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Lies Basis 📂Hilbert Space

Lies Basis

Definition1

Let’s assume that a normalized orthogonal basis {ek}k∈N\left\{ \mathbf{e}_{k} \right\}_{k \in \mathbb{N}} of a Hilbert space HH is given. If surjective U:H→HU : H \to H is a linear and bounded operator for all k∈Nk \in \mathbb{N}, then vk:=Uek\mathbf{v}_{k} := U \mathbf{e}_{k} implies that {vk}k∈N\left\{ \mathbf{v}_{k} \right\}_{k \in \mathbb{N}} forms a basis of HH, and the following holds.

v=∑k∈N⟨v,(U−1)∗ek⟩vk \mathbf{v} = \sum_{k \in \mathbb{N}} \left\langle \mathbf{v} , \left( U^{-1} \right)^{ \ast } \mathbf{e}_{k} \right\rangle \mathbf{v}_{k}

Explanation

While it is certainly good that we can specifically choose a basis as mentioned above when UU is given, one must be careful since the condition of UU must be equally favorable. Of course, the identity operator II easily satisfies this, but ultimately it leads to vk=ek\mathbf{v}_{k} = \mathbf{e}_{k}, leaving only a vacuous claim.

Proof

Since {ek}k=1∞\left\{ \mathbf{e}_{k} \right\}_{k=1}^{\infty} is a basis of VV, for any v∈V\mathbf{v} \in V that is not a1=⋯=0a_{1} = \cdots = 0 but {ak}k=1∞⊂C\left\{ a_{k} \right\}_{k=1}^{\infty} \subset \mathbb{C}, it is represented as follows.

v=∑k=1∞akek \mathbf{v} = \sum_{k =1}^{\infty} a_{k} \mathbf{e}_{k}

By the normal orthogonality of {ek}k=1∞\left\{ \mathbf{e}_{k} \right\}_{k=1}^{\infty}, ⟨ei,ei⟩=1\left\langle \mathbf{e}_{i} , \mathbf{e}_{i} \right\rangle = 1 and for i≠ji \ne j regarding ⟨ei,ej⟩=0\left\langle \mathbf{e}_{i} , \mathbf{e}_{j} \right\rangle = 0, we have

⟨v,v⟩=∑k=1∞ak2 \left\langle \mathbf{v} , \mathbf{v} \right\rangle = \sum_{k =1}^{\infty} a_{k}^{2}

Meanwhile, from v=∑k=1∞akek\mathbf{v} = \sum_{k =1}^{\infty} a_{k} \mathbf{e}_{k},

⟨v,v⟩=∑k=1∞ak⟨v,ek⟩ \left\langle \mathbf{v} ,\mathbf{v} \right\rangle = \sum_{k =1}^{\infty} a_{k} \left\langle \mathbf{v} , \mathbf{e}_{k} \right\rangle

Thus,

∑k=1∞ak2=∑k=1∞ak⟨v,ek⟩ \sum_{k =1}^{\infty} a_{k}^{2} = \sum_{k =1}^{\infty} a_{k} \left\langle \mathbf{v} , \mathbf{e}_{k} \right\rangle

Summing up,

∑k=1∞ak(ak−⟨v,ek⟩)=0 \sum_{k =1}^{\infty} a_{k} \left( a_{k} - \left\langle \mathbf{v} , \mathbf{e}_{k} \right\rangle \right) = 0

Therefore, for all k∈Nk \in \mathbb{N}, ak=⟨v,ek⟩a_{k} = \left\langle \mathbf{v} , \mathbf{e}_{k} \right\rangle must hold.

v=∑k=1∞⟨v,ek⟩ek \mathbf{v} = \sum_{k=1}^{\infty} \left\langle \mathbf{v} , \mathbf{e}_{k} \right\rangle \mathbf{e}_{k}

Taking the operator UU,

Uv=∑k=1∞⟨v,ek⟩vk U \mathbf{v} = \sum_{k=1}^{\infty} \left\langle \mathbf{v} , \mathbf{e}_{k} \right\rangle \mathbf{v}_{k}

Here, the adjoint operator (U−1)∗\left( U^{-1} \right)^{ \ast } of U−1U^{-1} uniquely exists due to the Riesz Representation Theorem.

v=U(U−1v)=∑k=1∞⟨U−1v,ek⟩vk=∑k=1∞⟨v,(U−1)∗ek⟩vk \begin{align*} \mathbf{v} =& U \left( U^{-1} \mathbf{v} \right) \\ =& \sum_{k=1}^{\infty} \left\langle U^{-1} \mathbf{v} , \mathbf{e}_{k} \right\rangle \mathbf{v}_{k} \\ =& \sum_{k=1}^{\infty} \left\langle \mathbf{v} , \left( U^{-1} \right)^{ \ast } \mathbf{e}_{k} \right\rangle \mathbf{v}_{k} \end{align*}

Thus, for all v∈H\mathbf{v} \in H and {vk=Uek}k=1∞\left\{ \mathbf{v}_{k} = U \mathbf{e}_{k} \right\}_{k=1}^{\infty}, there uniquely exists {ck:=⟨v,(U−1)∗ek⟩}k∈N⊂C\left\{ c_{k} := \left\langle \mathbf{v} , \left( U^{-1} \right)^{ \ast } \mathbf{e}_{k} \right\rangle \right\}_{k \in \mathbb{N}} \subset \mathbb{C} satisfying the following.

v=∑k=1∞ckvk \mathbf{v} = \sum_{k =1}^{\infty} c_{k} \mathbf{v}_{k}

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  1. Ole Christensen, Functions, Spaces, and Expansions: Mathematical Tools in Physics and Engineering (2010), p91 ↩︎