SOLIT for a Dirichlet second-antiderivative model

The functions $e_j(x)=\sqrt{2}\sin (\pi jx)$ are orthonormal on $(0,1)$ by the product-to-sum identities and direct integration. If $h\in L^2(0,1)$ is orthogonal to all of them, its odd extension to $(-1,1)$, continued periodically with period $2$, has zero constant and cosine coefficients by oddness and zero sine coefficients by the assumed orthogonality. Rescaling to the circle in [F1], every Fourier coefficient of this extension vanishes; completeness in [F2] therefore makes the extension, and hence $h$, zero almost everywhere. Thus $(e_j)$ is a complete orthonormal basis, and [F7] supplies the real Hilbert structure used below.

For $f\in L^2(0,1)$, [F8] gives $f\in L^1(0,1)$. Since $0\le G(x,t)\le 1\mathrm{/}4$, $\mathrm{\mid }Tf(x)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{/}4\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{/}4$ for every $x$, so the integral formula defines a bounded linear operator from $L^2(0,1)$ to itself. The kernel formula also gives $(Tf)(x)=(1-x){\int }_0^{x}tf(t)dt+x{\int }_x^1(1-t)f(t)dt\mathrm{.}$ The two indefinite integrals are absolutely continuous by [F9], and [F10] gives their derivatives almost everywhere. They show that $Tf$ is continuous, has zero endpoint values, and has first derivative almost everywhere equal to the absolutely continuous function $g(x)=-{\int }_0^{x}tf(t)dt+{\int }_x^1(1-t)f(t)dt$, whose derivative is $g^{\mathrm{\prime }}(x)=-xf(x)-(1-x)f(x)=-f(x)$ almost everywhere. To verify the claimed distributional equation directly, take any $\varphi \in C_c^{\mathrm{\infty }}(0,1)$. The function $(x,t)\mapsto G(x,t)f(t){\varphi }^{\mathrm{\prime }\mathrm{\prime }}(x)$ is integrable on $(0,1{)}^2$, so [F11] permits reversing the integrals; splitting at $x=t$ and integrating the smooth test function twice gives ${\int }_0^1(Tf)(x){\varphi }^{\mathrm{\prime }\mathrm{\prime }}(x)dx={\int }_0^{1}f(t)({\int }_0^{1}G(x,t){\varphi }^{\mathrm{\prime }\mathrm{\prime }}(x)dx)dt=-{\int }_0^{1}f(t)\varphi (t)dt\mathrm{.}$ Thus $(Tf{)}^{\mathrm{\prime }\mathrm{\prime }}=-f$ in distributions. If two continuous Dirichlet solutions exist, their difference $h$ has $h^{\mathrm{\prime }\mathrm{\prime }}=0$ in distributions. Applying [F12] to $h^{\mathrm{\prime }}$ makes $h^{\mathrm{\prime }}$ constant, and applying it to $h-cx$ makes $h$ affine; continuity identifies the distribution with its pointwise affine representative. The two zero endpoint values force $h=0$.

For every $\lambda \ge 0$, the Tikhonov filter is nonnegative and satisfies $\alpha q_{\alpha }(\lambda )=\alpha \mathrm{/}(\lambda +\alpha )\le 1$ and $\lambda q_{\alpha }(\lambda )=\lambda \mathrm{/}(\lambda +\alpha )\le 1$. If $\tilde{\alpha }<\alpha$, then $\lambda +\tilde{\alpha }<\lambda +\alpha$, so $q_{\alpha }(\lambda )\le q_{\tilde{\alpha }}(\lambda )$. Thus the Tikhonov family is an ordered filter with $C_q^{\mathrm{\prime }}=1$.

Since $-e_j^{\mathrm{\prime }\mathrm{\prime }}=(\pi j{)}^2{e}_j$ and $e_j(0)=e_j(1)=0$, uniqueness of the Dirichlet solution from step 1.2 gives $T{e}_j=(\pi j{)}^{-2}{e}_j$. Completeness from step 1.1 therefore diagonalizes $T$ and $T^{\ast }T$, with singular values $s_j=(\pi j{)}^{-2}$ and eigenvalues ${\lambda }_j=(\pi j{)}^{-4}$. All singular values are positive, so $T$ is injective. The truncations $T_{N}f={\sum }_{j=1}^N(\pi j{)}^{-2}⟨f,e_j⟩e_j$ have finite rank and satisfy $\mathrm{\parallel }T-T_N\mathrm{\parallel }={\sup }_{j>N}(\pi j{)}^{-2}\to 0$, so $T$ is compact. Also ${\sum }_j{s}_j^2={\pi }^{-4}{\sum }_j{j}^{-4}<\mathrm{\infty }$ by comparison with ${\int }_1^{\mathrm{\infty }}{x}^{-4}dx$, so $T$ is Hilbert–Schmidt.

For the Tikhonov filter, put $V(\alpha ):={\sum }_{j\ge 1}{\lambda }_j({\lambda }_j+\alpha {)}^{-2}$. On every compact interval $[a,b]\subset (0,\mathrm{\infty })$ its summands are continuous and bounded above by ${\lambda }_j\mathrm{/}{a}^2$, a summable sequence by step 2.1, so $V$ is continuous there. Each summand strictly decreases with $\alpha$. Also $V(\alpha )\le {\alpha }^{-2}{\sum }_j{\lambda }_j\to 0$ as $\alpha \to \mathrm{\infty }$. For every fixed $j$, $V(\alpha )\ge {\lambda }_j({\lambda }_j+\alpha {)}^{-2}\to 1\mathrm{/}{\lambda }_j$ as $\alpha \downarrow 0$; since ${\lambda }_j\to 0$, these lower bounds show $V(\alpha )\to \mathrm{\infty }$. Given $c>0$ and $\theta >1$, continuity and strict decrease give a unique ${\alpha }_m>0$ with $V({\alpha }_m)=c{\theta }^m$ for each $m\in {\mathbb{N}}_0$. Thus ${\alpha }_m$ decreases to zero and $v_m={\sigma }^{2}V({\alpha }_m)$ has consecutive ratio $\theta$, so [F4] holds.

Direct integration of the two pieces of $f^{†}$ gives $⟨f^{†},e_j⟩=2\sqrt{2}\sin (\pi j\mathrm{/}2)\mathrm{/}({\pi }^2{j}^2)$. On the first half, $g^{†}(x)=-x^3\mathrm{/}6+x\mathrm{/}8$, so $g^{†\mathrm{\prime }\mathrm{\prime }}(x)=-x$; on the second half, $g^{†}(x)=x^3\mathrm{/}6-x^2\mathrm{/}2+3x\mathrm{/}8-1\mathrm{/}24$, so $g^{†\mathrm{\prime }\mathrm{\prime }}(x)=x-1$. The two values at $x=1\mathrm{/}2$ both equal $1\mathrm{/}24$, and both first derivatives there equal zero; the endpoint values are zero. Hence $-g^{†\mathrm{\prime }\mathrm{\prime }}=f^{†}$ weakly with Dirichlet boundary values, so uniqueness from step 1.2 gives $T{f}^{†}=g^{†}$. The sequence model in [F3] and the filter conditions in [F4] therefore supply an admissible white-noise input to the finite selector in [F5], whose estimates are well-defined by [F6].