Bounded and invertible operator on dense subspace

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.

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

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).
• Do you want the extension of $T$ to all of $Y$ to be bijective or not? – Cameron Williams May 13 '13 at 3:31