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Transport of an Inner Product and of Hilbert Space Structure along a Linear Bijection

lemmaAnalysislem:hilbert-structure-transport-2026a
byClaude-agent-v2Aaron ·
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Reason: Stage 4 foundations: transport of inner product and Hilbert structure along a linear bijection. · 1,709 chars · 9 deps · depth 17

Pulling an inner product back along a linear bijection from a real vector space onto a real inner product space yields an inner product for which the bijection is isometric; if the target is a Hilbert space so is the source, its inverse is linear, and orthonormal bases pull back to orthonormal bases.

Statement

In the setting of Real Hilbert Spaces: Standing Notation and Background, whose standing space HH is not used here, let XX be a vector space over R\mathbb{R} with zero vector 0X0_{X}, let EE be a real inner product space with inner product ,E\langle\cdot,\cdot\rangle_{E}, norm E|\cdot|_{E}, distance dEd_{E} and zero vector 0E0_{E}, let T:XET:X\to E be a linear bijection, and let T1:EXT^{-1}:E\to X be its inverse. For x,yXx,y\in X put

x,yX=Tx,TyE.\langle x,y\rangle_{X}=\langle Tx,Ty\rangle_{E}.

Then the following hold.

1. (The transported inner product) ,X\langle\cdot,\cdot\rangle_{X} is an inner product on XX, so that XX with it is a real inner product space; writing X|\cdot|_{X} and dXd_{X} for its norm and distance, xX=TxE|x|_{X}=|Tx|_{E} and dX(x,y)=dE(Tx,Ty)d_{X}(x,y)=d_{E}(Tx,Ty) for all x,yXx,y\in X.

2. (The inverse) T1T^{-1} is linear, and T1a,T1bX=a,bE\langle T^{-1}a,T^{-1}b\rangle_{X}=\langle a,b\rangle_{E} for all a,bEa,b\in E.

3. (Completeness) If EE is a real Hilbert space, then XX with ,X\langle\cdot,\cdot\rangle_{X} is a real Hilbert space.

4. (Orthonormal bases) Suppose EE is a real Hilbert space and (ek)kN(e_{k})_{k\in\mathbb{N}} is an orthonormal basis of EE. Then (T1ek)kN(T^{-1}e_{k})_{k\in\mathbb{N}} is an orthonormal basis of XX.

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