Incomplete tensor product of one-mode Hilbert spaces with a reference vector

Use finite Borel representatives as in [F2]. By [F3], the nonnegative function $\mathrm{\mid }f(x){\mathrm{\mid }}^2{\prod }_{j=n+1}^m\mathrm{\mid }\mathrm{\Omega }(x_j){\mathrm{\mid }}^2$ is product-measurable, and Tonelli gives $\mathrm{\parallel }{J}_{n,m}f{\mathrm{\parallel }}_2^2=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2{\prod }_{j=n+1}^m\mathrm{\parallel }\mathrm{\Omega }{\mathrm{\parallel }}_2^2=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2<\mathrm{\infty }$. For $n=0$ the first factor is $\mathrm{\mid }f{\mathrm{\mid }}^2$, and $m=n$ is the identity map. Applying the same calculation to a difference shows independence of the representative of $f$; changing a representative of $\mathrm{\Omega }$ changes the product only on finitely many null-coordinate cylinders, which are null by Tonelli and sigma-finiteness. Thus $J_{n,m}$ is a well-defined linear isometry on classes. Substitution gives $J_{m,r}{J}_{n,m}=J_{n,r}$. Computing two vectors' inner product at a common index is independent of a larger common index, since linear isometries preserve inner products by polarization. This defines a positive-definite inner product on the compatible union.

For $n\ge j$, write $x=(z,y)$ with $y=x_j$ and the other coordinates denoted by $z$. This coordinate permutation preserves Euclidean measure: on Borel rectangles both measures give the product of the side measures, so [F3] applies in this order too. Tonelli gives $\int \mathrm{\parallel }f(z,\cdot ){\mathrm{\parallel }}_2^{2}dz=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2$, so almost every section lies in $L^2(\mathbb{R})$. Set $a_f(z)=\int f(z,y)\overline{\mathrm{\Phi }(y)}dy$ on those sections, and zero on the exceptional set. Cauchy--Schwarz gives $\mathrm{\mid }{a}_f(z){\mathrm{\mid }}^2\le \mathrm{\parallel }f(z,\cdot ){\mathrm{\parallel }}_2^2$. Tonelli applied to the positive and negative parts of the real and imaginary integrand makes $a_f$ measurable; its section integral is finite almost everywhere by this estimate. Define $(P_{j,n}^{\mathrm{\Phi }}f)(z,y)=a_f(z)\mathrm{\Phi }(y)$. Tonelli now gives $\mathrm{\parallel }{P}_{j,n}^{\mathrm{\Phi }}f{\mathrm{\parallel }}_2^2=\int \mathrm{\mid }{a}_f{\mathrm{\mid }}^2\le \mathrm{\parallel }f{\mathrm{\parallel }}_2^2$, and the same estimate for differences makes the map well-defined on classes. Since $\mathrm{\parallel }\mathrm{\Phi }{\mathrm{\parallel }}_2=1$, section integration gives $a_{P_{j,n}^{\mathrm{\Phi }}f}=a_f$, hence idempotence. The products $(P_{j,n}^{\mathrm{\Phi }}f)\bar{g}$ and $f\overline{P_{j,n}^{\mathrm{\Phi }}g}$ are integrable by Cauchy--Schwarz on ${\mathbb{R}}^n$; Fubini therefore yields $⟨P_{j,n}^{\mathrm{\Phi }}f,g⟩=\int a_f\overline{a_g}=⟨f,P_{j,n}^{\mathrm{\Phi }}g⟩$. Thus $P_{j,n}^{\mathrm{\Phi }}$ is an orthogonal projection. When $n=j=1$, the $z$-space is the one-point space and the same formulas hold.

If $r\ge n\ge j$, the formula in step 1.2 acts on one of the first $n$ coordinates, while $J_{n,r}$ appends only factors $\mathrm{\Omega }$ after coordinate $n$. Substitution in the integral formula gives $P_{j,r}^{\mathrm{\Phi }}{J}_{n,r}=J_{n,r}{P}_{j,n}^{\mathrm{\Phi }}$. For a vector initially represented in $H_n$ with $n<j$, first embed it in $H_j$; the compatibility identity shows that this and every later representation give the same image. Hence the projections define a contraction $P_j^{\mathrm{\Phi }}$ on the compatible union. It extends uniquely by continuity to the Hilbert completion, and the identities of step 1.2 pass to limits, so the extension remains an orthogonal projection.

The compatible union is dense in its Hilbert completion by construction. Steps 1.1 and 2.1 prove, respectively, the well-defined incomplete tensor product and every stated local-projection property.