Suppose $X$ is a normed space, I want to construct an unbounded one-to-one linear map from $X$ to itself, could anyone give some hint?
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1Take a normalized Hamel basis of $X$ containing $e_1,e_2,\ldots$. Map $e_n$ to $ne_n$ and the other basis vectors to themselves. – David Mitra Jul 31 '16 at 08:08
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It is not the same question, but looking at this post and some of other related posts might help you. – Martin Sleziak Jul 31 '16 at 09:05
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@DavidMitra I think that your comment could be posted as an answer. – Martin Sleziak Jul 31 '16 at 09:07
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You probably want this for infinite-dimensional $X$, right? Every linear mapping on a finite dimensional space is continuous – Martin Sleziak Jul 31 '16 at 09:07
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As Martin points out in the comments, you can only do this in infinite dimensional spaces. In such a space, there is a normalized (elements of norm $1$) Hamel basis with infinitely many elements (using the Axiom of Choice).
Take such a basis and pick a countably infinite subset $\{e_1,e_2,\ldots\}$ of it. Map $e_n$ to $ne_n$ and the other basis elements to themselves. Then extend linearly to define a map on the entire space. This gives the desired operator.
David Mitra
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I am thinking what is wrong with your example with finite case. how about choose countably set with norm 1 in finite dimension? – 89085731 Jul 31 '16 at 12:58
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1@89085731 In a finite dimensional space, an infinite set would not be linearly independent; assigning arbitrary values to each element would likely not give a well-defined linear map. – David Mitra Jul 31 '16 at 13:07