高々可算無限次元Hilbert空間の完全正規直交系と同値な条件

証明

$ 1\implies2

$ \forall\phi\in H:(\forall n\in\N:\Braket{\psi_n|\phi}=0)\implies\phi=0

$ \implies\forall\phi\in H:\left(\forall n\in\N:\Braket{\psi_n|\phi-\sum_{i\in\N}\Braket{\phi|\psi_i}\psi_i}=0\right)\implies\phi-\sum_{i\in\N}\Braket{\phi|\psi_i}\psi_i=0

$ \iff\forall\phi\in H:\left(\forall n\in\N:\Braket{\psi_n|\phi}-\Braket{\psi_n|\sum_{i\in\N}\Braket{\phi|\psi_i}\psi_i}=0\right)\implies\phi=\sum_{i\in\N}\Braket{\phi|\psi_i}\psi_i

$ \iff\forall\phi\in H:\left(\forall n\in\N:\Braket{\psi_n|\phi}-\sum_{i\in\N}\Braket{\psi_n|\Braket{\phi|\psi_i}\psi_i}=0\right)\implies\phi=\sum_{i\in\N}\Braket{\phi|\psi_i}\psi_i

$ \iff\forall\phi\in H:\left(\forall n\in\N:\Braket{\psi_n|\phi}=\sum_{i\in\N}\Braket{\phi|\psi_i}^*\Braket{\psi_n|\psi_i}\right)\implies\phi=\sum_{i\in\N}\Braket{\phi|\psi_i}\psi_i

$ \iff\forall\phi\in H:\left(\forall n\in\N:\Braket{\psi_n|\phi}=\sum_{i\in\N}\Braket{\psi_n|\psi_i}\Braket{\psi_i|\phi}\right)\implies\phi=\sum_{i\in\N}\Braket{\phi|\psi_i}\psi_i

$ \iff\forall\phi\in H:\left(\forall n\in\N:\Braket{\psi_n|\phi}=\Braket{\psi_n|\phi}\right)\implies\phi=\sum_{i\in\N}\Braket{\phi|\psi_i}\psi_i

$ \iff\forall\phi\in H:\top\implies\phi=\sum_{i\in\N}\Braket{\phi|\psi_i}\psi_i

$ \underline{\iff\forall\phi\in H:\phi=\sum_{i\in\N}\Braket{\phi|\psi_i}\psi_i\quad}_\blacksquare

$ 2\implies 3

$ \forall\phi\in H:\phi=\sum_{i\in\N}\Braket{\phi|\psi_i}\psi_i

$ \implies\forall\phi,\chi\in H:\Braket{\phi|\chi}=\Braket{\sum_{i\in\N}\Braket{\phi|\psi_i}\psi_i|\chi}

$ =\sum_{i\in\N}\Braket{\Braket{\phi|\psi_i}\psi_i|\chi}

$ =\sum_{i\in\N}\Braket{\phi|\psi_n}\Braket{\psi_n|\chi}

$ 3\implies4

$ \forall\phi,\chi\in H:\Braket{\phi|\chi}=\sum_{n\in\N}\Braket{\phi|\psi_n}\Braket{\psi_n|\chi}

$ \implies\forall\phi\in H:\Braket{\phi|\phi}=\sum_{n\in\N}\Braket{\phi|\psi_n}\Braket{\psi_n|\phi}

$ =\sum_{n\in\N}\Braket{\phi|\psi_i}\Braket{\phi|\psi_i}^*

$ =\sum_{n\in\N}|\Braket{\phi|\psi_i}|^2

$ \iff\forall\phi\in H:\lVert\phi\rVert^2=\sum_{n\in\N}|\Braket{\phi|\psi_i}|^2

$ 4\implies1

$ \forall\phi\in H:\lVert\phi\rVert^2=\sum_{n\in\N}|\Braket{\phi|\psi_i}|^2

$ \implies\forall\phi\in H:

$ \forall n\in\N:\Braket{\psi_n|\phi}=0

$ \implies \lVert\phi\rVert^2=\sum_{n\in\N}|\Braket{\phi|\psi_i}|^2

$ =\sum_{n\in\N}0^2

$ =0

$ \implies\phi=0

$ \implies\forall\phi\in H:(\forall n\in\N:\Braket{\psi_n|\phi}=0)\implies\phi=0

$ \forall a_\bullet\in l^2(K):\left\lVert\sum_{n\in\N}a_n\psi_n\right\rVert^2=\sum_{n\in\N}|a_n|^2

証明

$ \forall a_\bullet:

$ a_\bullet\in l^2(K)

$ \implies\forall\varepsilon>0\exist N\in\N\forall n>N\forall m>n:\left|\sum_{1\le i\le m}|a_i|^2-\sum_{1\le i\le n}|a_i|^2\right|<\varepsilon

$ \implies\forall\varepsilon>0\exist N\in\N\forall n>N\forall m>n:\sum_{n<i\le m}|a_i|^2<\varepsilon^2

$ \implies\forall\varepsilon>0\exist N\in\N\forall n>N\forall m>n:

$ \left\lVert\sum_{1\le i\le m}a_i\psi_i-\sum_{1\le i\le n}a_i\psi_i\right\rVert=\left\lVert\sum_{n< i\le m}a_i\psi_i\right\rVert

$ =\sqrt{\sum_{n<i\le m}|a_i|^2}

$ <\varepsilon

$ \implies\sum_{n\in\N}a_n\psi_n\in H

$ \implies\left\lVert\sum_{1\le n\le N}a_n\psi_n\right\rVert^2\to\left\lVert\sum_{n\in\N}a_n\psi_n\right\rVert^2\pod{N\to\infty}

$ \iff\sum_{1\le n\le N}|a_n|^2\to\left\lVert\sum_{n\in\N}a_n\psi_n\right\rVert^2\pod{N\to\infty}

$ \iff\left\lVert\sum_{n\in\N}a_n\psi_n\right\rVert^2=\sum_{n\in\N}|a_n|^2

$ \underline{\implies\forall a_\bullet\in l^2(K):\left\lVert\sum_{n\in\N}a_n\psi_n\right\rVert^2=\sum_{n\in\N}|a_n|^2\quad}_\blacksquare

可算無限個のときと有限個のときとをまとめて示す

References

同値な条件の一つとして3.を上げていた