The diagonal heat semigroup on $c_0$ is contractive and strongly continuous

For $t\ge 0$ and $j\in \mathbb{N}$, [F4] and [F5] give $0<\exp (-j^{2}t)\le \exp (0)=1$. Thus the coordinate formula is linear and $\mathrm{\parallel }U(t)x{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$ If $x\in c_0$ and $\epsilon >0$, choose $N$ so that $\mathrm{\mid }{x}_j\mathrm{\mid }<\epsilon$ for all $j\ge N$; then $\mathrm{\mid }(U(t)x{)}_j\mathrm{\mid }\le \mathrm{\mid }{x}_j\mathrm{\mid }<\epsilon$ for $j\ge N$. So $U(t)x\in c_0$, $U(t)$ is bounded, and [F6] gives $\mathrm{\parallel }U(t){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 1$.

Since $\exp (0)=1$, $U(0)=I_X$. For $s,t\ge 0$, [F3] gives $U(t)U(s)x=(\exp (-j^{2}t)\exp (-j^{2}s)x_j{)}_j=(\exp (-j^2(t+s))x_j{)}_j=U(t+s)x\mathrm{.}$ Thus the semigroup identities hold on every coordinate and hence on $X$.

Fix $x\in X$, $t_0\ge 0$, and $\epsilon >0$. By [F1], choose $N$ such that $\mathrm{\mid }{x}_j\mathrm{\mid }<\epsilon \mathrm{/}3$ for $j\ge N$. For those indices, $\mathrm{\mid }\exp (-j^{2}t)-\exp (-j^2{t}_0)\mathrm{\mid }\mathrm{\mid }{x}_j\mathrm{\mid }<\epsilon \mathrm{/}3$ for every $t\ge 0$, because both exponential factors lie in $[0,1]$. For the finitely many indices $j<N$, continuity in [F4] supplies a common $\delta >0$ such that the same coordinate differences are below $\epsilon \mathrm{/}3$ whenever $t\ge 0$ and $\mathrm{\mid }t-t_0\mathrm{\mid }<\delta$. Taking the supremum over all coordinates proves $\mathrm{\parallel }U(t)x-U(t_0)x{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon$. Therefore every orbit is continuous, so $U$ is strongly continuous.

The boundedness in step 1.1 holds for every $t$, so for each $T>0$, ${\sup }_{0\le t\le T}\mathrm{\parallel }U(t){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 1.$ Together with the Banach property [F2], the semigroup identities in step 2.1, and strong continuity in step 2.2, this proves the statement.