Banach algebra valued contour integral

Let $E$ be a complex Banach space (Banach space), let $\gamma :[a,b]\to \mathbb{C}$ be a piecewise $C^1$ contour (Rectifiable complex contours, reversal, concatenation, closedness, and orientation), and let $f:{\gamma }^{\ast }\to E$ be continuous. The construction applies in particular to $E=A$ for a unital complex Banach algebra (Unital Banach algebra); no algebra multiplication or unit is used.

For $a<b$ fix a finite subdivision $a=t_0<\cdots <t_m=b$ such that the restriction of $\gamma$ on each piece has a continuous derivative extension $v_k$ to its closed interval. Given a tagged partition $P=(s_j,{\xi }_j)$ refining these nodes, define $S(f,\gamma ,P,\xi )={\sum }_{j}f(\gamma ({\xi }_j))v_{k(j)}({\xi }_j)(s_j-s_{j-1})\in E,$ where $[s_{j-1},s_j]$ lies in the $k(j)$-th piece. At a node, use the derivative extension from this piece; the two adjacent intervals may therefore use different values. The contour integral is the norm limit ${\int }_{\gamma }f(z)dz={\lim }_{\mathrm{mesh}(P)\to 0}S(f,\gamma ,P,\xi )\mathrm{.}$ Existence and independence of all these choices are verified below. On a singleton parameter interval the integral is defined to be zero.

Equivalently it is the Bochner integral on the finite Lebesgue measure interval ${\int }_a^{b}f(\gamma (t)){\gamma }^{\mathrm{\prime }}(t)dt,$ where the finitely many corner values may be assigned arbitrarily. It satisfies $\parallel {\int }_{\gamma }f(z)dz\parallel \le L(\gamma ){\sup }_{z\in {\gamma }^{\ast }}\mathrm{\parallel }f(z)\mathrm{\parallel }\mathrm{.}$ Concatenation adds the integrals and reversal negates them. An increasing piecewise-$C^1$ bijection of compact parameter intervals whose inverse is also piecewise $C^1$ leaves the integral unchanged.

For a finite complex chain $\mathrm{\Gamma }={\sum }_{k<r}{m}_k{\gamma }_k$ whose nonzero terms are piecewise $C^1$ contours, and continuous $f:{\mathrm{\Gamma }}^{\ast }\to E$, define ${\int }_{\mathrm{\Gamma }}f(z)dz={\sum }_{\begin{array}{c}k<r\\ m_k\mathrm{\ne }0\end{array}}{m}_k{\int }_{{\gamma }_k}f(z)dz\mathrm{.}$ Zero-coefficient terms are omitted; the empty chain integrates to zero.

Existence and Bochner agreement. On the $k$-th closed piece put $F_k(t)=f(\gamma (t))v_k(t)$. This is uniformly continuous and bounded. Subdivide each piece into $2^n$ equal intervals and use its left endpoint values to obtain finite-valued measurable step functions $h_n$. Assign fixed values at the finitely many nodes. Uniform continuity on the finitely many pieces shows $h_n\to F$ uniformly away from these nodes; here $F$ denotes $f(\gamma ){\gamma }^{\mathrm{\prime }}$ with the chosen node values. In particular $F$ is strongly measurable, not merely scalar measurable. Each $h_n$ is integrable, and $\int \mathrm{\parallel }F-h_n\mathrm{\parallel }\to 0$ on this finite interval. Thus the definition of Bochner integration (Bochner-integrable function) supplies its integral, and Bochner integrability criterion gives independence of the approximants.

For arbitrary tagged refinements the corresponding step function differs from $F$ in norm by at most a common modulus $\omega (\mathrm{mesh}P)$ off the nodes. Hence its $L^1$ difference from $F$ is at most $(b-a)\omega (\mathrm{mesh}P)$. Comparing its simple integral with those of $h_n$, the triangle inequality for finite sums bounds the difference of integrals by the $L^1$ difference. Passing to the limit proves convergence of all tagged sums to the same Bochner value. Different finite subdivisions have a common refinement and the same a.e. function $F$; changing finitely many endpoint values changes neither integral. This proves all independence claims without a choice of an infinite family of tags.

Norm estimate. The triangle inequality gives $\mathrm{\parallel }S(f,\gamma ,P,\xi )\mathrm{\parallel }\le {\sup }_{{\gamma }^{\ast }}\mathrm{\parallel }f\mathrm{\parallel }{\sum }_j\mathrm{\mid }{v}_{k(j)}({\xi }_j)\mathrm{\mid }(s_j-s_{j-1})\mathrm{.}$ The scalar sums tend to the sum of the speed integrals on the pieces, which is $L(\gamma )$ by A continuous piecewise-$C^1$ path is rectifiable and its length is the sum of the speed integrals over its pieces. Taking the limit proves the bound. A constant contour and a zero integrand therefore have zero integral, as does a contour with singleton parameter interval.

Increment sums and parameter changes. The same value is the limit of $T(f,\gamma ,P,\xi )={\sum }_{j}f(\gamma ({\xi }_j))(\gamma (s_j)-\gamma (s_{j-1}))\mathrm{.}$ To see this, apply the real mean-value theorem (The mean value theorem, as the case $g(x)=x$ of Cauchy's: for $f$ continuous on $[a,b]$ with $a<b$ and differentiable on $(a,b)$ there is $c\in (a,b)$ with $f(b)-f(a)=f^{\mathrm{\prime }}(c)(b-a)$) separately to the two coordinates of $\gamma$ on an interval contained in one smooth piece. If $\eta$ is a common modulus of the derivative extensions, the difference between its complex increment and $v_{k(j)}({\xi }_j)(s_j-s_{j-1})$ is at most $\sqrt{2}\eta (\mathrm{mesh}P)(s_j-s_{j-1})$. Therefore $\mathrm{\parallel }T-S\mathrm{\parallel }\le \sqrt{2}\sup \mathrm{\parallel }f\mathrm{\parallel }(b-a)\eta (\mathrm{mesh}P)\to 0$. This also holds for partitions not containing the original nodes: inserting the finitely many nodes changes only intervals of total length at most $2m\mathrm{mesh}P$. The bounded derivative extensions bound the variation of $\gamma$ there by a constant times this length, so both their old and subdivided contributions tend to zero.

Under an increasing reparametrization as specified above, tagged partitions and tags map to tagged partitions and tags with exactly the same increment sums. Uniform continuity of the parameter map makes the image mesh tend to zero. Both contours remain piecewise $C^1$, so their integrals agree. Reversal reverses the order and the signs of the increments. For concatenation, split a partition at the joining parameter and use its two affine pieces; the increment sums split into the two sums. These facts prove the asserted reversal and concatenation identities. Finite linearity in chains follows from their definition. No claim is made here for a reparametrization taking a contour outside the piecewise-$C^1$ domain.

Scalar consistency. When $E=\mathbb{C}$, expansion into real and imaginary parts turns the increment sums into the four Riemann–Stieltjes sums in The complex line integral over a rectifiable path as a componentwise Riemann–Stieltjes integral. Thus the limits agree on the common piecewise-$C^1$ domain. Taking finite sums gives agreement with Integration over a complex chain and the index of a chain. The zero Banach space is allowed and all its integrals are zero. The construction uses completeness, uniform continuity and explicitly prescribed finite subdivisions; it makes no new choice assumption.