Banach repair and constructed split coordinates

Proof

Put \(r_m=df^{(m)}\). Additivity of \(d\) gives

Each \(r_m\) is the residual of an actual candidate, so the stated hypothesis applies at every iteration. Induction yields, for \(j\ge0\),

For \(n>m\), the finite triangle inequality therefore gives

Since \({\left\lVert r_m\right\rVert}\le q^m{\left\lVert df\right\rVert}\) tends to zero, this is a Cauchy estimate. Completeness of \(E\) supplies a limit, denoted \(P_Af\). Passing \(n\to\infty\) in the same norm inequality proves the required tail estimate.

Continuity of \(d\) implies \(d(P_Af)=\lim_m r_m=0\). Protection holds before taking a limit: \(pA=0\) gives \(pf^{(m+1)}=pf^{(m)}\), hence \(pf^{(m)}=pf\) for every \(m\). Continuity of \(p\) gives \(p(P_Af)=pf\). This explains why continuity of the observable belongs to this analytic statement.

For linearity, write \(L=I-Ad\), so \(f^{(m)}=L^mf\). For fixed \(f,g\) and scalars \(\lambda,\mu\),

Continuity of addition and scalar multiplication allows passage to the norm limit and proves linearity of \(P_A\). The estimate with \(m=0\) gives

so the linear map is bounded. If \(df=0\), every iterate equals \(f\), hence \(P_Af=f\). Its image is contained in \(\ker d\) and it fixes that kernel, proving surjectivity onto its stated image and \(P_A^2=P_A\). For \(q=0\) the residual vanishes after the first update and the same inequalities remain valid.

Back to statement · Permalink · Source line 128

Proof

Let \(\theta=SB_X<1\). The number \(K=S/(1-\theta)\) is nonnegative and satisfies

Fix \(r\in R\) and define its coordinates recursively as in (8.3). The initial coordinate has norm zero. If \({\left\lVert(Ar)_n\right\rVert}\le K{\left\lVert r\right\rVert}_\infty\), submultiplicativity of operator norms gives

Induction proves the estimate at every coordinate, so the recursively defined family actually belongs to the bounded product \(E\).

The coordinate recurrence preserves addition and scalar multiplication, starting from the zero initial coordinate. Induction therefore proves these laws for the family map \(A\). Its uniform estimate proves \({\left\lVert A\right\rVert}\le K\), giving a continuous linear map with the claimed domain and codomain.

For every \(n\), apply the target restriction to the recurrence. The equation \(\rho^Y_ns_n=I\) gives

Coordinate equality yields the operator identity \(\partial A=I_R\). In particular this identity holds for every bounded residual, not just residuals already known to lie in the image.

Set \(P=I-A\partial\). Then

The expression \(2A\partial\) abbreviates addition of the same operator twice and uses no division by two. Every exact \(f\) is fixed because \(\partial f=0\). Since \((Ar)_0=0\), the seed of \(Pf\) is \(f_0\). Finally \(f-Pf=A\partial f\), and the bound on \(A\) gives the asserted residual estimate. The denominators above are justified exactly by the strict propagation gap.

Back to statement · Permalink · Source line 189

Proof

Put \(B_n=I-s_n\rho^Y_n\). Since \(\rho^Y_ns_n=I\), direct multiplication gives

Moreover, \(B_ny=y\) whenever \(\rho^Y_ny=0\). Thus \(B_n\) really projects onto the target kernel. Its norm is at most \(1+SB_Y\). Coordinate zero of \(p_L\) is the identity, and every later coordinate applies \(B_n\), so \(p_L^2=p_L\) and \({\left\lVert p_L\right\rVert}\le1+SB_Y\). In particular the formula defines a map on the bounded product, not just on arbitrary coordinate families.

The recurrence for \(Ar\) gives zero seed and places \((Ar)_{n+1}\) in the image of \(s_n\). The identities just proved make all its innovations zero. Therefore \(p_LA=0\), and

We next prove separation on the exact space. If \(f\) is exact, its restriction equation rewrites as

Two exact families with the same seed and innovations agree first at level zero and then at every later level by this recurrence. More generally, if two arbitrary families have the same observable and differential, their difference is exact with zero observable, so separation on the exact space makes their difference zero.

The proposed chart has the claimed target because \(p_L(p_Lf)=p_Lf\). Given \(s\in\mathcal S_L\) and \(r\in R\), set \(f=Ps+Ar\). The preceding identities and \(\partial A=I\) give

This proves one round trip. For the other, \(f\) and \(P(p_Lf)+A\partial f\) have the same observable and residual, so the joint separation just proved identifies them. All maps are linear and bounded on their indicated subspaces and finite products, establishing a continuous linear equivalence.

Finally, \(Pf\) has coordinates \((p_Lf,0)\). Any exact answer with the same observable has those same two coordinates, hence equals \(Pf\). This uniqueness is relative to the complete observable of seed and innovations. It is not obtained by assuming that the seed alone determines an exact growing-target family.

Back to statement · Permalink · Source line 244

Proof

At level zero the value is one. Exactness forces every inherited row at the next level to be one half of its previous value. Preservation of the innovation forces the new row to remain the corresponding identity row. Induction gives the displayed diagonal.

Back to statement · Permalink · Source line 285