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)

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)

I see. Well, it does not upset me. I've read with pleasure red herring, and I can add more examples: a finitely additive measure is generally not a measure, but every measure is a finitely additive measure; etc. Let us change the article accordingly. Boris Tsirelson (talk) 12:02, 3 March 2017 (CET)

And, by the way, Wikipedia says: "The term "unbounded operator" can be misleading, since

• "unbounded" should sometimes be understood as "not necessarily bounded";
• "operator" should be understood as "linear operator" (as in the case of "bounded operator");
• the domain of the operator is a linear subspace, not necessarily the whole space;
• this linear subspace is not necessarily closed; often (but not always) it is assumed to be dense;
• in the special case of a bounded operator, still, the domain is usually assumed to be the whole space."

Boris Tsirelson (talk) 18:39, 4 March 2017 (CET)

Please let's not base anything on what Wikipedia says. Seriously. If there's a point at issue here, let's get someone who knows the area well to advise. Richard Pinch (talk) 08:41, 5 March 2017 (CET)

That is, you feel unsure about "the standard practice", do you? Boris Tsirelson (talk) 12:30, 5 March 2017 (CET)

I don't know what you mean by that, but I do not think it is related to what I said. Please explain. Richard Pinch (talk) 21:54, 6 March 2017 (CET)

Sorry for being unclear. Yes, I mean your phrase "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 $" above. As for me, your phrase and the quote from wikipedia conform. However, you seem to object to the latter. I wonder, why. Boris Tsirelson (talk) 21:16, 7 March 2017 (CET)

I didn't say anything of the kind – Leonard Huang did, though – so your phrase "your phrase" remains unclear. I object to Wikipedia in principle: it is not a reliable source. Richard Pinch (talk) 23:15, 7 March 2017 (CET)

Oops! I am sorry for my inadvertence. Now I understand. Yes, you are right, in general. But in particular (revealing the secret) those Wikipedia phrases are written (mostly) by myself. Boris Tsirelson (talk) 06:21, 8 March 2017 (CET)

I see, that explains it! But why not just state them in your own voice? Richard Pinch (talk) 08:37, 8 March 2017 (CET)