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corrected typos and added explicatory comments (nota bene, to MY response)
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edit approved May 17, 2018 at 16:48
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edit approved May 17, 2018 at 16:48
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Firstly, with regard to the above comment, I imagine that the claim is that the spaces are isometric in a natural way and this is indeed the case. For the sake of simplicity, I will assume that $M$ is injective and is multiplication by the positive sequence $(\lambda_n)$ on $\ell^2$. Then a simple calculation shows that both spaces consist of the weighted $\ell^2$-spaces of sequences $(\xi_n)$ for which $\sum \lambda_n \xi_n^2 <\infty$. One can use the spectral theorem to reduce to this case, at least for operators with discrete spectrum. The general case is much the same except that,rather than sequences, one has to consider multiplication operators on $L^2$-spaces.

I should perhaps add that one can prove this directly by con sidering the map, $x \mapsto M^{(1/2)} (x)$, factoring over the quotient by its kernel, and extending it to the completions. This dispenses with the recourse to the spectral theorem, but only apparently since it is there implicitly in the use of the square root. I think the above version is more transparent but that is a matter of taste.

Firstly, with regard to the above comment, I imagine that the claim is that the spaces are isometric in a natural way and this is indeed the case. For the sake of simplicity, I will assume that $M$ is injective and is multiplication by the positive sequence $(\lambda_n)$ on $\ell^2$. Then a simple calculation shows that both spaces consist of the weighted $\ell^2$-spaces of sequences $(\xi_n)$ for which $\sum \lambda_n \xi_n^2 <\infty$. One can use the spectral theorem to reduce to this case, at least for operators with discrete spectrum. The general case is much the same except that,rather than sequences, one has to consider multiplication operators on $L^2$-spaces.

Firstly, with regard to the above comment, I imagine that the claim is that the spaces are isometric in a natural way and this is indeed the case. For the sake of simplicity, I will assume that $M$ is injective and is multiplication by the positive sequence $(\lambda_n)$ on $\ell^2$. Then a simple calculation shows that both spaces consist of the weighted $\ell^2$-spaces of sequences $(\xi_n)$ for which $\sum \lambda_n \xi_n^2 <\infty$. One can use the spectral theorem to reduce to this case, at least for operators with discrete spectrum. The general case is much the same except that,rather than sequences, one has to consider multiplication operators on $L^2$-spaces.

I should perhaps add that one can prove this directly by con sidering the map, $x \mapsto M^{(1/2)} (x)$, factoring over the quotient by its kernel, and extending it to the completions. This dispenses with the recourse to the spectral theorem, but only apparently since it is there implicitly in the use of the square root. I think the above version is more transparent but that is a matter of taste.

corrected typos
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edit approved May 17, 2018 at 15:25
Community Bot
edit approved May 17, 2018 at 15:25
Community Bot
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Firstly, with regard to the above comment, I imagine that the claim is that the spaces are isometric in a natural way and this is indeed the case. For the sake of simplicity, I will assume that $M$ is injective and is multiplication by the positive sequence $(\lambda_n)$ on $\ell^2$. Then a simple calculation shows that both spaces consist of the weighted $\ell^2$-spaces conisting of sequences $(\xi_n)$ for which $\sum \lambda \xi^2 <\infty$$\sum \lambda_n \xi_n^2 <\infty$. One can use the spectral theorem to reduce to this case, at least for operators with discrete spectrum. The general case is much the same except that,rather than sequences, one has to consider multiplication operators on $L^2$-spaces.

Firstly, with regard to the above comment, I imagine that the claim is that the spaces are isometric in a natural way and this is indeed the case. For the sake of simplicity, I will assume that $M$ is injective and is multiplication by the positive sequence $(\lambda_n)$ on $\ell^2$. Then a simple calculation shows that both spaces consist of the weighted $\ell^2$-spaces conisting of sequences $(\xi_n)$ for which $\sum \lambda \xi^2 <\infty$. One can use the spectral theorem to reduce to this case, at least for operators with discrete spectrum. The general case is much the same except that,rather than sequences, one has to consider multiplication operators on $L^2$-spaces.

Firstly, with regard to the above comment, I imagine that the claim is that the spaces are isometric in a natural way and this is indeed the case. For the sake of simplicity, I will assume that $M$ is injective and is multiplication by the positive sequence $(\lambda_n)$ on $\ell^2$. Then a simple calculation shows that both spaces consist of the weighted $\ell^2$-spaces of sequences $(\xi_n)$ for which $\sum \lambda_n \xi_n^2 <\infty$. One can use the spectral theorem to reduce to this case, at least for operators with discrete spectrum. The general case is much the same except that,rather than sequences, one has to consider multiplication operators on $L^2$-spaces.

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solway
solway
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Firstly, with regard to the above comment, I imagine that the claim is that the spaces are isometric in a natural way and this is indeed the case. For the sake of simplicity, I will assume that $M$ is injective and is multiplication by the positive sequence $(\lambda_n)$ on $\ell^2$. Then a simple calculation shows that both spaces consist of the weighted $\ell^2$-spaces conisting of sequences $(\xi_n)$ for which $\sum \lambda \xi^2 <\infty$. One can use the spectral theorem to reduce to this case, at least for operators with discrete spectrum. The general case is much the same except that,rather than sequences, one has to consider multiplication operators on $L^2$-spaces.