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Difference between revisions of "Talk:Unbounded operator"

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(Reply to Boris Tsirelson’s comment.)
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"The simplest example of an unbounded operator is the differentiation operator $ \dfrac{\mathrm{d}}{\mathrm{d}{t}} $, defined on the set $ {C^{1}}([a,b]) $ of all continuously differentiable functions into the space $ C([a,b]) $ of all continuous functions on $ a \leq t \leq b $, because the operator $ \dfrac{\mathrm{d}}{\mathrm{d}{t}} $ takes the bounded set $ \{ t \mapsto \sin(n t) \}_{n \in \mathbb{N}} $ to the unbounded set $ \{ t \mapsto n \cos(n t) \}_{n \in \mathbb{N}} $" — really? The set $ \{ t \mapsto \sin(n t) \}_{n \in \mathbb{N}} $ is unbounded in the space $ {C^{1}}([a,b]) $; and the operator is bounded from the space $ {C^{1}}([a,b]) $ to the space $ C([a,b]) $. True, the article mentions " the set $ {C^{1}}([a,b]) $", not  "the space $ {C^{1}}([a,b]) $"; but it is not said that this set is treated as a subset of the space $ C([a,b]) $. The text may be misleading. [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 06:52, 1 March 2017 (CET)
 
"The simplest example of an unbounded operator is the differentiation operator $ \dfrac{\mathrm{d}}{\mathrm{d}{t}} $, defined on the set $ {C^{1}}([a,b]) $ of all continuously differentiable functions into the space $ C([a,b]) $ of all continuous functions on $ a \leq t \leq b $, because the operator $ \dfrac{\mathrm{d}}{\mathrm{d}{t}} $ takes the bounded set $ \{ t \mapsto \sin(n t) \}_{n \in \mathbb{N}} $ to the unbounded set $ \{ t \mapsto n \cos(n t) \}_{n \in \mathbb{N}} $" — really? The set $ \{ t \mapsto \sin(n t) \}_{n \in \mathbb{N}} $ is unbounded in the space $ {C^{1}}([a,b]) $; and the operator is bounded from the space $ {C^{1}}([a,b]) $ to the space $ C([a,b]) $. True, the article mentions " the set $ {C^{1}}([a,b]) $", not  "the space $ {C^{1}}([a,b]) $"; but it is not said that this set is treated as a subset of the space $ C([a,b]) $. The text may be misleading. [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 06:52, 1 March 2017 (CET)
  
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:Hi Boris. Yes, I agree that the original text is misleading. I can change ‘set’ to ‘space’ so that $ ': {C^{1}}([a,b]) \to C([a,b]) $ is an unbounded linear operator from one normed vector space to another. I must admit that I’m not entirely happy with the definition of ‘unbounded operator’ being offered here. Standard practice, I believe, is to call any densely defined linear operator $ S $ from one normed vector space $ X $ to another $ Y $ an ‘unbounded linear operator’ even if $ S $ has a bounded extension $ T: X \to Y $. [[User:Leonard Huang|Leonard Huang]] ([[User talk:Leonard Huang|talk]]) 07:27, 3 March 2017 (CET)
  
Hi Boris. Yes, I agree that the original text is misleading. I can change ‘set’ to ‘space’ so that $ ': {C^{1}}([a,b]) \to C([a,b]) $ is an unbounded linear operator from one normed vector space to another. I must admit that I’m not entirely happy with the definition of ‘unbounded operator’ being offered here. Standard practice, I believe, is to call any densely defined linear operator $ S $ from one normed vector space $ X $ to another $ Y $ an ‘unbounded linear operator’ even if $ S $ has a bounded extension $ T: X \to Y $.
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::Really? I did not know. But wait, the whole $X$ is also a dense linear subset of $X$. Do you mean that (usual, defined everywhere) bounded operators are a special case of unbounded operators? [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 09:15, 3 March 2017 (CET)
  
[[User:Leonard Huang|Leonard Huang]] ([[User talk:Leonard Huang|talk]]) 07:27, 3 March 2017 (CET)
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:::Hi Boris. Yes, that is (unfortunately) so — a bounded globally defined linear operator between normed vector spaces (Banach spaces or Hilbert spaces mostly) is, by definition, an unbounded linear operator. The [https://ncatlab.org/nlab/show/unbounded+operator nLab article on unbounded operators] calls this an example of a red herring. [[User:Leonard Huang|Leonard Huang]] ([[User talk:Leonard Huang|talk]]) 11:37, 3 March 2017 (CET)
 
 
:Really? I did not know. But wait, the whole $X$ is also a dense linear subset of $X$. Do you mean that (usual, defined everywhere) bounded operators are a special case of unbounded operators? [[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 09:15, 3 March 2017 (CET)
 

Revision as of 10:37, 3 March 2017

"The simplest example of an unbounded operator is the differentiation operator $ \dfrac{\mathrm{d}}{\mathrm{d}{t}} $, defined on the set $ {C^{1}}([a,b]) $ of all continuously differentiable functions into the space $ C([a,b]) $ of all continuous functions on $ a \leq t \leq b $, because the operator $ \dfrac{\mathrm{d}}{\mathrm{d}{t}} $ takes the bounded set $ \{ t \mapsto \sin(n t) \}_{n \in \mathbb{N}} $ to the unbounded set $ \{ t \mapsto n \cos(n t) \}_{n \in \mathbb{N}} $" — really? The set $ \{ t \mapsto \sin(n t) \}_{n \in \mathbb{N}} $ is unbounded in the space $ {C^{1}}([a,b]) $; and the operator is bounded from the space $ {C^{1}}([a,b]) $ to the space $ C([a,b]) $. True, the article mentions " the set $ {C^{1}}([a,b]) $", not "the space $ {C^{1}}([a,b]) $"; but it is not said that this set is treated as a subset of the space $ C([a,b]) $. The text may be misleading. Boris Tsirelson (talk) 06:52, 1 March 2017 (CET)

Hi Boris. Yes, I agree that the original text is misleading. I can change ‘set’ to ‘space’ so that $ ': {C^{1}}([a,b]) \to C([a,b]) $ is an unbounded linear operator from one normed vector space to another. I must admit that I’m not entirely happy with the definition of ‘unbounded operator’ being offered here. Standard practice, I believe, is to call any densely defined linear operator $ S $ from one normed vector space $ X $ to another $ Y $ an ‘unbounded linear operator’ even if $ S $ has a bounded extension $ T: X \to Y $. Leonard Huang (talk) 07:27, 3 March 2017 (CET)
Really? I did not know. But wait, the whole $X$ is also a dense linear subset of $X$. Do you mean that (usual, defined everywhere) bounded operators are a special case of unbounded operators? Boris Tsirelson (talk) 09:15, 3 March 2017 (CET)
Hi Boris. Yes, that is (unfortunately) so — a bounded globally defined linear operator between normed vector spaces (Banach spaces or Hilbert spaces mostly) is, by definition, an unbounded linear operator. The nLab article on unbounded operators calls this an example of a red herring. Leonard Huang (talk) 11:37, 3 March 2017 (CET)
How to Cite This Entry:
Unbounded operator. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Unbounded_operator&oldid=40215