Real and complex Hilbert source powers and fixed-reference Tikhonov filter bounds

If $X=\{0\}$ or $A=0$, all asserted powers and products involving $A$ are zero, and the bounds and kernel conclusions follow directly. In the remaining case $S=A^{\ast }A$ is positive self-adjoint and $\mathrm{\parallel }S\mathrm{\parallel }=\mathrm{\parallel }A{\mathrm{\parallel }}^2>0$ by the adjoint identities.

For real $X$, construct $X_c=X\times X$ with $(a+ib)(x,y)=(ax-by,bx+ay)$ and $⟨(x,y),(u,v){⟩}_c=⟨x,u⟩+⟨y,v⟩+i(⟨y,u⟩-⟨x,v⟩)$. Expansion verifies conjugate symmetry, complex linearity in the first argument and norm squared $\mathrm{\parallel }x{\mathrm{\parallel }}^2+\mathrm{\parallel }y{\mathrm{\parallel }}^2$; completeness follows by taking the two component limits of a Cauchy sequence. Thus this is a complex Hilbert space. Construct $Y_c$ similarly. The extensions $A_c(x,y)=(Ax,Ay)$ and $S_c(x,y)=(Sx,Sy)$ have the same operator norms as $A,S$: the upper bound follows by summing squared component bounds, and real vectors give the reverse inequality. The inner-product identity gives $A_c^{\ast }=(A^{\ast }{)}_c$ and $S_c=A_c^{\ast }{A}_c$. This construction uses the Hilbert sum norm, not a rotation-supremum Banach norm.

In the nonzero complex space, or in step 2.1's complexification, apply [F1]–[F2]. For $\nu >0$, $t^{\nu }$ extends continuously to zero: for $0<t<{\epsilon }^{1\mathrm{/}\nu }$, the positive derivative and [F8], followed by the power laws, give $t^{\nu }<\epsilon$. Hence [F3] supplies real polynomials converging uniformly to $t^{\nu }$ on $[0,\mathrm{\parallel }S\mathrm{\parallel }]$. The calculus maps them to a norm-convergent sequence of self-adjoint operators. In the real case each polynomial in $S_c$ preserves $j(X)=\{(x,0)\}$, a closed real subspace, so the limit preserves it too. Restriction defines $S^{\nu }$ over the real field, independent of approximants by the norm estimate from [F1]. Polynomial identities and norm bounds pass to these restrictions.

For $h\in \ker A$, $Sh=0$, so every polynomial satisfies $p(S)h=p(0)h$. Taking step 3.1's approximants gives $S^{\nu }h=0$. Self-adjointness then gives $⟨S^{\nu }w,h⟩=⟨w,S^{\nu }h⟩=0$. Also the calculus norm gives $\mathrm{\parallel }{S}^{\nu }\mathrm{\parallel }\le \mathrm{\parallel }S{\mathrm{\parallel }}^{\nu }=\mathrm{\parallel }A{\mathrm{\parallel }}^{2\nu }$.

The calculus identifies $T_{\alpha }$ with $(t+\alpha {)}^{-1}$: multiplying this function by $t+\alpha$ gives the identity, so uniqueness of the inverse in [F7] applies. For $0<\gamma \le 1$ and $u\ge 0$, $u^{\gamma }\le 1+u$ (use $u\le 1$ and $u\ge 1$ separately). Scaling $t=\alpha u$ therefore gives $\alpha t^{\nu }\mathrm{/}(t+\alpha )\le {\alpha }^{\nu }$ for $0<\nu \le 1$. Applying the isometric calculus and its real restriction proves the first filter bound.

Put $Q=\alpha T_{\alpha }{S}^{\nu }$. By [F4], $\mathrm{\parallel }AQ{\mathrm{\parallel }}^2=\mathrm{\parallel }{Q}^{\ast }SQ\mathrm{\parallel }$. Its scalar function is $[\alpha t^{\nu +1\mathrm{/}2}\mathrm{/}(t+\alpha ){]}^2$. Since $\nu +1\mathrm{/}2\le 1$, the same scalar estimate as step 4.2 bounds this by ${\alpha }^{2\nu +1}$, proving the second bound. The noise bound is [F7]. Finally $(A{A}^{\ast }+\alpha I)A=A(S+\alpha I)$ yields $A{T}_{\alpha }{A}^{\ast }=(A{A}^{\ast }+\alpha I{)}^{-1}A{A}^{\ast }$. Apply the calculus to $A{A}^{\ast }$ on $Y$, or directly use zero if $Y=\{0\}$: its function $t\mathrm{/}(t+\alpha )$ is at most one. This proves the last bound in either field.