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*}
$\square$