5.3 Riemann-Stieltjes Sums and Integrals

Definition 5.3.1 (Riemann-Stieltjes Sum).label Let $[a,b]\suf\R$, $E,F,H$ be TVSs over $K\in \curl{\R,\com}$, $G:[a,b]\to F$ and suppose there exists a continuous bilinear map $E\ti F\ni (x,y)\mto xy\in H$. Let $f:[a,b]\to E$ and $(P=\curl{x_j}^{n}_{j=0},c=\curl{c_j}^{n}_{j=1})\in \scr{P}_{t}([a,b])$ then

\begin{align*}S(P,c,f,G)=\sums{n}{j=1}f(c_{j})\braks{G(x_j)-G(x_{j-1})}\end{align*}

is the Riemann-Stieltjes sum of $f$ with respect to $G$ and $(P,c)$.

Definition 5.3.2 (Riemann-Stieltjes Integral).label Let $[a,b]\suf\R$, $E,F,H$ be TVSs over $K\in \curl{\R,\com}$, $G:[a,b]\to F$ and suppose there exists a continuous bilinear map $E\ti F\ni (x,y)\mto xy\in H$. Let $f:[a,b]\to E$ then $f$ is Riemann-Stieltjes integrable with respect to $G$ if the limit

\begin{align*}\integral{a}{b}fdG=\integral{a}{b}f(t)G(dt)=\limit{(P,c)\in\scr{P}_t([a,b])}S(P,c,f,G)\end{align*}

exists. In which case we call the Riemann-Stieltjes integral of $G$. The set $RS([a,b],G)$ of all Riemann-Stieltjes integrable functions with respect to $G$ is a vector space.

Lemma 5.3.3 (Summation by Parts).label Let $[a,b]\suf\R$, $E,F,H$ be TVSs over $K\in \curl{\R,\com}$, $f:[a,b]\to E,G:[a,b]\to F$, $(P,c)\in\scr{P}_{t}([a,b])$ and suppose there exists a continuous bilinear map $E\ti F\ni (x,y)\mto xy\in H$ then

\begin{align*}S(P,c,f,G)+S(P',c',G,f)=f(b)G(b)-f(a)G(a)\end{align*}

for any $P'=\curl{y_j}_{j=0}^{n+1}=[a,c_{1},...,c_{n},b]$ and $c'=\curl{d_j}_{j=1}^{n+1}=[x_{0},...,x_{n}]$.

Proof. We denote $c_{0}=a$ and $c_{n+1}=b$ then reindexing gives

\begin{align*}S(P,c,f,G)&=\sums{n}{j=1}f(c_{j})[G(x_{j})-G(x_{j-1})]\\&=\sums{n}{j=1}f(c_{j})G(x_{j})-\sums{n}{j=1}f(c_{j})G(x_{j-1})\\&=f(c_{n})G(x_{n})-f(c_{0})G(x_{0})+\sums{n-1}{j=0}f(c_{j})G(x_{j})-\sums{n}{j=1}f(c_{j})G(x_{j-1})\\&=f(c_{n})G(x_{n})-f(c_{0})G(x_{0})+\sums{n}{j=1}f(c_{j-1})G(x_{j-1})-\sums{n}{j=1}f(c_{j})G(x_{j-1})\\&=f(c_{n})G(x_{n})-f(c_{0})G(x_{0})-\sums{n}{j=1}G(x_{j-1})[f(c_{j})-f(c_{j-1})]\\&=f(c_{n+1})G(x_{n})-f(c_{0})G(x_{0})-\sums{n+1}{j=1}G(x_{j-1})[f(c_{j})-f(c_{j-1})]\\&=f(b)G(b)-f(a)G(a)-S(P',c',G,f)\end{align*}

$\square$

Theorem 5.3.4 (Integration by Parts).label Let $[a,b]\suf\R$, $E,F,H$ be TVSs over $K\in \curl{\R,\com}$, $f:[a,b]\to E,G:[a,b]\to F$ and suppose there exists a continuous bilinear map $E\ti F\ni (x,y)\mto xy\in H$ then $f\in RS([a,b],G)\iff G\in RS([a,b],f)$ in which case

\begin{align*}\integral{a}{b}fdG+\integral{a}{b}Gdf=f(b)G(b)-f(a)G(a)\end{align*}

Proof. Suppose that $f\in RS([a,b],G)$ then for each $U\in \cali{N}_{K}(0)$ there exists $P_{0}=\curl{x_j}_{0}^{n}\in \scr{P}([a,b])$ such that $S(P,c,f,G)-\integral{a}{b}fdG\in U$ for all $(P,c)\in \scr{P}_{t}([a,b])$ with $P\geq P_{0}$. Let $Q_{0}=[x_{0},x_{1},x_{1},...,x_{n},x_{n}]$ then for any $(Q,d)\in \scr{P}_{t}([a,b])$ satisfying $Q\geq Q_{0}$ we have by Lemma 5.3.3 there exists $(Q',d')\in\scr{P}_{t}([a,b])$ such that

\begin{align*}S(Q,d,G,f)+S(Q',d',f,G)&=f(b)G(b)-f(a)G(a)\\ f(b)G(b)-f(a)G(a)-\integral{a}{b}fdG-S(Q,d,G,f)&=S(Q',d',f,G)-\integral{a}{b}fdG\end{align*}

Moreover by construction $Q'\geq P_{0}$ so we have by first observation

\begin{align*}\parens{f(b)G(b)-f(a)G(a)-\integral{a}{b}fdG}-S(Q,d,G,f)&=S(Q',d',f,G)-\integral{a}{b}fdG\in U\end{align*}

as desired.$\square$

Theorem 5.3.5 (Locally Convex Fundamental Theorem of Calculus).label Let $E$ be a locally convex space over $K\in \curl{\R,\com}$ and $f\in C([a,b];E)$ then

\begin{align*}F:[a,b]\to E&&x\mto \integral{a}{x}f(t)dt\end{align*}

is $C^{1}$ with $F'=f$. In particular for any $G:[a,b]\to E$ such that $G\in C^{1}((a,b);E)$ and $G'=f$ on $(a,b)$ we have

\begin{align*}G(y)-G(x)=\integral{x}{y}f(t)dt&&\fall x,y\in [a,b]\end{align*}

Proof. For any continuous seminorm $\rho:E\to [0,\infty)$, observe that for any $x\in [a,b]$ and $h>0$ such that $x+h\in [a,b]$ we have

\begin{align*}\rho\parens{\frac{F(x+h)-F(x)-f(x)}{h}}&=\rho\parens{\frac{1}{h}\integral{x}{x+h}f(t)-f(x)dt}\\&\leq \integral{x}{x+h}\rho\parens{\frac{f(t)-f(x)}{h}}dt\\&\leq \SUP{t\in B(x,\abs{h})}\rho\parens{f(t)-f(x)}\end{align*}

which tends to $0$ as $h\to 0$ by continuity of $f$, showing $F\in C^{1}([a,b];E)$ and $F'(x)=f(x)$. Suppose $G:[a,b]\to E$ is continuous and $G$ differentiable on $(a,b)$ with $G'(x)=f(x)$ then

\begin{align*}H(x)\define G(x)-F(x)\in C^{1}([a,b];E)\end{align*}

and

\begin{align*}H'(x)=G'(x)-F'(x)=0&&\fall x\in [a,b]\end{align*}

So for every continuous seminorm $\rho$

\begin{align*}\der{}{x}\rho(H(x))&=\rho(H'(x))=0\\ \implies \rho(H(x))&=\rho(H(a))\\ \implies H(x)&=H(a)=G(a)-F(a)=G(a)\\ \implies G(x)-G(a)&=F(x)=\integral{a}{x}f(t)dt\\ \implies G(y)-G(x)&=F(y)-F(x)=\integral{x}{y}f(t)dt\end{align*}

$\square$

Theorem 5.3.6 (Mean Value Theorem).label Let $E$ be a locally convex space and $f\in C^{1}([a,b];E)$, then for any continuous seminorm $\rho:E\to [0,\infty)$ and $x<y\in [a,b]$

\begin{align*}\rho\parens{f(y)-f(x)}\leq \integral{x}{y}\rho\parens{Df(t)}dt\leq (y-x)\SUP{t\in[x,y]}\rho(Df(t))\end{align*}

Proof. Since $f\in C^{1}([a,b];E)$ we have by Locally Convex FTC 5.3.5 that $f(b)-f(a)=\integral{a}{b}Df(t)dt$. Then by subadditivity and homogeneity of $\rho$ applied to RS sums and Definition 5.3.2 as the limit, we have

\begin{align*}\rho(f(y)-f(x))=\rho\parens{\integral{x}{y}Df(t)dt}\leq \integral{x}{y}\rho(Df(t))dt&\leq \integral{x}{y}\SUP{s\in[x,y]}\rho(Df(s))dt\\&=(y-x)\SUP{s\in[x,y]}\rho(Df(s))\end{align*}

$\square$

Proposition 5.3.7.label Let $X$ be a compact metrizable space, $Y$ a locally convex space over $K\in \curl{\R,\com}$ then $C(X;Y)=UC(X;Y)$.

Proof. Let $f\in C(X;Y)$ and $\rho$ a continuous seminorm on $Y$. Consider

\begin{align*}g:X\ti X\to K&&(x,y)\mto \rho(f(x)-f(y))\end{align*}

then $g$ is continuous as a composition of continuous functions. Since $X$ is compact, so is $X\ti X$ thus $g$ is uniformly continuous as a map between metrizable spaces. Since $\rho$ was arbitrary we conclude $f$ is uniformly continuous.$\square$

Theorem 5.3.8 (Change of Variables).label Let $[a,b]\suf\R$, $E,F,H$ be locally convex spaces over $K\in \curl{\R,\com}$, with $G\in C^{1}([a,b];F)$ and suppose there exists a continuous bilinear map $E\ti F\ni (x,y)\mto xy\in H$. For any bounded $f\in RS([a,b],G;E)$

\begin{align*}\integral{a}{b}f(t)G(dt)=\integral{a}{b}f(t)DG(t)dt\end{align*}

Proof. Let $\rho_{H}:H\to [0,\infty)$ be a continuous seminorm. Pick similarly $\rho_{E},\rho_{F}$ such that for any $x\in E$ and $y\in F$ we have

\begin{align*}\rho_{H}(xy)\leq\rho_{E}(x)\rho_{F}(y)\end{align*}

Since $f$ is bounded

\begin{align*}M\define \SUP{t\in[a,b]}\rho_{E}(f(t))<\infty\end{align*}

For any $(P=\curl{x_j}^{n}_{0},c=\curl{c_j}^{n}_{1})\in\scr{P}_{t}([a,b])$ such that $\s(P)\leq \de$ to be determined later

\begin{align*}\rho_{H}\parens{S(P,c,f,G)-S(P,c,fDG,\text{Id})}&=\rho_{H}\parens{\sums{n}{j=1}f(c_j)\parens{G(x_j)-G(x_{j-1})-(x_j-x_{j-1})DG(c_j)}}\\&\leq \sums{n}{j=1}\rho_{E}(f(c_{j}))\rho_{F}(G(x_{j})-G(x_{j-1})-(x_{j}-x_{j-1})DG(c_{j}))\\&\leq M\sums{n}{j=1}\rho_{F}(G(x_{j})-G(x_{j-1})-(x_{j}-x_{j-1})DG(c_{j}))\end{align*}

Fix $j$ and define

\begin{align*}H_{j}:[x_{j-1},x_{j}]\to F&&t\mto G(t)-G(c_{j})-(t-c_{j})DG(c_{j})\end{align*}

then $H_{j}\in C^{1}([x_{j-1},x_{j}],F)$ and by Mean Value Theorem 5.3.6

\begin{align*}\rho_{F}\parens{G(x_j)-G(x_{j-1})-(x_j-x_{j-1})DG(c_j)}&=\rho_{F}\parens{H_j(x_j)-H_j(x_{j-1})}\\&\leq \integral{x_{j-1}}{x_j}\rho_{F}(DH(t))dt\\&\leq (x_{j}-x_{j-1})\SUP{t\in [x_{j-1},x_j]}\rho_{F}(DH(t))\\&=(x_{j}-x_{j-1})\SUP{t\in [x_{j-1},x_j]}\rho_{F}(DG(t)-DG(c_{j}))\end{align*}

Observe $DG\in C([a,b];F)=UC([a,b];F)$ by Proposition 5.3.7 so $\fall\e>0,\exists\de>0$ such that after refining the partition if necessary we have

\begin{align*}\rho_{H}\parens{S(P,c,f,G)-S(P,c,fDG,\text{Id})}\leq M \sums{n}{j=1}(x_{j}-x_{j-1})\e=M (b-a)\e\end{align*}

as desired.$\square$

The Eigenspace

5.3 Riemann-Stieltjes Sums and Integrals

Direct References

The Eigenspace

Direct References