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How can we construct an example of isometric linear operator $T: H \rightarrow H$ which is not unitary but maps the Hilbert space $H$ onto a proper closed subspace of $H.$

My attempt I found this proof here Prob. 9, Sec. 3.10 in Kreyszig's functional analysis book: The image of ann isometric non-unitary operator on a Hilbert space

But in the given link answer there is no example .

My Question is that how can we take an example that satisfied the given above statement

jasmine
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    Try exploring the space $\ell^2$, the square summable sequences. Try some linear maps that keep all the same terms of the given sequence, while introducing some $0$s. – user780985 May 03 '20 at 09:05

1 Answers1

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Credit : user780985

take $f : l^2 \rightarrow l^2$ defined by $f( x_1,x_2,......) = (0 ,x_1,x_2,x_3,...)$

where $x= (x_n) \in l^2$

This is not unitary since $f(x_1,x_2,......) \notin (1, 1/2,1/3 ....)$

i mean to say that $f$ is not onto

lastly used this logics

Unitary = onto + isometry

jasmine
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