The square-root filter has an explicit conditioned finite-horizon error bound

Put ${\delta }_{k-1}^P=\mathrm{\parallel }{\widehat{P}}_{k-1\mathrm{\mid }k-1}-P_{k-1\mathrm{\mid }k-1}{\mathrm{\parallel }}_2$ and ${\delta }_{k-1}^m=\mathrm{\parallel }{\widehat{m}}_{k-1\mathrm{\mid }k-1}-m_{k-1\mathrm{\mid }k-1}{\mathrm{\parallel }}_2$. By [F1], [F5] and the prediction defect, $\mathrm{\parallel }{\widehat{P}}_k^{-}-P_k^{-}{\mathrm{\parallel }}_2\le f_k^2{\delta }_{k-1}^P+e_k^P,\mathrm{\parallel }{\widehat{m}}_k^{-}-m_k^{-}{\mathrm{\parallel }}_2\le f_k{\delta }_{k-1}^m+e_k^m\mathrm{.}$

Let $C_H=\left[\begin{array}{c}{H}_k\\ I\end{array}\right]$. Direct block multiplication gives $J_k({\widehat{P}}_k^{-})-J_k(P_k^{-})=C_H({\widehat{P}}_k^{-}-P_k^{-})C_H^T$. For every $x$, $\mathrm{\parallel }{C}_{H}x{\mathrm{\parallel }}_2^2=\mathrm{\parallel }{H}_{k}x{\mathrm{\parallel }}_2^2+\mathrm{\parallel }x{\mathrm{\parallel }}_2^2\le (1+h_k^2)\mathrm{\parallel }x{\mathrm{\parallel }}_2^2$, so [F5] and the joint defect imply $\mathrm{\parallel }{\widehat{J}}_k-J_k(P_k^{-}){\mathrm{\parallel }}_2\le (1+h_k^2)(f_k^2{\delta }_{k-1}^P+e_k^P)+e_k^J\mathrm{.}$ If ${\delta }_{k-1}^P\le {\mathrm{\Delta }}_{k-1}$, the right side is at most ${\epsilon }_k$. Each of its observation, cross, and state blocks therefore has operator error at most ${\epsilon }_k$: every block is obtained by left- and right-multiplying by coordinate projections of operator norm at most one, so [F5] and submultiplicativity give the bound.

Assume ${\delta }_{k-1}^P\le {\mathrm{\Delta }}_{k-1}$ and suppress $k$. Write $E_S=\widehat{S}-S$, $E_B=\widehat{B}-B$, and $E_D=\widehat{D}-P^{-}$. Step 2.1 bounds their operator norms by $\epsilon \le \mu \mathrm{/}2$. Cauchy–Schwarz gives $\mathrm{\mid }{x}^T{E}_{S}x\mathrm{\mid }\le \epsilon \mathrm{\parallel }x{\mathrm{\parallel }}_2^2$, hence $\widehat{S}⪰(\mu -\epsilon )I\succ 0$. The same inequality applied to $y=Sx$ and $y=\widehat{S}x$ gives $\mathrm{\parallel }{S}^{-1}{\mathrm{\parallel }}_2\le {\mu }^{-1}$ and $\mathrm{\parallel }{\widehat{S}}^{-1}{\mathrm{\parallel }}_2\le \beta :=(\mu -\epsilon {)}^{-1}\le 2\mathrm{/}\mu$. The resolvent identity ${\widehat{S}}^{-1}-S^{-1}=-S^{-1}{E}_S{\widehat{S}}^{-1}$ then bounds the inverse difference by ${\mu }^{-1}\beta \epsilon$. Expanding the gain and Schur differences as in [F4], now applying operator-norm triangle and product inequalities throughout, yields $\mathrm{\parallel }\widehat{B}{\widehat{S}}^{-1}-B{S}^{-1}{\mathrm{\parallel }}_2\le \beta \epsilon +b{\mu }^{-1}\beta \epsilon \le C^K\epsilon ,$ $\mathrm{\parallel }{\widehat{P}}_{k\mathrm{\mid }k}-P_{k\mathrm{\mid }k}{\mathrm{\parallel }}_2\le \epsilon +2b\beta \epsilon +b^2{\mu }^{-1}\beta \epsilon +\beta {\epsilon }^2\le C^P\epsilon \mathrm{.}$ Here [F2] identifies the represented posterior as the Schur complement, and $\beta {\epsilon }^2\le \epsilon$. Finally, $\mathrm{\parallel }\widehat{B}{\widehat{S}}^{-1}{\mathrm{\parallel }}_2\le (b+\epsilon )\beta \le 1+2b\mathrm{/}\mu =K^{\mathrm{\#}}$. These estimates use operator block bounds directly, without a Frobenius conversion.

The innovation residual and computed-innovation defect give $\mathrm{\parallel }{\widehat{v}}_k-v_k{\mathrm{\parallel }}_2\le h_k\mathrm{\parallel }{\widehat{m}}_k^{-}-m_k^{-}{\mathrm{\parallel }}_2+e_k^v\mathrm{.}$ Let $z_k^{\ast }={\widehat{L}}_{11,k}^{-1}{\widehat{v}}_k$. By [F6] and the ${\widehat{S}}_k⪰({\mu }_k\mathrm{/}2)I$ bound from step 3.1, $\mathrm{\parallel }{\widehat{z}}_k-z_k^{\ast }{\mathrm{\parallel }}_2\le \sqrt{2\mathrm{/}{\mu }_k}{\rho }_k$, as included in the Statement. Because ${\widehat{L}}_{11,k}{\widehat{z}}_k={\widehat{v}}_k+{\xi }_k$ with $\mathrm{\parallel }{\xi }_k{\mathrm{\parallel }}_2\le {\rho }_k$, [F2] gives ${\widehat{L}}_{21,k}{\widehat{z}}_k={\widehat{B}}_k{\widehat{S}}_k^{-1}({\widehat{v}}_k+{\xi }_k)$. Subtract the exact mean equation in [F1], insert and subtract ${\widehat{B}}_k{\widehat{S}}_k^{-1}{v}_k$, and use step 3.1. The result is $\mathrm{\parallel }{\widehat{m}}_{k\mathrm{\mid }k}-m_{k\mathrm{\mid }k}{\mathrm{\parallel }}_2\le (1+h_k{K}_k^{\mathrm{\#}})(f_k{\delta }_{k-1}^m+e_k^m)+C_k^K{\epsilon }_k{r}_k+K_k^{\mathrm{\#}}(e_k^v+{\rho }_k)+e_k^u\mathrm{.}$

Add the covariance bound in step 3.1 to the mean bound in step 4.1. Using ${\delta }_{k-1}^P,{\delta }_{k-1}^m\le {\mathrm{\Delta }}_{k-1}$ and the definition of ${\epsilon }_k$ yields $\mathrm{\parallel }{\widehat{m}}_{k\mathrm{\mid }k}-m_{k\mathrm{\mid }k}{\mathrm{\parallel }}_2+\mathrm{\parallel }{\widehat{P}}_{k\mathrm{\mid }k}-P_{k\mathrm{\mid }k}{\mathrm{\parallel }}_2\le A_k{\mathrm{\Delta }}_{k-1}+{\eta }_k={\mathrm{\Delta }}_k\mathrm{.}$ Induction from the stated ${\mathrm{\Delta }}_0$ proves the bound for each $k$. Repeated substitution of the scalar recurrence gives the displayed product-and-sum formula at $T$. Since each assumption and step concerns only the finite indices $1,\dots ,T$, no uniform-in-time conclusion follows.