As is usual, if E as a vector space (over R) we will denote by
E^{∗} = Hom(E, R) its algebraic dual space. We will indifferently
use both notations p(v) or hp, vi to denote the value of v ∈ E
under the linear form p ∈ E^{∗}.

Definition 1.2.1 (Supporting Linear Form). We say that the
linear form p ∈ E^{∗} is a supporting linear form at x_{0} ∈ U for the
function f : U → R, defined on U ⊂ E, if we have

∀x ∈ U, f (x) − f (x0) ≥ p(x − x_{0}) = hp, x − x_{0}i.

We will denote by SLF_{x}(f ) the set of supporting linear form at x
for f , and by SLF(f ) the graph

SLF(f ) = ∪_{x∈U}{x} × SLFx(f ) ⊂ U × E^{∗}.

In the literature, the linear form p is also called
subderiva-tive of f at x_{0} or even sometimes subgradient. We prefer to call
it supporting linear form to avoid confusion with the notion of
subdifferential that we will introduce in another chapter.

Example 1.2.2. a) If f : R → R, t 7→ |t| then SLF_{0}(f ) = [−1, 1],
for t > 0, SLF_{t}(f ) = {1}, and for t < 0, SLF_{t}(f ) = {−1}.

b) If g : R → R, t 7→ t^{3} then SLFt(g) = ∅, for everyt ∈ R.

The following Proposition is obvious.

Proposition 1.2.3. The set SLF_{x}(f ) is a convex subset of E^{∗}.
Moreover, if we endow E^{∗}with the topology of simple convergence
on E (”weak topology”) then SLF_{x}(f ) is also closed.

Here is the relation between supporting linear form and deriva-tive.

Proposition 1.2.4. Let f : U → R be a function defined on an open subset U of the normed space E.

a) If f is differentiable at some given x ∈ U then SLF_{x}(f ) ⊂
{Df (x)}, i.e. it is either empty or equal to the singleton {Df (x)}.

b) If E = R^{n}, and all partial derivatives ∂f /∂x_{i}(x), i = 1, . . . , n,
at some given x ∈ U , then SLF_{x}(f ) is either empty or reduced to
the single linear form (a_{1}, . . . , a_{n}) 7→Pn

i=1a_{i}∂f /∂x_{i}(x).

Proof. a) If SLF_{x}(f ) 6= ∅, let p be a supporting linear form of f
at x. If v ∈ E is fixed, for all ǫ > 0 small we have x + ǫv ∈ U and
thus f (x + ǫv) − f (x) ≥ ǫp(v). Dividing by ǫ and taking the limit
as ǫ goes to 0 in this last inequality, we find Df (x)(v) ≥ p(v). For
linear forms this implies equality, because a linear form which is

≥ 0 everywhere has to be 0.

b) We denote by (e_{1}, . . . , e_{n}) the canonical base in R^{n}. Let us
consider a point x = (x_{1}, . . . , x_{n}) ∈ R^{n} where all partial
deriva-tives exist. This implies that the function of one variable h 7→

f (x_{1}, . . . , x_{i−1}, h, x_{i+1}, . . . , x_{n}) is differentiable at x_{i}, hence by part
a), if p ∈ SLF_{x}(f ), we have p(e_{i}) = ∂f /∂x_{i}(x). Since this is true
for every i = 1, . . . , n, therefore the map p must be (a_{1}, . . . , a_{n}) 7→

P_{n}

i=1a_{i}∂f /∂x_{i}(x).

We have not imposed any continuity in the definition of a sup-porting linear form for a function f . This is indeed the case under very mild conditions on f , as we will see presently.

Proposition 1.2.5. Let U is an open subset of the topological
vector space E, and let f : U → R be a function. Suppose that
f is bounded from above on a neighborhood of x_{0} ∈ U , then any
supporting linear form of f at x_{0} is continuous.

Proof. Let V be a neighborhood of 0 such that V = −V , and f is
defined and bounded from above by K < +∞ on x_{0}+ V . Since V

9 is symmetrical around 0, for each v ∈ V , we have

p(v) ≤ f (x0+ v) − f (x0) ≤ 2K

−p(v) = p(−v) ≤ f (x0− v) − f (x0) ≤ 2K,

hence the linear form p is thus bounded on a nonempty open sub-set, it is therefore continuous.

As is customary, if E is a topological vector space, we will
denote by E^{′} ⊂ E^{∗} the topological dual space of E, namely E^{′} is
the subset formed by the continuous linear forms.Of course E^{′} =
E^{∗} if E is finite-dimensional. If E is a normed space, with norm
k·k, then E^{′} is also a normed space for the usual norm

kpk = sup{p(v) | v ∈ E, kvk ≤ 1}.

In the case of continuous map, we can improve Proposition 1.2.3.

Proposition 1.2.6. Suppose that f : U → R is a continuous
function defined on the topological vector space E. If we endow E^{′}
with the topology of simple convergence on E (”weak topology”),
then the graph SLF(f ) is a closed subset of U × E^{′}.

The proof of this Proposition is obvious.

Exercise 1.2.7. Let f : U → R be a locally bounded function defined on the open subset U of the normed space E. (Recall that locally bounded means that each point in U has a neighborhood on which the absolute value of f is bounded)

a) Show that for every x ∈ U , we can find a constant K, and a
neighborhood V such that for every y ∈ V and every p ∈ SLF_{y}(f )
we have kpk ≤ K. [Indication: see the proof of Theorem 1.4.1]

b) If E is finite dimensional, and f is continuous, show the
following continuity property: for every x ∈ U , and every
neigh-borhood W of SLF_{x}(f ) in E^{′} = E^{∗}, we can find a neighborhood V
of x such that for every y ∈ V we have SLF_{y}(f ) ⊂ W .

As we will see now the notion of linear supporting form is tailored for convex functions.

Proposition 1.2.8. If the function f : U → R, defined on the convex subset U of the vector space E, admits a supporting linear form at every x ∈ U , then f is convex.

Proof. Let us suppose that x_{0} = y + (1 − t)z with y, z ∈ U and
t ∈ [0, 1]. If p is a supporting linear form at x_{0}, we have

f (y) − f (x_{0}) ≥ p(y − x_{0}) and f (z) − f (x_{0}) ≥ p(z − x_{0}),
hence

tf (y) + (1 − t)f (z) − f (x_{0}) ≥ p(t(y − x_{0}) + (1 − t)(z − x_{0}))

= p(ty + (1 − t)z − x_{0}) = 0.

The following theorem is essentially equivalent to the Hahn-Banach Theorem.

Theorem 1.2.9. Let U be a convex open subset of the locally convex topological vector space E. If f : U → R is continuous and convex, then we can find a supporting linear form for f at each x ∈ U .

Proof. As f is continuous and convex, the set O = {(x, t) | x ∈ U , f (x) < t}

is open, non-empty, and convex in E × R. Since (x_{0}, f (x_{0})) is not
in O, by the Hahn-Banach Theorem, see [RV73, Theorem C, page
84] or [Rud91, Theorem, 3.4, page 59], there exists a continuous
and non identically zero linear form α : E × R → R and such that

∀(x, t) ∈ O, α(x, t) > α(x0, f (x_{0})).

We can write α(x, t) = p_{0}(x) + k_{0}t, with p_{0}: E → R a continuous
linear form and k_{0} ∈ R. Since α(x0, t) > α(x_{0}, f (x_{0})) for all
t > f (x_{0}), we see that k_{0} > 0. If we define ˜p_{0} = k^{−1}_{0} p_{0}, we get

˜

p_{0}(x)+t ≥ ˜p_{0}(x_{0})+f (x_{0}), for all t > f (x), therefore f (x)−f (x_{0}) ≥
(−˜p_{0})(x − x_{0}). The linear form −˜p_{0} is the supporting linear form
we are looking for.

The following Proposition is a straightforward consequence of Theorem 1.2.9 and Proposition 1.2.4

11
Proposition 1.2.10. Let f : U → R be a continuous convex
function defined on an open convex subset U of the normed space
E. If f is differentiable at x_{0} then the derivative Df (x_{0}) is the
only supporting linear form of f at x_{0}. In particular, we have

∀x ∈ U, f (x) − f (x0) ≥ Df (x_{0})(x − x_{0}).

Corollary 1.2.11. Let f : U → R be a continuous convex
func-tion defined on an open convex subset U of a normed space. If f
is differentiable at x_{0}, then x_{0} is a global minimum if and only if
Df (x_{0}) = 0.

Proof. Of course, if the derivative exists at minimum it must be
0, this is true even if f is not convex. The converse, which uses
convexity, follows from the inequality f (y) − f (x_{0}) ≥ Df (x_{0})(y −
x_{0}) = 0 given by Proposition 1.2.10 above.

Corollary 1.2.12. If U ⊂ R^{n} is open and convex and f : U → R
is a convex function, then, for almost all x, the function f admits
a unique supporting linear form at x.

Proof. This is a consequence of Proposition 1.2.10 above and Rade-macher’s Theorem 1.1.10.

Exercise 1.2.13. Let U be an open and convex subset of R^{n}.
Suppose that f : U → R is convex and continuous. Show that
if f admits a unique supporting linear formp_{0} at x_{0} then Df (x_{0})
exists, and is equal to p_{0}. [Indication: For each x ∈ U \ 0, choose
px∈ SLFx(f ), and prove that

p_{(}x − x_{0}) ≤ f (x) − f (x_{0}) ≤ p_{x}(x − x_{0}).

Conclude using exercise 1.2.7.