We know that \(\sigma \) is a convex function if and only if \(\operatorname {epi}\sigma \) is a nonempty convex set in \(\mathbb R^n\times \mathbb R\) (Proposition B.1.1.6). Therefore, we just have to prove the equivalence between positive homogeneity and \(\operatorname {epi}\sigma \) being a cone.

Let \(\sigma \) be positively homogeneous. For \((x,r)\in \operatorname {epi}\sigma \), the relation \(\sigma (x)\le r\) gives

so \(\operatorname {epi}\sigma \) is a cone. Conversely, if \(\operatorname {epi}\sigma \) is a cone in \(\mathbb R^n\times \mathbb R\), the property \((x,\sigma (x))\in \operatorname {epi}\sigma \) implies \((tx,t\sigma (x))\in \operatorname {epi}\sigma \), i.e.

From Remark 1.1.2, this is just positive homogeneity.

For \(x_1,x_2\in \mathbb {R}^n\) and \(t_1,t_2{\gt}0\), set \(t:=t_1+t_2{\gt}0\); we have

and (1.1.4) is proved.

[(1.1.4) \(\implies \) (1.1.5)] A function satisfying (1.1.4) is obviously subadditive (take \(t_1=t_2=1\)) and satisfies (take \(x_1=x_2=x\), \(t_1=t_2=1/2t\))

i.e. it is positively homogeneous.

[(1.1.5)\(\Rightarrow \text{sublinearity}\)] Take \(t_1,t_2{\gt}0\) with \(t_1+t_2=1\) and apply successively subadditivity and positive homogeneity:

hence \(\sigma \) is convex.

Take \(x_2=-x_1\) in (1.1.3) and remember that \(\sigma (0)\ge 0\).

With \(x_1,\dots ,x_m\) as stated, each \(-x_j\) is in \(\text{dom }\sigma \). Let \(x:=\sum _{j=1}^m t_j x_j\) be an arbitrary linear combination of \(x_1,\dots ,x_m\); we must prove that \(\sigma (x)=\sum _{j=1}^m t_j\sigma (x_j)\). Set

and obtain (as usual, \(\sum _\emptyset =0\)):

In summary, we have proved \(\sigma (x)\le \sum _{j=1}^m t_j\sigma (x_j)\le -\sigma (-x)\le \sigma (x).\) □

In view of subadditivity, we just have to prove “\(\ge \)” in (1.1.10). Start from the identity \(y=x+y-x\); apply successively subadditivity and (1.1.9) to obtain

Nonnegativity and positive homogeneity are obvious from the definition of \(\gamma _C\); also, \(\gamma _C(0)=0\) because \(0\in C\). We prove convexity via a geometric interpretation of (1.2.2). Let

be the convex conical hull of \(C\times \{ 1\} \subset \mathbb {R}^n\times \mathbb {R}\). It is convex (beware that \(K_C\) need not be closed) and \(\gamma _C\) is clearly given by

Thus, \(\gamma _C\) is the lower-bound function of §B.1.3(g), constructed on the convex set \(K_C\); this establishes the convexity of \(\gamma _C\), hence its sublinearity.

Now we prove

This will imply the first part in (iii), thanks to positive homogeneity. Then the second part will follow because of (A.2.2.2): \(C_\infty =\cap \{ rC:\ r{\gt}0\} \) and closedness of \(\gamma _C\) will also result from (iii) via Proposition B.1.2.2.

So, to prove (1.2.3), observe first that \(x\in C\) implies from (1.2.2) that certainly \(\gamma _C(x)\le 1\). Conversely, let \(x\) be such that \(\gamma _C(x)\le 1\); we must prove that \(x\in C\). For this we prove that \(x_k:= (1-1/k)x\in C\) for \(k=1,2,\dots \) (and then, the desired property will come from the closedness of \(C\)). By positive homogeneity, \(\gamma _C(x_k)=(1-1/k)\gamma _C(x)\le 1\), so there is \(\lambda _k\in [0,1]\) such that \(x_k=\lambda _k C\), or equivalently \(x_k/\lambda _k\in C\). Because \(C\) is convex and contains the origin, \(\lambda _k (x_k/\lambda _k)+(1-\lambda _k)0 = x_k\) is in \(C\), which is what we want.

[(ii)] Assume \(0\in \operatorname {int}C\). There is \(\varepsilon {\gt}0\) such that for all \(x\neq 0\), \(x_\varepsilon := x/\| x\| \in C\); hence \(\gamma _C(x_\varepsilon )\le 1\) because of (1.2.3). We deduce by positive homogeneity

this inequality actually holds for all \(x\in \mathbb {R}^n\) (\(\gamma _C(0)=0\)) and \(\gamma _C\) is a finite function.

Conversely, suppose \(\gamma _C\) is finite everywhere. By continuity (Theorem B.3.1.2), \(\gamma _C\) has an upper bound \(L{\gt}0\) on the unit ball:

where the last implication comes from (iii). In other words, \(B(0,1/L)\subset C\).

Concerning convexity and closedness, everything is known from §B.2. Note in passing that a closed sublinear function is zero (hence finite) at zero. As for positive homogeneity, it is straightforward.

Concerning convexity and closedness, everything is known from §B.2. Note in passing that a closed sublinear function is zero (hence finite) at zero. As for positive homogeneity, it is straightforward.

Once again, the only thing to prove for (i) is positive homogeneity. Actually, it suffices to multiply \(x\) and each \(x_j\) by \(t{\gt}0\) in a formula giving \(\operatorname {co}\big(\inf _j\sigma _j\big)(x)\), say (B.2.5.3).

By definition, computing \(\operatorname {co}\big(\min _j\sigma _j\big)(x)\) amounts to solving the minimization problem in the \(m\) couples of variables \((x_j,\alpha _j)\in \operatorname {dom}\sigma _j\times \mathbb {R}\)

In view of positive homogeneity, the variables \(\alpha _j\) play no role by themselves: the relevant variables are actually the products \(\alpha _j x_j\) and (1.3.1) can be written – denoting \(\alpha _j x_j\) again by \(x_j\):

We recognize the infimal convolution of the \(\sigma _j\)’s.

Clearly \(\Delta (\sigma _1,\sigma _2){\lt}+\infty \) and \(\Delta (\sigma _1,\sigma _2)=\Delta (\sigma _2,\sigma _1)\). Now positive homogeneity of \(\sigma _1\) and \(\sigma _2\) gives for all \(x\neq 0\)

In addition, \(\sigma _1(0)=\sigma _2(0)=0\), so

and \(\Delta (\sigma _1,\sigma _2)=0\) if and only if \(\sigma _1=\sigma _2\).

As for the triangle inequality, we have for arbitrary \(\sigma _1,\sigma _2,\sigma _3\) in \(\Phi \)

so there holds

which is the required inequality.

First, the (finite) function \(\sigma \) is of course sublinear whenever it is the pointwise limit of sublinear functions. The equivalence between (i) and (ii) comes from the general Theorem B.3.1.4 on the convergence of convex functions.

Now, (ii) clearly implies (iii). Conversely \(\Delta (\sigma _k,\sigma )\to 0\) is the uniform convergence on the unit ball, hence on any ball of radius \(L{\gt}0\) (the maxmind in (1.3.2) is positively homogeneous), hence on any compact set.

This results from Proposition 1.3.1(ii) (a linear form is closed and convex!). Observe in particular that a support function is null (hence \({\lt} +\infty \)) at the origin.

Let \(S\) be bounded, say \(S\subset B(0,L)\) for some \(L{\gt}0\). Then

which implies \(\sigma _S(x)\le L\| x\| \) for all \(x\in \mathbb {R}^n\).

Conversely, finiteness of the convex \(\sigma _S\) implies its continuity on the whole space (Theorem B.3.1.2), hence its local boundedness: for some \(L\),

If \(s\neq 0\), we can take \(x=s/\| s\| \) in the above relation, which implies \(\| s\| \le L\).

The continuity [resp. linearity, hence convexity] of the function \(\langle s,\cdot \rangle \), which is maximized over \(S\), implies that \(\sigma _S=\sigma _{\operatorname {cl} S}\) [resp. \(\sigma _S=\sigma _{\operatorname {co} S}\)]. Knowing that \(\overline{\operatorname {co}}\, S=\operatorname {cl}\, \operatorname {co}\, S\) (Proposition A.1.4.2), (2.2.1) follows immediately.

First, the equivalence between all the choices for \(X\) is clear enough; in particular due to positive homogeneity. Because “\(\Rightarrow \)” is Proposition 2.2.1, we have to prove “\(\Leftarrow \)” only, with \(X=\mathbb {R}^n\) say.

So suppose that \(s\notin \overline{\operatorname {co}}S\). Then \(\{ s\} \) and \(\overline{\operatorname {co}}S\) can be strictly separated (Theorem A.4.1.1): there exists \(d_0\in \mathbb {R}^n\) such that

where the last equality is (2.2.1). Our result is proved by contradiction.

Let first \(s\in S\). We have already seen in Definition 2.1.4 that

If the breadth of \(S\) along \(d\) is zero, we obtain a pair of equalities: for such \(d\), there holds

an equality which extends by affine combination to any element \(s\in \operatorname {aff}S\).

Conversely, let \(s\) satisfy (2.2.3). A first case is when the only \(d\) described in (2.2.3) is \(d = 0\); as a consequence of our observations in Definition 2.1.4, there is no affine hyperplane containing \(S\), i.e. \(\operatorname {aff}S = \mathbb {R}^n\) and there is nothing to prove. Otherwise, there does exist a hyperplane \(H\) containing \(S\); it is defined by

for some \(d_H \neq 0\). We proceed to prove \(\langle s,\cdot \rangle \le \sigma _H\).

In fact, the breadth of \(S\) along \(d_H\) is certainly 0, hence \(\langle s,d_H\rangle = \sigma _S(d_H)\) because of (2.2.3), while (2.2.6) shows that \(\sigma _S(d_H) = \sigma _H(d_H)\). On the other hand, it is obvious that \(\sigma _H(d) = +\infty \) if \(d\) is not collinear to \(d_H\). In summary, we have proved \(\langle s,d\rangle \le \sigma _H(d)\) for all \(d\), i.e. \(s \in H\). We conclude that our \(s\) is in any affine manifold containing \(S\): \(s \in \operatorname {aff}S\).

[(iii)] In view of positive homogeneity, we can normalize \(d\) in (2.2.5). For \(s \in \operatorname {int} S\), there exists \(\varepsilon {\gt} 0\) such that \(s + \varepsilon d \in S\) for all \(d\) in the unit sphere \(\widetilde{B}\). Then, from the very definition (2.1.1),

Conversely, let \(s \in \mathbb {R}^n\) be such that

which implies, because \(\sigma _S\) is closed and the unit sphere is compact:

Thus

Now take \(u\) with \(\| u\| {\lt} \varepsilon \). From the Cauchy-Schwarz inequality, we have for all \(d \in \widetilde{B}\)

and this implies \(s + u \in S\) because of Theorem 2.2.2: \(s \in \operatorname {int} S\) and (iii) is proved.

[(ii)] Look at Fig. 2.2.2 again: decompose \(\mathbb {R}^n = V \oplus U\), where \(V\) is the subspace parallel to \(\operatorname {aff}S\) and \(U = V^\perp \). In the decomposition \(d = d_V + d_U\), \(\langle \cdot ,d_U\rangle \) is constant over \(S\), so \(S\) has 0-breadth along \(d_U\) and

for any \(s \in S\). With these notations, a direction described as in (2.2.4) is a \(d\) such that

Then, (ii) is just (iii) written in the subspace \(V\).

Recall from §A.3.2 that, if \(K_1\) and \(K_2\) are two closed convex cones, then \(K_1\subset K_2\) if and only if \((K_1)^\circ \supset (K_2)^\circ \).

Let \(p\in S_\infty \). Fix \(s_0\) arbitrary in \(S\) and use the fact that \(S_\infty = \bigcap _{t{\gt}0} t(S-s_0)\) ( §A.2.2); for all \(t{\gt}0\), we can find \(s_t\in S\) such that \(p=t(s_t-s_0)\). Now, for \(q\in \operatorname {dom}\sigma _S\), there holds

and letting \(t\downarrow 0\) shows that \(\langle p,q\rangle \le 0\). In other words, \(\operatorname {dom}\sigma _S\subset (S_\infty )^\circ \); then \(\overline{\operatorname {dom}\sigma _S}\subset (S_\infty )^\circ \) since the latter is closed.

Conversely, let \(q\in (\operatorname {dom}\sigma _S)^\circ \), which is a cone, hence \(tq\in (\operatorname {dom}\sigma _S)^\circ \) for any \(t{\gt}0\). Thus, given \(s_0\in S\), we have for arbitrary \(p\in \operatorname {dom}\sigma _S\)

so \(s_0+tq\in S\) by virtue of Theorem 2.2.2. In other words: \(q\in (S-s_0)/t\) for all \(t{\gt}0\) and \(q\in S_\infty \).

Being convex, \(\sigma \) is minorized by some affine function (Proposition B.1.2.1): for some \((s,r)\in \mathbb {R}^n\times \mathbb {R}\),

Because \(\sigma (0)=0\), the above \(r\) is nonnegative. Also, by positive homogeneity,

Letting \(t\to +\infty \), we see that \(\sigma \) is actually minorized by a linear function:

Now observe that the minorization (3.1.3) is sharper than (3.1.2): when expressing the closed convex \(\sigma \) as the supremum of all the affine functions minorizing it (Proposition B.1.2.8), we can restrict ourselves to linear functions. In other words

in the above index-set, we just recognize \(S_\sigma \).

The case \(X=\mathbb {R}^n\) is just Theorem 3.1.1. The other cases are then clear.

Observe from Definition 3.1.3 that the face exposed by \(d\neq 0\) does not depend on \(\| d\| \). This establishes the equivalence between the first two choices for \(X\). As for the third choice, it is due to the fact that \(F_C(d)=\varnothing \) if \(d\notin \operatorname {dom}\sigma _C\).

Now, if \(x\) is interior to \(C\) and \(d\neq 0\), then \(x+\varepsilon d\in C\) and \(x\) cannot be a maximizer of \(\langle \cdot ,d\rangle \); \(x\) is not in the face exposed by \(d\). Conversely, take \(x\) on the boundary of \(C\). Then \(N_C(x)\) contains a nonzero vector \(d\); by Proposition 3.1.4, \(x\in F_C(d)\).

It is a particular case of the results 3.2.4 and 3.2.5 below.

We know that \(\gamma _C\) (which, by Theorem 1.2.5(i), is closed, sublinear and nonnegative) is the support function of some closed convex set containing the origin, say \(D\); from (3.1.1),

As seen in (1.2.4), \(\operatorname {epi}\gamma _C\) is the closed convex conical hull of \(C\times \{ 1\} \); we can use positive homogeneity to write

In view of Theorem 1.2.5(iii), the above index-set is just \(C\); in other words, \(D=C^\circ \).

Left as an exercise; the assumptions are present to make sure that every nonzero vector in \(\mathbb {R}^n\) does expose a face in each set.

Apply the equivalence stated in Corollary 3.1.2:

Call \(S\) the closed convex set \(\operatorname {cl}(t_1S_1+t_2S_2)\). By definition, its support function is

In the above expression, \(s_1\) and \(s_2\) run independently in their index sets \(S_1\) and \(S_2\), \(t_1\) and \(t_2\) are positive, so

The support function of \(S:=\bigcup _{j\in J}S_j\) is

This implies (ii) since \(\sigma _S=\sigma _{\overline{\operatorname {co}}\, S}\).

The set \(S:=\bigcap _j S_j\) being nonempty, it has a support function \(\sigma _S\). Now, from Corollary 3.1.2,

where the last equivalence comes directly from the Definition B.2.5.3 of a closed convex hull. Again Corollary 3.1.2 tells us that the closed sublinear function \(\overline{\operatorname {co}}(\inf \sigma _j)\) is just the support function of \(S\).

Just write the definitions

and use Proposition 2.2.1 to obtain the result.

Our assumption is tailored to guarantee \(A\sigma \in \operatorname {Conv}\mathbb {R}^n\) (Theorem B.2.4.2). The positive homogeneity of \(A\sigma \) is clear: for \(d\in \mathbb {R}^n\) and \(t{\gt}0\),

Thus, the closed sublinear function \(\operatorname {cl}(A\sigma )\) supports some set \(S'\); by definition, \(s\in S'\) if and only if

but this just means

i.e. \(A^*s\in S\), because \(\langle s,Ap\rangle =\langle A^*s,p\rangle \).

As mentioned in §0.5.1, for all \(r\ge 0\), the property

simply means \(S'\subset S+B(0,r)\).

Now, the support function of \(B(0,1)\) is \(\| \cdot \| \) — see (2.3.1). Calculus rules on support functions therefore tell us that (3.3.6) is also equivalent to

which in turn can be written

In summary, we have proved

and symmetrically

the result follows.

Calculus with support functions tells us that our definition (0.5.2) of outer semi-continuity is equivalent to

and division by \(\lVert d\rVert \) shows that this is exactly upper semi-continuity of the support function for \(\lVert d\rVert =1\). Same proof for inner/lower semi-continuity.