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)