Continuous finite-dimensional ordered-simplex integrals

First consider a continuous curve $g:[a,b]\to V$. In a real basis $e_1,\dots ,e_q$, its coordinate functions are integrable by [F1], and define its integral by their integrals. For a complex basis use the underlying real basis $e_1,i{e}_1,\dots ,e_q,i{e}_q$; this is exactly the real-and-imaginary convention. The zero-dimensional space is immediate. In either case the triangle inequality gives $\mathrm{\parallel }{\sum }_j{c}_j{e}_j\mathrm{\parallel }\le {\sum }_j\mathrm{\mid }{c}_j\mathrm{\mid }\mathrm{\parallel }{e}_j\mathrm{\parallel }$. Thus coordinate convergence implies convergence in any chosen norm, without assuming norm equivalence. By [F2], tagged sums of $g$ converge in that norm to its coordinate integral. The reverse triangle inequality and the same finite-coordinate bound show that $s\mapsto \mathrm{\parallel }g(s)\mathrm{\parallel }$ is continuous, so it too is integrable. For each tagged partition, the norm of the vector sum is at most the corresponding sum of the scalar norms. Taking limits by [F2] proves $\mathrm{\parallel }{\int }_a^{b}g\mathrm{\parallel }\le {\int }_a^b\mathrm{\parallel }g\mathrm{\parallel }$ when $a<b$; for $a=b$ both integrals are zero. A coordinate change or a linear map has entries given by a finite matrix, so scalar linearity [F2] proves that integration commutes with it, and hence proves basis independence. This also proves linearity, and [F2] proves that a nonnegative real curve has nonnegative integral. All statements concern finite lists of coordinates, and require only finitely many simultaneous estimates.

Suppose $G_{k-1}$ is continuous on $D_{n-k+1}(T)$, as is true initially for $G_0=F$. For each $y\in D_{n-k}(T)$, the curve $v\mapsto G_{k-1}(v,y)$ is continuous for $0\le v\le b_k(y)$ and is integrable by step 1.1. Set $b=b_k$ and

The evaluation map $(u,y)\mapsto (ub(y),y)$ is continuous and lands in $D_{n-k+1}(T)$, because $0\le ub(y)\le b(y)$ and the coordinates of $y$ are ordered. Thus $H$ is coordinate-continuous. The parameter domain $[0,1]\times D_{n-k}(T)$ is closed and bounded in a finite-dimensional Euclidean space, hence compact by [F4]; when there are no remaining coordinates it is just $[0,1]$. The finite-coordinate inequality of step 1.1 shows that $H$ is also continuous into any chosen norm on $V$. Apply [F4] with the Euclidean metric on the parameter domain and that norm metric on $V$. Uniform continuity gives, as $y\to y_0$ within its parameter domain,

Indeed the Euclidean distance from $(u,y)$ to $(u,y_0)$ is $\mathrm{\mid }y-y_0\mathrm{\mid }$, independently of $u$. This conclusion also holds coordinate by coordinate if no norm was initially specified, by using the explicitly defined coordinate norm ${\sum }_j\mathrm{\mid }{c}_j\mathrm{\mid }$.

If $b(y)>0$, apply [F3] in each real coordinate with $\varphi (u)=ub(y)$ and $f(v)=G_{k-1}(v,y)$ on $[0,b(y)]$. All its scalar continuity and differentiability hypotheses hold, and ${\varphi }^{\mathrm{\prime }}$ is constant and integrable. This gives

If $b(y)=0$, both sides are zero by [F1], so the same identity holds. By linearity and step 1.1, for any fixed $y_0$ its difference at $y$ satisfies

The first term tends to zero by step 2.1 and the second by continuity of $b$. Thus $G_k$ is continuous, including at parameters with $b=0$. If there are no remaining coordinates this is continuity on a singleton. Induction proves existence and continuity at every recursive stage. When $T=0$ and $n\ge 1$, already the first integration has equal endpoints, so all later integrals are zero; the specified $n=0$ value needs no integration.

Apply the linearity, linear-map commutation and basis independence of step 1.1 successively at each stage. Every resulting coordinate curve is continuous by step 3.1, so these operations are legitimate at every stage and prove the corresponding claims for $I_n$. Likewise a nonnegative real $F$ has nonnegative $G_1$, then nonnegative $G_2$, and so on by scalar monotonicity. No exchange of integration order or assertion of differentiability is involved.

Suppose $\mathrm{\parallel }F\mathrm{\parallel }\le C$. At the first stage step 1.1 and scalar monotonicity give $\mathrm{\parallel }{G}_1(y)\mathrm{\parallel }\le C{b}_1(y)$. If $k\ge 2$ and the previous bound is $\mathrm{\parallel }{G}_{k-1}(s_k,y)\mathrm{\parallel }\le C{s}_k^{k-1}\mathrm{/}(k-1)!$, the same integral inequality bounds $\mathrm{\parallel }{G}_k(y)\mathrm{\parallel }$ by ${\int }_0^{b_k(y)}C{v}^{k-1}\mathrm{/}(k-1)!dv=C{b}_k(y{)}^k\mathrm{/}k!$. The monomial primitive and integral evaluation follow from [F1] and [F5] when the endpoint is positive; equal endpoints give the same value zero. Induction gives the stated final bound. For scalar $F=1$ the same recursion is an equality at every stage, yielding $I_n(1)=T^n\mathrm{/}n!$. For $n=0$ both assertions follow directly from $I_0(F)=F(())$ and the factor convention.