Gruber P. Convex and Discrete Geometry

Characterization of Support Functions
The next result reveals the simple nature of support functions.
4 Support and Separation 57
Theorem 4.3. Let h : E d → R. Then the following statements are equivalent:
(i) h is the support function of a (unique) convex body C, i.e. h = hC.
(ii) h has the following properties:
h(λu) = λh(u) for u ∈ E d ,λ≥ 0
h(u + v) ≤ h(u) + h(v) for u,v ∈ E d
Statement (ii) means that h is positively homogeneous of degree 1 and subadditive
and thus, in particular, convex.
Proof. (i)⇒(ii) Let h = hC, where C is a suitable convex body. Then
hC(λu) = sup{λu · x : x ∈ C} =λ sup{u · x : x ∈ C} =λhC(u),
hC(u + v) = sup{(u + v) · x : x ∈ C}
≤ sup{u · x : x ∈ C}+sup{v · x : x ∈ C} =hC(u) + hC(v)
for u,v ∈ E d ,λ≥ 0.
(ii)⇒(i) Define
(5) C = � x : v · x ≤ h(v) for all v ∈ E d� = �
{x : v · x ≤ h(v)}.
v∈E d
Being an intersection of closed halfspaces, C is closed and convex. Taking v =
±b1,...,±bd, where {b1,...,bd} is the standard basis of E d , shows that C is
bounded. If C �= ∅, the definition (5), of C, implies that hC ≤ h (note (4)). Thus, to
finish the proof, it is sufficient to show that
(6) C �= ∅and h ≤ hC.
Let u ∈ E d \{o}. By (ii), the epigraph epi h, ofh, is a closed convex cone in
E d+1 = E d × R with apex at the origin (o, 0), directed upwards and with nonempty
interior. By Theorem 4.1, there is a support hyperplane H of epi h at the point
� u, h(u) � ∈ bd epi h, where � u, h(u) � �= (o, 0). Since epi h is a convex cone with
non-empty interior and apex (o, 0), it follows that H supports epi h also at (o, 0).
(Fig. 4.2)
The exterior normal vectors of H at (o, 0) point below E d . Thus we may choose
such a vector of the form (x, −1). Hence H = {(v, s) : v · x − s = 0}. Then
H − ={(v, s) : v · x ≤ s} ⊇epi h and thus, in particular, v · x ≤ h(v) for each point
(v, h(v)) ∈ bd epi h and therefore v · x ≤ h(v) for all v ∈ E d . Hence x ∈ C, bythe
definition of C in (5), and thus C �= ∅. Since H is a support hyperplane of epi h at
(u, h(u)), we have u · x = h(u). For x ∈ C, the definition of the support function
hC implies that hC(u) ≥ u · x. Thus hC(u) ≥ h(u). The proof of (6) is complete,
concluding the proof of the implication (ii)⇒(i). ⊓⊔
Warning. We warn the reader: the above proof cannot be trivialized, since, a priori,
it is not clear that the boundary hyperplanes of the halfspaces appearing in (5) all
touch C.