SOLIT for the periodic heat equation (reported simulation)

Interleaving the constant coordinate and cosine–sine pairs is a linear isometry onto real ${\mathrm{\ell }}^2({\mathbb{N}}_{>0})$, so [F7] gives the stated real Hilbert structure. With normalized Haar measure $dx\mathrm{/}(2\pi )$ and characters $e_n(x)=e^{inx}$, use the real modes $1$, $C_n=(e_n+e_{-n})\mathrm{/}\sqrt{2}=\sqrt{2}\cos (nx)$ and $S_n=(e_n-e_{-n})\mathrm{/}(i\sqrt{2})=\sqrt{2}\sin (nx)$ for $n\ge 1$. Character orthonormality in [F1] gives their orthonormality. They are complete in real $L^2$: a real function orthogonal to all these modes is orthogonal to every character, hence is zero by [F1]. Thus their coefficient map is an isometry onto the stated coefficient space. The bounded truth lies in $L^2$ and has such coordinates. On each positive frequency, ${\mathrm{\partial }}_{xx}{C}_n=-n^2{C}_n$ and ${\mathrm{\partial }}_{xx}{S}_n=-n^2{S}_n$, so solving ${\mathrm{\partial }}_{t}u={\mathrm{\partial }}_{xx}u$ multiplies both coefficients by $e^{-n^2\bar{t}}$; the constant mode is unchanged. Therefore the operator in the Example is the periodic heat forward map in normalized Fourier coordinates.

Every multiplier of $T_{\bar{t}}$ is positive, so $T_{\bar{t}}$ is injective. The positive-frequency multipliers tend to zero, so finite-frequency truncations converge in operator norm and $T_{\bar{t}}$ is compact. Moreover, ${\sum }_{n\ge 1}2e^{-2n^2\bar{t}}<\mathrm{\infty }$ by comparison with the geometric series ${\sum }_{n\ge 1}2e^{-2n\bar{t}}$, so the operator is Hilbert–Schmidt. Squaring the multipliers gives the stated eigenvalues of $T_{\bar{t}}^{\ast }{T}_{\bar{t}}$.

For spectral cut-off, the inverse-eigenvalue contribution at frequency $n$ is $2e^{2n^2\bar{t}}$. Since $V_{n-1}\le (2n-1)e^{2(n-1{)}^2\bar{t}}$,

Hence no finite ${\theta }_2$ bounds all consecutive ratios in this sequence, as claimed.

For every $\lambda \ge 0$, $q_{\alpha }(\lambda )=(\lambda +\alpha {)}^{-1}$ is nonnegative, $\alpha q_{\alpha }(\lambda )\le 1$, and $\lambda q_{\alpha }(\lambda )\le 1$. If $\tilde{\alpha }<\alpha$, then $\lambda +\tilde{\alpha }<\lambda +\alpha$, so $q_{\alpha }(\lambda )\le q_{\tilde{\alpha }}(\lambda )$. Thus it is an ordered filter with $C_q^{\mathrm{\prime }}=1$ by [F6]. Its variance is $V(\alpha )=1\mathrm{/}(1+\alpha {)}^2+2{\sum }_{n\ge 1}{e}^{-2n^2\bar{t}}(e^{-2n^2\bar{t}}+\alpha {)}^{-2}$. On each compact interval $\alpha \ge a>0$, the summands are continuous and bounded by $2a^{-2}{e}^{-2n^2\bar{t}}$, a summable sequence, so $V$ is continuous there by uniform convergence. It is strictly decreasing. For $\alpha \ge 1$, $V(\alpha )\le {\alpha }^{-2}(1+2{\sum }_{n\ge 1}{e}^{-2n^2\bar{t}})\to 0$ as $\alpha \to \mathrm{\infty }$. It tends to infinity as $\alpha \downarrow 0$: for every $N$, its first $N$ positive-frequency terms tend to $2{\sum }_{n=1}^N{e}^{2n^2\bar{t}}$, and these partial sums are unbounded in $N$. The intermediate value theorem therefore selects recursively a decreasing sequence with any prescribed fixed ratio ${\theta }_1>1$ between consecutive variance values; its upper ratio is the same ${\theta }_1$. The estimator and each finite comparison are well-defined by [F3]–[F5].