Compact support expands at speed at most c

Reduction to a point outside the domain of influence: if $K=\mathrm{\varnothing }$, all data and the source vanish, so [F1] applied at every $(x,t)$ with $t\in (0,T)$ gives $u=0$ there; at $t=0$, continuity and the Cauchy displacement limit give $u(\cdot ,0)=u_0=0$, so the support inclusion holds throughout $[0,T)$. Otherwise let $t\in (0,T)$ and $x\mathrm{\notin }K+{\bar{B}}_{ct}(0)$, so $\mathrm{dist}(x,K)>ct$; then the base ball of $K^{-}(x,t)$ is $B_{ct}(x)$ by [F3], and $B_{ct}(x)\cap K=\mathrm{\varnothing }$: if $z\in B_{ct}(x)\cap K$ then $\mathrm{dist}(x,K)\le \mathrm{\mid }x-z\mathrm{\mid }<ct$, contradicting $\mathrm{dist}(x,K)>ct$; hence the initial data vanish on $B_{ct}(x)$ by [F2]: $u_0=u_1=0$ there; and the source vanishes on the cone: if $(y,s)\in K^{-}(x,t)$ had $\mathrm{dist}(y,K)\le cs$, then $\mathrm{dist}(x,K)\le \mathrm{\mid }x-y\mathrm{\mid }+\mathrm{dist}(y,K)\le c(t-s)+cs=ct$ by [F3], a contradiction, so $\mathrm{dist}(y,K)>cs$ and $(y,s)\mathrm{\notin }\mathrm{supp}f$, i.e. $f(y,s)=0$.

Conclusion: by step 1.1 the data and source of ${\mathrm{\square }}_{c}u=f$ vanish in the cone $K^{-}(x,t)$, so [F1] applied to $u$ gives $u(x,t)=0$; as $x\mathrm{\notin }K+{\bar{B}}_{ct}(0)$ was arbitrary, every point outside $K+{\bar{B}}_{ct}(0)$ has $u(\cdot ,t)=0$, and the containing set is closed by [F3], whence $\mathrm{supp}u(\cdot ,t)\subseteq K+{\bar{B}}_{ct}(0)$ for every $t\in (0,T)$; at $t=0$, continuity and the Cauchy displacement limit give $u(\cdot ,0)=u_0$, so the inclusion is exactly the support hypothesis on $u_0$.