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Gruber P. Convex and Discrete Geometry

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13 The Space of <strong>Convex</strong> Bodies 241<br />

To see this let C ∈ C. By property (a), s(C) = s( 1 l C) +···+s( 1 l C) or s( 1 l C) =<br />

1<br />

l s(C) for l = 1, 2,... This in turn implies that s( k l C) = k l s(C) for k, l = 1, 2,...<br />

Hence s(λC) = λs(C) for λ ≥ 0 by property (c). The proof of (5) is complete. Now<br />

the construction of H is as follows: let h ∈ Hd . Then h = h0 +···+hm where<br />

hn ∈ Hd n . By Proposition 13.1(vi) there is a constant α>0 such that each of the<br />

functions hn + α is (the restriction to Sd−1 of) a support function of a convex body,<br />

say Cn. Now define<br />

(6) H(h) = s(C0) +···+s(Cm) − (m + 1)s(αB d ).<br />

It is easy to see that H(h) does not depend on the particular choice of α as long as all<br />

functions hn + α are support functions of convex bodies. Thus H(h) is well defined<br />

<strong>and</strong> maps H d into E d .Next,<br />

(7) H satisfies the assumptions of (1).<br />

The definition of H in (6) together with (5) <strong>and</strong> property (a) imply that H is linear.<br />

By its definition H satisfies the equality H(rh) = rH(h) for all h ∈ H d <strong>and</strong> all<br />

rotations r. The proof of (7) is complete.<br />

In the third step the aim is to show the following proposition.<br />

(8) Let C ∈ C be such that hC = h0 +···+hm with suitable hn ∈ Hd n . Then<br />

s(C) = sC.<br />

Apply (6) with h = hC <strong>and</strong> note property (a). Then<br />

H(hC) + (m + 1)s(αB d ) = s(C0) +···+s(Cm) = s(C0 +···+Cm).<br />

From hC + (m + 1)α = (h0 + α)+···+(hm + α) it follows that C + (m + 1)αB d =<br />

C0 +···+Cm. Thus property (a) shows that<br />

s(C) + (n + 1)s(αB d ) = s(C) + s � (n + 1)αB d� = s � C + (n + 1)αB d�<br />

= s(C0 +···+Cn).<br />

It follows that H(hC) = s(C). Since by (7) H satisfies the assumptions in (1),<br />

Proposition (1) then shows that<br />

�<br />

(9) s(C) = H(hC) = β hC(u)udσ(u).<br />

S d−1<br />

To determine β, lete = (1, 0,...,0) ∈ E d <strong>and</strong> let t : E d → E d be the translation<br />

x → x + e for x ∈ E d . Then property (b) yields<br />

s({e}) + s({e}) = s({e}+{e}) = s � t({e}) � = ts({e}) = s({e}) + e.<br />

Hence s({e}) = e <strong>and</strong>, applying (9) in the special case where C ={e} <strong>and</strong> thus<br />

h{e}(u) = e · u ∈ Hd 1 , we find that<br />

�<br />

�<br />

e = β (e · u)udσ(u) = βe u 2 1 dσ(u) = βκde<br />

S d−1<br />

S d−1

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