8 at 5:43
Hanno
2,091425
2,091425
asked Jan 27 '18 at 17:50
PhysicsMathsLovePhysicsMathsLove
1,189314
asked Jan 27 '18 at 17:50
PhysicsMathsLovePhysicsMathsLove
1,189314
asked Jan 27 '18 at 17:50
PhysicsMathsLovePhysicsMathsLove
1,189314
1,189314
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hints: Another way to look at the operator norm is to see what the operator does to a unit vector:
$$
|A|_{op}=sup{|Af|: |f|_{Bbb X}=1}
$$
Think about how large $|Af|$ could possibly be if we require that $|f|_{Bbb X}=1$.
Some more pointers: $A$ takes in a function and spits out a real number. Which real number it spits out depends on the function, of course. For instance, $f(x)=x^2$ gives $Af=1$, and $g(x)=sin(x)$ gives $Ag=cos(0.5)$.
But if $f$ is a function such that $|Af|=1$, what can you say about $|f|_{Bbb X}$? How small could it be? (This specific hint is more directed at the operator norm definition given in the question.)
Finally, since you were wondering, the expression for $|Av|$ is $|v'(0.5)|$.
answered Jan 27 '18 at 17:54
ArthurArthur
112k7107191
$endgroup$
$begingroup$
Is the expression for $||Af||_{mathbb{X}} = ||Af||_{infty} + ||Af'||_{infty} = ||f'(frac{1}{2})||_{infty} + ||f''(frac{1}{2})||_{infty}$? If so, I have no idea where to go from here, even with the restriction $||f||_{mathbb{X}} = 1$, as we are taking infinity norms now...
$endgroup$
– PhysicsMathsLove
Jan 27 '18 at 17:59
$begingroup$
No. $Af$ is a real number and the corresponding norm is the usual absolute value.
$endgroup$
– Paul K
Jan 27 '18 at 17:59
$begingroup$
I am confused... Please can you add an answer?
$endgroup$
– PhysicsMathsLove
Jan 27 '18 at 18:00
$begingroup$
@PhyaicsMathsLove I added another paragraph for you to think about.
$endgroup$
– Arthur
Jan 27 '18 at 18:07
$begingroup$
It is apparent that $|Af|leq|f|_{X}$, so $|A|leq 1$. But I wonder if $|A|=1$. To do this, let $epsilon>0$ be small, I fail to find some $f_{epsilon}$ with $|f|_{epsilon}leq 1$ and such that $|Af_{epsilon}|>1-epsilon$.
$endgroup$
– user284331
Jan 27 '18 at 22:55
|
show 3 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hints: Another way to look at the operator norm is to see what the operator does to a unit vector:
$$
|A|_{op}=sup{|Af|: |f|_{Bbb X}=1}
$$
Think about how large $|Af|$ could possibly be if we require that $|f|_{Bbb X}=1$.
Some more pointers: $A$ takes in a function and spits out a real number. Which real number it spits out depends on the function, of course. For instance, $f(x)=x^2$ gives $Af=1$, and $g(x)=sin(x)$ gives $Ag=cos(0.5)$.
But if $f$ is a function such that $|Af|=1$, what can you say about $|f|_{Bbb X}$? How small could it be? (This specific hint is more directed at the operator norm definition given in the question.)
Finally, since you were wondering, the expression for $|Av|$ is $|v'(0.5)|$.
answered Jan 27 '18 at 17:54
ArthurArthur
112k7107191
$endgroup$
$begingroup$
Is the expression for $||Af||_{mathbb{X}} = ||Af||_{infty} + ||Af'||_{infty} = ||f'(frac{1}{2})||_{infty} + ||f''(frac{1}{2})||_{infty}$? If so, I have no idea where to go from here, even with the restriction $||f||_{mathbb{X}} = 1$, as we are taking infinity norms now...
$endgroup$
– PhysicsMathsLove
Jan 27 '18 at 17:59
$begingroup$
No. $Af$ is a real number and the corresponding norm is the usual absolute value.
$endgroup$
– Paul K
Jan 27 '18 at 17:59
$begingroup$
I am confused... Please can you add an answer?
$endgroup$
– PhysicsMathsLove
Jan 27 '18 at 18:00
$begingroup$
@PhyaicsMathsLove I added another paragraph for you to think about.
$endgroup$
– Arthur
Jan 27 '18 at 18:07
$begingroup$
It is apparent that $|Af|leq|f|_{X}$, so $|A|leq 1$. But I wonder if $|A|=1$. To do this, let $epsilon>0$ be small, I fail to find some $f_{epsilon}$ with $|f|_{epsilon}leq 1$ and such that $|Af_{epsilon}|>1-epsilon$.
$endgroup$
– user284331
Jan 27 '18 at 22:55
|
show 3 more comments
$begingroup$
Hints: Another way to look at the operator norm is to see what the operator does to a unit vector:
$$
|A|_{op}=sup{|Af|: |f|_{Bbb X}=1}
$$
Think about how large $|Af|$ could possibly be if we require that $|f|_{Bbb X}=1$.
Some more pointers: $A$ takes in a function and spits out a real number. Which real number it spits out depends on the function, of course. For instance, $f(x)=x^2$ gives $Af=1$, and $g(x)=sin(x)$ gives $Ag=cos(0.5)$.
But if $f$ is a function such that $|Af|=1$, what can you say about $|f|_{Bbb X}$? How small could it be? (This specific hint is more directed at the operator norm definition given in the question.)
Finally, since you were wondering, the expression for $|Av|$ is $|v'(0.5)|$.
answered Jan 27 '18 at 17:54
ArthurArthur
112k7107191
$endgroup$
$begingroup$
Is the expression for $||Af||_{mathbb{X}} = ||Af||_{infty} + ||Af'||_{infty} = ||f'(frac{1}{2})||_{infty} + ||f''(frac{1}{2})||_{infty}$? If so, I have no idea where to go from here, even with the restriction $||f||_{mathbb{X}} = 1$, as we are taking infinity norms now...
$endgroup$
– PhysicsMathsLove
Jan 27 '18 at 17:59
$begingroup$
No. $Af$ is a real number and the corresponding norm is the usual absolute value.
$endgroup$
– Paul K
Jan 27 '18 at 17:59
$begingroup$
I am confused... Please can you add an answer?
$endgroup$
– PhysicsMathsLove
Jan 27 '18 at 18:00
$begingroup$
@PhyaicsMathsLove I added another paragraph for you to think about.
$endgroup$
– Arthur
Jan 27 '18 at 18:07
$begingroup$
It is apparent that $|Af|leq|f|_{X}$, so $|A|leq 1$. But I wonder if $|A|=1$. To do this, let $epsilon>0$ be small, I fail to find some $f_{epsilon}$ with $|f|_{epsilon}leq 1$ and such that $|Af_{epsilon}|>1-epsilon$.
$endgroup$
– user284331
Jan 27 '18 at 22:55
|
show 3 more comments
$begingroup$
Hints: Another way to look at the operator norm is to see what the operator does to a unit vector:
$$
|A|_{op}=sup{|Af|: |f|_{Bbb X}=1}
$$
Think about how large $|Af|$ could possibly be if we require that $|f|_{Bbb X}=1$.
Some more pointers: $A$ takes in a function and spits out a real number. Which real number it spits out depends on the function, of course. For instance, $f(x)=x^2$ gives $Af=1$, and $g(x)=sin(x)$ gives $Ag=cos(0.5)$.
But if $f$ is a function such that $|Af|=1$, what can you say about $|f|_{Bbb X}$? How small could it be? (This specific hint is more directed at the operator norm definition given in the question.)
Finally, since you were wondering, the expression for $|Av|$ is $|v'(0.5)|$.
answered Jan 27 '18 at 17:54
ArthurArthur
112k7107191
$endgroup$
Hints: Another way to look at the operator norm is to see what the operator does to a unit vector:
$$
|A|_{op}=sup{|Af|: |f|_{Bbb X}=1}
$$
Think about how large $|Af|$ could possibly be if we require that $|f|_{Bbb X}=1$.
Some more pointers: $A$ takes in a function and spits out a real number. Which real number it spits out depends on the function, of course. For instance, $f(x)=x^2$ gives $Af=1$, and $g(x)=sin(x)$ gives $Ag=cos(0.5)$.
But if $f$ is a function such that $|Af|=1$, what can you say about $|f|_{Bbb X}$? How small could it be? (This specific hint is more directed at the operator norm definition given in the question.)
Finally, since you were wondering, the expression for $|Av|$ is $|v'(0.5)|$.
answered Jan 27 '18 at 17:54
ArthurArthur
112k7107191
edited Jan 27 '18 at 18:07
answered Jan 27 '18 at 17:54
ArthurArthur
112k7107191
answered Jan 27 '18 at 17:54
ArthurArthur
112k7107191
answered Jan 27 '18 at 17:54
ArthurArthur
112k7107191
112k7107191
$begingroup$
Is the expression for $||Af||_{mathbb{X}} = ||Af||_{infty} + ||Af'||_{infty} = ||f'(frac{1}{2})||_{infty} + ||f''(frac{1}{2})||_{infty}$? If so, I have no idea where to go from here, even with the restriction $||f||_{mathbb{X}} = 1$, as we are taking infinity norms now...
$endgroup$
– PhysicsMathsLove
Jan 27 '18 at 17:59
$begingroup$
No. $Af$ is a real number and the corresponding norm is the usual absolute value.
$endgroup$
– Paul K
Jan 27 '18 at 17:59
$begingroup$
I am confused... Please can you add an answer?
$endgroup$
– PhysicsMathsLove
Jan 27 '18 at 18:00
$begingroup$
@PhyaicsMathsLove I added another paragraph for you to think about.
$endgroup$
– Arthur
Jan 27 '18 at 18:07
$begingroup$
It is apparent that $|Af|leq|f|_{X}$, so $|A|leq 1$. But I wonder if $|A|=1$. To do this, let $epsilon>0$ be small, I fail to find some $f_{epsilon}$ with $|f|_{epsilon}leq 1$ and such that $|Af_{epsilon}|>1-epsilon$.
$endgroup$
– user284331
Jan 27 '18 at 22:55
|
show 3 more comments
$begingroup$
Is the expression for $||Af||_{mathbb{X}} = ||Af||_{infty} + ||Af'||_{infty} = ||f'(frac{1}{2})||_{infty} + ||f''(frac{1}{2})||_{infty}$? If so, I have no idea where to go from here, even with the restriction $||f||_{mathbb{X}} = 1$, as we are taking infinity norms now...
$endgroup$
– PhysicsMathsLove
Jan 27 '18 at 17:59
$begingroup$
No. $Af$ is a real number and the corresponding norm is the usual absolute value.
$endgroup$
– Paul K
Jan 27 '18 at 17:59
$begingroup$
I am confused... Please can you add an answer?
$endgroup$
– PhysicsMathsLove
Jan 27 '18 at 18:00
$begingroup$
@PhyaicsMathsLove I added another paragraph for you to think about.
$endgroup$
– Arthur
Jan 27 '18 at 18:07
$begingroup$
It is apparent that $|Af|leq|f|_{X}$, so $|A|leq 1$. But I wonder if $|A|=1$. To do this, let $epsilon>0$ be small, I fail to find some $f_{epsilon}$ with $|f|_{epsilon}leq 1$ and such that $|Af_{epsilon}|>1-epsilon$.
$endgroup$
– user284331
Jan 27 '18 at 22:55
$begingroup$
Is the expression for $||Af||_{mathbb{X}} = ||Af||_{infty} + ||Af'||_{infty} = ||f'(frac{1}{2})||_{infty} + ||f''(frac{1}{2})||_{infty}$? If so, I have no idea where to go from here, even with the restriction $||f||_{mathbb{X}} = 1$, as we are taking infinity norms now...
$endgroup$
– PhysicsMathsLove
Jan 27 '18 at 17:59
$begingroup$
Is the expression for $||Af||_{mathbb{X}} = ||Af||_{infty} + ||Af'||_{infty} = ||f'(frac{1}{2})||_{infty} + ||f''(frac{1}{2})||_{infty}$? If so, I have no idea where to go from here, even with the restriction $||f||_{mathbb{X}} = 1$, as we are taking infinity norms now...
$endgroup$
– PhysicsMathsLove
Jan 27 '18 at 17:59
$begingroup$
No. $Af$ is a real number and the corresponding norm is the usual absolute value.
$endgroup$
– Paul K
Jan 27 '18 at 17:59
$begingroup$
No. $Af$ is a real number and the corresponding norm is the usual absolute value.
$endgroup$
– Paul K
Jan 27 '18 at 17:59
$begingroup$
I am confused... Please can you add an answer?
$endgroup$
– PhysicsMathsLove
Jan 27 '18 at 18:00
$begingroup$
I am confused... Please can you add an answer?
$endgroup$
– PhysicsMathsLove
Jan 27 '18 at 18:00
$begingroup$
@PhyaicsMathsLove I added another paragraph for you to think about.
$endgroup$
– Arthur
Jan 27 '18 at 18:07
$begingroup$
@PhyaicsMathsLove I added another paragraph for you to think about.
$endgroup$
– Arthur
Jan 27 '18 at 18:07
$begingroup$
It is apparent that $|Af|leq|f|_{X}$, so $|A|leq 1$. But I wonder if $|A|=1$. To do this, let $epsilon>0$ be small, I fail to find some $f_{epsilon}$ with $|f|_{epsilon}leq 1$ and such that $|Af_{epsilon}|>1-epsilon$.
$endgroup$
– user284331
Jan 27 '18 at 22:55
$begingroup$
It is apparent that $|Af|leq|f|_{X}$, so $|A|leq 1$. But I wonder if $|A|=1$. To do this, let $epsilon>0$ be small, I fail to find some $f_{epsilon}$ with $|f|_{epsilon}leq 1$ and such that $|Af_{epsilon}|>1-epsilon$.
$endgroup$
– user284331
Jan 27 '18 at 22:55
|
show 3 more comments
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