Why is identity map on a separable Hilbert space not compact? False proof.

Mathematics Asked on January 5, 2022

Let $e_n$ be the orthonormal basis. Then the projection map onto $H$ is defined by $sum(x,e_k)e_k$. What is stopping us from taking a finite part of this series. This gives us a compact operators, that converge to the identity map. Clearly something is wrong here not all Hilbert spaces are finite dimensional.

compact operators functional analysis hilbert spaces

One Answer

Beacause that sequence of operators does not converge to the identity: if $ninBbb N$,$$leftlVert e_{n+1}-sum_{j=1}^nlangle e_{n+1},e_krangle e_krightrVert=lVert e_{n+1}rVert=1$$and therefore$$leftlVertoperatorname{Id}-sum_{k=1}^nlanglecdot,e_krangle e_krightrVertgeqslant1.$$

Answered by José Carlos Santos on January 5, 2022

Why is identity map on a separable Hilbert space not compact? False proof.

One Answer

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