Theorem 19.7.321551B679B8B7

Proof. The uniqueness of the extension is clear once existence is known. Moreover, (19.7.34) follows easily from theorem 19.2.5 and theorem 19.7.5 by continuity. It remains to argue the existence of the continuous extension. We first construct an extension and prove its continuity.

Step 1. We construct the desired map (19.7.33). Let $S\in \overline{\mathcal{S}'(C)}_{>0}$. We wish to construct a convex body $\Delta (S)\in \mathcal{K}_ n$.

Let $S^ i\in \mathcal{S}'(C)$ be a sequence that converges to $S$ such that

For each $i,j\geq 0$, we introduce

Then by 19.7.19,

Take $i_0>0$ large enough so that, for $i\geq i_0$, we have $\operatorname{vol}S^ i>2^{-1}\operatorname{vol}S$ and $2^{2-i}< \operatorname{vol}S$. Then

It follows that $\operatorname{vol}S^{i,j}> 2^{-1}\operatorname{vol}S-2^{1-i}>0$ whenever $i\geq i_0$. In particular, by 19.7.25, $S^{i,j}\in \mathcal{S}'(C)$ for $i\geq i_0$.

By 19.7.29, for $i\geq i_0$, there exists $T^ i\in \mathcal{S}'(C)$ such that $S^{i,j}\to T^ i$ as $j\to \infty $. Moreover,

Therefore, $T^ i\to S$. We then define

In other words, we have defined

Indeed, by the construction in the proof of 19.7.29 and the cone equality in 19.7.25,

Thus the bodies $\Delta (T^ i)$ are increasing in $i$ and are all contained in the compact slice $\{ x\in \mathbb {R}^ n:(x,1)\in C\} $. In particular, $\Delta (S)$ is a convex body and

This is an honest limit: if $\Delta $ is the limit of a subsequence of $\Delta (S^ i)$, then $\Delta (S)\subseteq \Delta $ by (19.7.8). Moreover,

Comparing the volumes, we find that equality holds. So by theorem 19.7.4,

Next we claim that $\Delta (S)$ as defined above does not depend on the choice of the sequence $S^ i$. In fact, suppose that $S'^ i\in \mathcal{S}'(C)$ is another sequence satisfying the same conditions as $S^ i$. By 19.7.19, we have

Since $\operatorname{vol}S>0$ and the volume is continuous on $\overline{\mathcal S}(C)$, we have $\operatorname{vol}R^ i>0$ for all large $i$. Thus $R^ i\in \mathcal S'(C)$ by 19.7.25. Moreover,

Passing simultaneously to subsequences of $S^ i$ and $S'^ i$, we may assume

Then the construction above applies to $R^ i=S^{i+1}\cap S'^{i+1}$. Since $R^ i\subseteq S^{i+1}$ and $R^ i\subseteq S'^{i+1}$, every Hausdorff limit of $\Delta (R^ i)$ is contained in the corresponding limits of $\Delta (S^ i)$ and $\Delta (S'^ i)$. The volumes of all three limits are $\operatorname{vol}S$, hence the inclusions are equalities. The same is true with $S'^ i$ in place of $S^ i$. So we conclude that $\Delta (S)$ as in (19.7.35) does not depend on the choices we made.

Step 2. It remains to prove the continuity of $\Delta $ defined in Step 1. Suppose that $S^ i\in \overline{\mathcal{S}'(C)}_{>0}$ is a sequence with limit $S\in \overline{\mathcal{S}'(C)}_{>0}$. We want to show that

We first reduce to the case where $S^ i\in \mathcal{S}'(C)$. By (19.7.35), for each $i$, we can choose $T^ i\in \mathcal{S}'(C)$ such that $d_{\mathrm{sg}}(S^ i,T^ i)<2^{-i}$ and $d_{\mathrm{Haus}}(\Delta (S^ i),\Delta (T^ i))<2^{-i}$. If we have shown $\Delta (T^ i)\xrightarrow {d_{\mathrm{Haus}}}\Delta (S)$, then (19.7.36) follows immediately.

Next we reduce to the case where $d_{\mathrm{sg}}(S^ i,S^{i+1})\leq 2^{-i}$. Thanks to theorem 19.7.4, in order to prove (19.7.36), it suffices to show that each subsequence of $\Delta (S^ i)$ admits a subsequence that converges to $\Delta (S)$. Hence, we may pass to a sufficiently fast subsequence and reduce to the required case.

After these reductions, (19.7.36) is nothing but (19.7.35). $\square$