Theorem 5.3.5: Locally Convex Fundamental Theorem of Calculus

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 linear functional $\el\in E^{*}$ we have

\begin{align*}\der{}{x}\el(H(x))&=\el(H'(x))=0,\fall x\in (a,b)\end{align*}

so $\el(H(\cd))\text{ is constant on }(a,b)$ by classical mean value theorem and hence constant on $[a,b]$ by continuity of $\el$ and $H$. By Hahn-Banach Theorem $H$ is constant on $[a,b]$.

\begin{align*}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*}

$\square$

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