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Is differentiation as a map discontinuous?
Differentiation Operator a Contraction MappingDifferentiable Operator Continuous?Discontinuous mapping between function spacesDifferentiation operator is closed?Density of continuosly differentiable function in space of continuous functionsisomorphism between $C[0,1]$ and $C^1[0,1]$completeness of a functional spaceCharacterizing discontinuous derivativesDoes the Frétchet derivative exist of this minimizer map?Is this linear mapping discontinuous?
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I came across the statement below:
Let $C([0,1])$ be the space of all continuous functions over the interval $[0,1]$ equipped with the Supremum norm. Assume $A$ is a map on the space of all differentiable functions whose derivative is continuous into $C([0,1])$. Also, $A$ is differentiation in the sense that it maps a functions to its derivative. The map $A$ (differentiation) is discontinuous.
It's written that the last sentence is well-known but I can't make any sense of it. How can I reach at such a conclusion? Actually, I am looking for an explicit counterexample.
Any help would be highly appreciated.
real-analysis calculus general-topology functional-analysis
asked 8 hours ago
RajaeiRajaei
2766 bronze badges
$endgroup$
add a comment
|
$begingroup$
I came across the statement below:
Let $C([0,1])$ be the space of all continuous functions over the interval $[0,1]$ equipped with the Supremum norm. Assume $A$ is a map on the space of all differentiable functions whose derivative is continuous into $C([0,1])$. Also, $A$ is differentiation in the sense that it maps a functions to its derivative. The map $A$ (differentiation) is discontinuous.
It's written that the last sentence is well-known but I can't make any sense of it. How can I reach at such a conclusion? Actually, I am looking for an explicit counterexample.
Any help would be highly appreciated.
real-analysis calculus general-topology functional-analysis
asked 8 hours ago
RajaeiRajaei
2766 bronze badges
$endgroup$
$begingroup$
What it means is that $A$ is an unbounded linear operator from $C^1([a,b])$ to $C([a,b])$ (both spaces having the supremum norm).
$endgroup$
– Lord Shark the Unknown
8 hours ago
add a comment
|
$begingroup$
I came across the statement below:
Let $C([0,1])$ be the space of all continuous functions over the interval $[0,1]$ equipped with the Supremum norm. Assume $A$ is a map on the space of all differentiable functions whose derivative is continuous into $C([0,1])$. Also, $A$ is differentiation in the sense that it maps a functions to its derivative. The map $A$ (differentiation) is discontinuous.
It's written that the last sentence is well-known but I can't make any sense of it. How can I reach at such a conclusion? Actually, I am looking for an explicit counterexample.
Any help would be highly appreciated.
real-analysis calculus general-topology functional-analysis
asked 8 hours ago
RajaeiRajaei
2766 bronze badges
$endgroup$
I came across the statement below:
Let $C([0,1])$ be the space of all continuous functions over the interval $[0,1]$ equipped with the Supremum norm. Assume $A$ is a map on the space of all differentiable functions whose derivative is continuous into $C([0,1])$. Also, $A$ is differentiation in the sense that it maps a functions to its derivative. The map $A$ (differentiation) is discontinuous.
It's written that the last sentence is well-known but I can't make any sense of it. How can I reach at such a conclusion? Actually, I am looking for an explicit counterexample.
Any help would be highly appreciated.
real-analysis calculus general-topology functional-analysis
real-analysis calculus general-topology functional-analysis
asked 8 hours ago
RajaeiRajaei
2766 bronze badges
asked 8 hours ago
RajaeiRajaei
2766 bronze badges
asked 8 hours ago
RajaeiRajaei
2766 bronze badges
asked 8 hours ago
RajaeiRajaei
2766 bronze badges
asked 8 hours ago
RajaeiRajaei
2766 bronze badges
2766 bronze badges
$begingroup$
What it means is that $A$ is an unbounded linear operator from $C^1([a,b])$ to $C([a,b])$ (both spaces having the supremum norm).
$endgroup$
– Lord Shark the Unknown
8 hours ago
add a comment
|
$begingroup$
What it means is that $A$ is an unbounded linear operator from $C^1([a,b])$ to $C([a,b])$ (both spaces having the supremum norm).
$endgroup$
– Lord Shark the Unknown
8 hours ago
$begingroup$
What it means is that $A$ is an unbounded linear operator from $C^1([a,b])$ to $C([a,b])$ (both spaces having the supremum norm).
$endgroup$
– Lord Shark the Unknown
8 hours ago
$begingroup$
What it means is that $A$ is an unbounded linear operator from $C^1([a,b])$ to $C([a,b])$ (both spaces having the supremum norm).
$endgroup$
– Lord Shark the Unknown
8 hours ago
add a comment
|
2 Answers
2
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$begingroup$
For a counterexample, take the sequence
$$frac sin nx n$$
These are all continuously differentiable, but the sequence converges to $0$ and the sequence of derivatives doesn't converge at all. The derivative of the limit is not equal to the limit of the derivatives, so it is not continuous.
answered 8 hours ago
Matt SamuelMatt Samuel
41.6k6 gold badges39 silver badges73 bronze badges
$endgroup$
add a comment
|
$begingroup$
It depends on your understanding of $C^1([0,1])$, the space of all differentiable functions whose derivative is continuous. It is a linear subspace of $C([0,1])$. If you give $C^1([0,1])$ the norm inherited from $C([0,1])$, i.e. the supremum norm, then $A$ is not continuous (see Matt Samuel's answer). But you can also give $C^1([0,1])$ the norm
$$lVert f rVert^(1) = lVert f rVert + lVert f' rVert .$$
With respect to this norm $A$ is trivially continuous.
answered 8 hours ago
Paul FrostPaul Frost
18.7k3 gold badges12 silver badges42 bronze badges
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2 Answers
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2 Answers
2
active
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$begingroup$
For a counterexample, take the sequence
$$frac sin nx n$$
These are all continuously differentiable, but the sequence converges to $0$ and the sequence of derivatives doesn't converge at all. The derivative of the limit is not equal to the limit of the derivatives, so it is not continuous.
answered 8 hours ago
Matt SamuelMatt Samuel
41.6k6 gold badges39 silver badges73 bronze badges
$endgroup$
add a comment
|
$begingroup$
For a counterexample, take the sequence
$$frac sin nx n$$
These are all continuously differentiable, but the sequence converges to $0$ and the sequence of derivatives doesn't converge at all. The derivative of the limit is not equal to the limit of the derivatives, so it is not continuous.
answered 8 hours ago
Matt SamuelMatt Samuel
41.6k6 gold badges39 silver badges73 bronze badges
$endgroup$
add a comment
|
$begingroup$
For a counterexample, take the sequence
$$frac sin nx n$$
These are all continuously differentiable, but the sequence converges to $0$ and the sequence of derivatives doesn't converge at all. The derivative of the limit is not equal to the limit of the derivatives, so it is not continuous.
answered 8 hours ago
Matt SamuelMatt Samuel
41.6k6 gold badges39 silver badges73 bronze badges
$endgroup$
For a counterexample, take the sequence
$$frac sin nx n$$
These are all continuously differentiable, but the sequence converges to $0$ and the sequence of derivatives doesn't converge at all. The derivative of the limit is not equal to the limit of the derivatives, so it is not continuous.
answered 8 hours ago
Matt SamuelMatt Samuel
41.6k6 gold badges39 silver badges73 bronze badges
answered 8 hours ago
Matt SamuelMatt Samuel
41.6k6 gold badges39 silver badges73 bronze badges
answered 8 hours ago
Matt SamuelMatt Samuel
41.6k6 gold badges39 silver badges73 bronze badges
answered 8 hours ago
Matt SamuelMatt Samuel
41.6k6 gold badges39 silver badges73 bronze badges
41.6k6 gold badges39 silver badges73 bronze badges
add a comment
|
add a comment
|
$begingroup$
It depends on your understanding of $C^1([0,1])$, the space of all differentiable functions whose derivative is continuous. It is a linear subspace of $C([0,1])$. If you give $C^1([0,1])$ the norm inherited from $C([0,1])$, i.e. the supremum norm, then $A$ is not continuous (see Matt Samuel's answer). But you can also give $C^1([0,1])$ the norm
$$lVert f rVert^(1) = lVert f rVert + lVert f' rVert .$$
With respect to this norm $A$ is trivially continuous.
answered 8 hours ago
Paul FrostPaul Frost
18.7k3 gold badges12 silver badges42 bronze badges
$endgroup$
add a comment
|
$begingroup$
It depends on your understanding of $C^1([0,1])$, the space of all differentiable functions whose derivative is continuous. It is a linear subspace of $C([0,1])$. If you give $C^1([0,1])$ the norm inherited from $C([0,1])$, i.e. the supremum norm, then $A$ is not continuous (see Matt Samuel's answer). But you can also give $C^1([0,1])$ the norm
$$lVert f rVert^(1) = lVert f rVert + lVert f' rVert .$$
With respect to this norm $A$ is trivially continuous.
answered 8 hours ago
Paul FrostPaul Frost
18.7k3 gold badges12 silver badges42 bronze badges
$endgroup$
add a comment
|
$begingroup$
It depends on your understanding of $C^1([0,1])$, the space of all differentiable functions whose derivative is continuous. It is a linear subspace of $C([0,1])$. If you give $C^1([0,1])$ the norm inherited from $C([0,1])$, i.e. the supremum norm, then $A$ is not continuous (see Matt Samuel's answer). But you can also give $C^1([0,1])$ the norm
$$lVert f rVert^(1) = lVert f rVert + lVert f' rVert .$$
With respect to this norm $A$ is trivially continuous.
answered 8 hours ago
Paul FrostPaul Frost
18.7k3 gold badges12 silver badges42 bronze badges
$endgroup$
It depends on your understanding of $C^1([0,1])$, the space of all differentiable functions whose derivative is continuous. It is a linear subspace of $C([0,1])$. If you give $C^1([0,1])$ the norm inherited from $C([0,1])$, i.e. the supremum norm, then $A$ is not continuous (see Matt Samuel's answer). But you can also give $C^1([0,1])$ the norm
$$lVert f rVert^(1) = lVert f rVert + lVert f' rVert .$$
With respect to this norm $A$ is trivially continuous.
answered 8 hours ago
Paul FrostPaul Frost
18.7k3 gold badges12 silver badges42 bronze badges
answered 8 hours ago
Paul FrostPaul Frost
18.7k3 gold badges12 silver badges42 bronze badges
answered 8 hours ago
Paul FrostPaul Frost
18.7k3 gold badges12 silver badges42 bronze badges
answered 8 hours ago
Paul FrostPaul Frost
18.7k3 gold badges12 silver badges42 bronze badges
18.7k3 gold badges12 silver badges42 bronze badges
add a comment
|
add a comment
|
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$begingroup$
What it means is that $A$ is an unbounded linear operator from $C^1([a,b])$ to $C([a,b])$ (both spaces having the supremum norm).
$endgroup$
– Lord Shark the Unknown
8 hours ago