Talk:Unbounded operator

"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)

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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 $.

Leonard Huang (talk) 07:27, 3 March 2017 (CET)