**Definition:** Let be a vector space. An affine hyperplane is a level set of some linear functional , ie. with .

In particular, such an hyperplane disconects into two half-spaces and , and we say that separates two subsets if they are included in two different half-spaces.

**Theorem:** *(separation)* Let be a vector space and be two nonempty disjoint convex subsets. If is open then there exists an affine hyperplane separating and .

**Sketch of proof.** Let and . Define and , and consider the following Minkowski functionnal:

Notice that is well-defined because is open. Then, define the linear functional:

If , clearly . If , then since implies . Therefore .

Now, using Hahn-Banach theorem, can be extended into a linear functional satisfying , hence

That is,

**Definition:** Let be a normed vector space, be a closed convex subset and . A supporting hyperplane at is an affine hyperplane separating and .

Our main geometrical results are:

**Theorem 1:** Let be a normed space and be a closed subset whose interior is nonempty. Then is convex if, and only if, admits a supporting hyperplane at any .

**Proof.** Suppose that is convex and let . Applying separation theorem, there exists an affine hyperplane separating and . Then is a supporting hyperplane at .

Conversely, suppose that admits a supporting hyperplane at any point of and suppose by contradiction that is not convex. Therefore, there exist and such that . Let and let be the triangle whose vertices are , and .

If is a supporting hyperplane at , then is a straight line separating and and therefore and . We deduce that itself separates and (two elements of ), a contradiction.

**Theorem 2:** Let be a normed vector space and be a closed convex subset whose interior is nonempty. Then the intersection of the half-spaces containing coincides with .

**Proof.** Let be the given intersection. Clearly, . Moreover, if then separation theorem can be used to separate and so that . So .

Notice that the converse is true: an intersection of closed convex sets is closed and convex.

The result allowing us to link convex sets and convex functions is the following (the (classic and easy) proof is left to the reader):

**Property 1:** Let be a vector space and be a function. Then is convex if, and only if, is convex.

In the following, we only work in finite dimension, and a useful fact is that convex functions are continuous:

**Property 2:** Let be a finite-dimensional normed vector space and be a convex function. Then is continuous.

**Proof.** Without loss of generality, suppose that . Let be a sequence converging to some and let be a cube containing the sequence whose vertices are .

If then we can write with and . Therefore,

Thus, is bounded above by on .

For all , there exist such that and . Therefore, there exist such that

hence by convexity,

To conclude that , and so that is continuous, it is sufficient to notice that . Indeed, and

The same argument works for .

Now we use theorem 1 and theorem 2 to prove two results in convex analysis:

**Definition:** Let be a normed vector space and be a function. We say that is affine if there exist and such that .

**Theorem 3:** Let be a finite-dimensional normed space and be a function. Then is convex if, and only if, there exists a family of affine functions such that .

**Proof.** Suppose convex. According to properties 1 and 2, is a closed convex subset whose interior is nonempty. Therefore, using theorem 2, there exists a family of affine hyperplanes where and such that

where is a half-space determined by .

Without loss of generality, we may suppose that

(Otherwise, replace by .) According to Riesz representation theorem, there exist and such that . Thus,

If , we get a contradiction as . If , we deduce that is bounded above hence and finally , a contradiction with the fact that is a hyperplane. Therefore, and we may suppose without loss of generality that (otherwise, replace with ). So

Hence

Let . We just proved that .

Let . Then so there exists an affine hyperplane () separating and , that is

Therefore, hence as . Finally, we deduce that .

The converse is clear, since the upper envelope of a family of convex functions is convex.

**Theorem:** *(Jensen’s inequality)* Let be a convex function and . Then

**Lemma 1:** Let be finite-dimensional normed vector space, be a convex function and . Then there exists such that

**Proof.** According to properties 1 and 2, is a closed convex set whose interior is nonempty. Using theorem 1, admits a supporting hyperplane at .

Without loss of generality, we may suppose that:

(Otherwise, replace with .) According to Riesz representation theorem, there exist and such that , hence

If , we get a contradiction as . If , then is bounded above hence and finally which is impossible since is a hyperplane. Therefore, .

Without loss of generality, we may suppose that (otherwise, replace with ). We finally deduce

**Proof of Jensen’s inequality.** Let be the linear functional given by the previous lemma for at the point , ie.

In particular,

By integrating:

Hence

Theorem 3 can be generalized in infinite dimension; however, has to be supposed lower semi-continuous to assure be closed. Also, lemma 1 suggests to introduce the subdifferential:

Then, lemma 1 just says that . The subdifferential generalizes the differential in the sense that, if is differentiable, . Lemma 1 can also be generalized in infinite dimension, but needs to be supposed a point of continuity of .

For more information about the previous generalizations, see for example Francis Clarke’s book, *Function Analysis, Calculus of Variations and Optimal Control*.