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Who can give me an operator like this or show it doesn't exist: Operator T: X-->Y, is a bijection from normed linear space X to normed linear space Y. X, Y are equipped with the same norm, and X is a proper dense subset of Y. Both T and inverse of T are bounded.

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    $\begingroup$ What about the identity with $X=Y$? Well, you say, $T$ is a 'bijection', do you mean, onto $Y$? $\endgroup$ – Berci May 13 '13 at 2:18
  • $\begingroup$ Thank you for your response. X should be a real subspace of Y. Bijection means one-to-one and onto. $\endgroup$ – L. Xu May 13 '13 at 2:28
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    $\begingroup$ I don't understand what you mean by "$X$ is contained and dense in $Y$." Do you mean via some other map $S : X \to Y$? $\endgroup$ – Qiaochu Yuan May 13 '13 at 2:43
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    $\begingroup$ I mean X is a dense subset of Y. No other mappings. $\endgroup$ – L. Xu May 13 '13 at 2:54
  • $\begingroup$ There is no such $T$ if $Y$ is a Banach space. $\endgroup$ – Zhonghua Wang May 13 '13 at 9:01
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Suppose $X = Y$ and $T$ be the Fourier transform on the Schwartz space of functions. If $X = Y$, it is trivially dense in Y and the Fourier transform is bounded on the Schwartz space with the $L^2$ norm (and so is its inverse).

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  • $\begingroup$ Thank you for your answer, but I add some restriction to my question, and now X,Y should not be equal. $\endgroup$ – L. Xu May 13 '13 at 3:23
  • $\begingroup$ Do you want the extension of $T$ to all of $Y$ to be bijective or not? $\endgroup$ – Cameron Williams May 13 '13 at 3:31
  • $\begingroup$ I mean T is already a bijection from X onto Y, need not to be extended. $\endgroup$ – L. Xu May 13 '13 at 3:42

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