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

Question

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

José Carlos Santos · 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.

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