Positive L1 weights, discrete comparison, and residual stability

For fixed $n$, the function $w(s)=(t_n-s{)}^{-\alpha }$ is positive and increasing on $[0,t_n)$, and $b_{n,k}=\mathrm{\Gamma }(1-\alpha {)}^{-1}{h}_k^{-1}{\int }_{t_{k-1}}^{t_k}w(s)ds$ by [F1] and [F4]. The factor $\mathrm{\Gamma }(1-\alpha {)}^{-1}$ is positive by [F7]. The average of an increasing function on an earlier interval is at most its value at the shared endpoint, which is at most its average on the next interval. Each integral is finite because $\alpha <1$. Thus the weights are positive and nondecreasing in $k$.

Put $\varphi (t)=t^{\alpha }$. By [F5], ${\varphi }^{\mathrm{\prime }}(s)=\alpha s^{\alpha -1}$ for $s>0$, and this derivative is decreasing because its derivative is $\alpha (\alpha -1)s^{\alpha -2}<0$ and [F8] applies. It is in $L^1(0,T)$ by [F9], with integral $T^{\alpha }$. Its indefinite integral is $\varphi (t)-\varphi (0)=t^{\alpha }$, so $\varphi$ is absolutely continuous by [F6] and its Caputo derivative is defined. In this proof, $D_C^{\alpha }\varphi (t_n)$ denotes the pointwise integral representative in [F2]; the displayed Beta-kernel calculation below proves that this value is finite at every grid node. On an interval $[a,b]=[t_{k-1},t_k]$ its secant slope ${\bar{\varphi }}_k^{\mathrm{\prime }}$ is the average of ${\varphi }^{\mathrm{\prime }}$, hence $e_k(s):={\bar{\varphi }}_k^{\mathrm{\prime }}-{\varphi }^{\mathrm{\prime }}(s)$ is increasing and ${\int }_a^b{e}_k(s)ds=0$. Since $w(s)=(t_n-s{)}^{-\alpha }$ is also increasing, their pairwise differences have the same sign. Expanding the double integral and using ${\int }_a^b{e}_k=0$ gives ${\int }_a^b{e}_k(s)w(s)ds=\frac{1}{2(b-a)}{\int }_a^b{\int }_a^b(e_k(s)-e_k(r))(w(s)-w(r))drds\ge 0.$ These integrals are finite: $e_k,w\in L^1(a,b)$, and the only possible singular product ${\varphi }^{\mathrm{\prime }}(s)w(s)$ occurs when $n=k=1$; there it is the integrable Beta kernel $\alpha s^{\alpha -1}(t_1-s{)}^{-\alpha }$ from [F3]. Expanding the absolute value of the double-integral integrand bounds it by the sum of products of the finite integrals $\int \mathrm{\mid }{e}_k\mathrm{\mid }$, $\int w$, and $\int \mathrm{\mid }{e}_k\mathrm{\mid }w$, which justifies the expansion. Summing over intervals and using [F2] yields ${\delta }_{\mathcal{T}}^{\alpha }\varphi (t_n)\ge D_C^{\alpha }\varphi (t_n)$. Direct substitution $s=t_{n}y$ and [F3]–[F4] give $D_C^{\alpha }(t^{\alpha })(t_n)=\frac{\alpha }{\mathrm{\Gamma }(1-\alpha )}{t}_n^{0}B(\alpha ,1-\alpha )=\mathrm{\Gamma }(1+\alpha )\mathrm{.}$ Thus, using $\mathrm{\Gamma }(1+\alpha )>0$ from [F7], $q^n:=t_n^{\alpha }\mathrm{/}\mathrm{\Gamma }(1+\alpha )$ satisfies $q^0=0$ and ${\delta }_{\mathcal{T}}^{\alpha }{q}^n\ge 1$.

Expanding [F1] by collecting coefficients gives ${\delta }_{\mathcal{T}}^{\alpha }{z}^n=b_{n,n}{z}^n+{\sum }_{k=1}^{n-1}(b_{n,k}-b_{n,k+1})z^k-b_{n,1}{z}^0\mathrm{.}$ If some $z^n<0$, let $m$ be the first such index. Then $z^k\ge 0$ for $k<m$, the middle coefficients in this display are nonpositive by step 1.1, and $-b_{m,1}{z}^0\le 0$. The terms $(b_{m,m}+\lambda )z^m$ are strictly negative because $b_{m,m}>0$ and $\lambda \ge 0$. Therefore ${\delta }_{\mathcal{T}}^{\alpha }{z}^m+\lambda z^m<0$, contradicting the hypothesis. This proves comparison.

For the residual equation, set $z_{\pm }^n:=R{q}^n\pm e^n$. Then $z_{\pm }^0=0$ and ${\delta }_{\mathcal{T}}^{\alpha }{z}_{\pm }^n+\lambda z_{\pm }^n=R({\delta }_{\mathcal{T}}^{\alpha }{q}^n+\lambda q^n)\pm r^n\ge R-\mathrm{\mid }{r}^n\mathrm{\mid }\ge 0.$ Comparison from step 2.1 gives $z_{\pm }^n\ge 0$, hence $\mathrm{\mid }{e}^n\mathrm{\mid }\le R{q}^n\le R{T}^{\alpha }\mathrm{/}\mathrm{\Gamma }(1+\alpha )$ for every $n$. This proves residual stability.

For each $n$, expanding the operator from [F1] gives ${\delta }_{\mathcal{T}}^{\alpha }{U}^n=b_{n,n}{U}^n+{\sum }_{k=1}^{n-1}(b_{n,k}-b_{n,k+1})U^k-b_{n,1}{U}^0\mathrm{.}$ By step 1.1, $b_{n,n}>0$, so with $\lambda \ge 0$ the coefficient $b_{n,n}+\lambda$ is positive. Rearranging ${\delta }_{\mathcal{T}}^{\alpha }{U}^n+\lambda U^n=F^n$ therefore gives exactly the recurrence in [F1], with a unique value at step $n$ once the earlier values are fixed. Induction over $n=1,\dots ,N$ proves well-definedness and uniqueness of the full sequence and the stated equation.