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Given a real vector space of finite dimensnion \(E\), a subset \(C \subseteq E\) is convex when the line segment between any two points remains in \(C\). Formally, \(C\) is convex if for all \(x, y \in C\):
\[\forall \lambda \in [0, 1] : (1 - \lambda) x + \lambda y \in C\]It is easy to see that the above is equivalent to \(C\) being stable by (finite) convex combinations:
\[\forall \block{x_i}_i \in C, \block{\lambda_i}_i \in \RR^+, \sum_i \lambda_i = 1: \sum_i \lambda_i x_i \in C\]The space of convex weights \(\lambda \in \RR^{n+}, \mathbb{1}^T\lambda = 1\) is usually called the \(n\)-dimensional unit simplex, denoted by \(\Delta^n\).
Given a convex subset \(X \subseteq E\), a function \(f: X \to \RR\) is convex when its epigraph is convex:
\[\forall x, y \in X: f\block{\block{1 - \lambda} x + \lambda y} \leq (1 - \lambda) f(x) + \lambda f(y)\]That is: the line segment between \(f(x)\) and \(f(y)\) remains above the graph of \(f\). Equivalently:
\[f\block{x + \lambda(y - x)} \leq f(x) + \lambda \block{f(y) - f(x)}\]For any \(0 < \lambda \leq 1\) we have:
\[\frac{f\block{x + \lambda (y - x)} - f(x)}{\lambda} \leq f(y) - f(x)\]Taking limits when \(\lambda\) goes to zero gives:
\[\lim_{\lambda \downarrow 0} \frac{f\block{x + \lambda (y - x)} - f(x)}{\lambda} \leq f(y) - f(x)\]Therefore, if \(f\) is differentiable at \(x\), we obtain:
\[\dd f(x).(y - x) \leq f(y) - f(x)\]Or, equivalently:
\[\inner{\nabla f(x), y - x} \leq f(y) - f(x)\]In other words, \(f\) always stays above its tangent space at \(x\). This condition is actually sufficient for convexity. Letting \(z = (1 - \lambda) x + \lambda y\), we obtain:
\[\begin{align} (1 - \lambda) f(x) &\geq (1 - \lambda)\block{f(z) + \dd f(z).(x - z)} \\ \lambda f(y) &\geq \lambda \block{f(z) + \dd f(z).(y - z)}\\ \end{align}\]and summing both equations provides the convexity of \(f\):
\[(1 - \lambda) f(x) + \lambda f(y) \geq f(z) + \dd f(z).\underbrace{\block{(1 - \lambda) x + \lambda y - z}}_{0}\]Besides, one can easily check that \(\dd f\) is monotone, that is:
\[\block{\dd f(y) - \dd f(x)}(y - x) \geq 0\]As it turns out, this condition is also sufficient for the convexity of \(f\). Let \(g(\lambda) = f\block{x + \lambda(y - x)}\), we have:
\[\begin{align} g(0) &= f(x) \\ g(1) &= f(y) \\ \end{align}\]and by the mean value theorem, there exists some \(0 < \lambda < 1\) such that:
\[\underbrace{g'(\lambda)}_{\dd f(z).(y - x)} = f(y) - f(x)\]where once again \(z = (1 - \lambda) x + \lambda y\). Assuming monotonicity of \(\dd f\) we obtain:
\[\block{\dd f(z) - \dd f(x)}\underbrace{(z - x)}_{\lambda\block{y - x}} \geq 0\]and since \(\lambda > 0\) this implies \(\block{\dd f(z) - \dd f(x)}(y - x) \geq 0\). Putting everything together:
\[f(y) - f(x) \geq \dd f(z).(y - x) \geq \dd f(x).(y - x)\]hence by the previous argument \(f\) is convex.
When \(f\) is twice continuously differentiable, the monotonicity of \(\dd f\) implies the following, for \(\lambda > 0\):
\[\frac{\dd f\block{x + \lambda (y - x)} - \dd f(x)}{\lambda}.(y - x) \geq 0\]Taking limits as \(\lambda \downarrow 0\), we obtain by the defintion of the second derivative:
\[\dd^2 f(x)(y - x, y - x) \geq 0\]Alternatively, using the Hessian matrix \(\nabla^2 f\):
\[(y - x)^T \nabla^2 f(x) (y - x) \geq 0\]In other words, the Hessian of \(f\) is positive semi-definite.
Let us assume \(x_1, x_2 \in C\) both minimize a strict convex function \(f\) over \(C\) with \(x_1 \neq x_2\). In particular, we have \(f\block{x_1} = f\block{x_2} = f^\star\). The strict convexity of \(f\) gives, for any \(0 < \lambda < 1\):
\[f(\underbrace{(1- \lambda) x_1 + \lambda x_2}_{\neq x_1, \neq x_2}) < (1 - \lambda) f\block{x_1} + \lambda f\block{x_2} = f^\star\]which contradicts the fact that \(x_1, x_2\) minimize \(f\). Therefore the minimizer, should it exist, must be unique. A similar argument can show that when \(f\) is merely convex, every point on the segment \([x_1, x_2]\) must also minimize \(f\).
A subset \(\cone{K}\) of a vector space \(E\) is a cone if it is stable by positive scalar multiplication:
\[\forall x \in \cone{K}, \lambda > 0\quad \lambda x \in \cone{K}\]A cone is pointed when \(0 \in \cone{K}\). A non-empty closed cone is always pointed. One generally deals with closed convex cones so that the projection \(\pi_\cone{K}\) onto \(\cone{K}\) is well-defined. A convex pointed cone \(K\) induces a preorder defined as:
\[x \leq_\cone{K} y \iff y - x \in \cone{K}\]This preorder can be made into a partial order by requiring \(\cone{K}\) to be flat, that is stable by negation (i.e. it contains lines).
When \(E\) is an Euclidean space, the inner product provides1 a generalization of the orthogonal complement for cones, called the dual cone. Given a subset \(X \subseteq E\), its dual cone is defined as the set:
\[X^* = \left\{y \in E : \inner{y, x} \geq 0 \quad \forall x \in X\right\}\]The dual cone is obviously a cone, and it is easy to check that it is convex even though \(X\) might not be. As an intersection of closed half-spaces, it is also closed. The negative of the dual cone \(-X^*\) is called the polar (or negative-dual) cone of \(X\), usually denoted \(X^\circ\) (or \(X^-\)).
The Moreau decomposition generalizes the direct sum decomposition between a linear subspace and its orthogonal complement to a convex cone and its (negative) dual.
Suppose we want to minimize a function \(f: E \to \RR\) over some set \(C\). For a given \(x \in C\) to be a local minimizer means that \(f\) can only increase locally around \(x\) in \(C\). If \(f\) is differentiable, this means that the derivative of \(f\) in any admissible direction \(v\) should be positive:
\[\dd f(x).v = \lim_{\epsilon \downarrow 0}\ \frac{f(x + \epsilon v) - f(x)}\epsilon \geq 0\]Now the set of admissible directions of \(C\) at \(x\) is obviously a subset of the full tangent space \(T_x(E)\) of \(E\) at \(x\), and should restrict the set of tangent vectors to the ones that somehow remain in \(C\) (to the first order).
In the general case, the correct definition is a bit technical (the Bouligand tangent cone) but in the case where \(C\) is convex, it suffices to consider the set of directions that intersect \(C\) near \(x\):
\[T_x(C) = \left\{v \in T_x\block{E} :\ \exists \epsilon > 0 : x + \lambda v \in C\quad\forall \lambda \in [0, \epsilon] \right\}\]This subset is obviously a cone, called the tangent cone to \(C\) at \(x\). It can be easily checked that for convex \(C\), the tangent cone at \(x\) is given by:
\[T_x(C) = \RR^+\block{C - x}\]which makes the cone structure even more explicit. The optimality condition above can be rewritten as:
\[\dd f(x).v \geq 0\quad \forall v \in T_x(C)\]Or, using the gradient of \(f\):
\[\inner{\nabla f(x), v} \geq 0\quad \forall v \in T_x(C)\]which is to say that \(\nabla f(x)\) should belong to the dual of the tangent cone at \(x\), called the (negative) normal cone \(N_x(C)\):
\[\nabla f(x) \in \block{T_x(C)}^* \triangleq -N_x(C)\]From our earlier expression for \(T_x(C)\), we obtain its dual as:
\[-N_x(C) = \block{T_x(C)}^* = \left\{y \in E : \inner{y, z - x} \geq 0\quad \forall z \in C \right\}\]When \(C=\cone{K}\) is a closed convex cone, this means that:
\[y \in -N_x(\cone{K}) \iff \inner{y, z - x} \geq 0 \quad \forall z \in \cone{K}\]Taking \(z = 0\) and \(z = 2 x\) yields \(\inner{y, x} = 0\) hence \(y \in x^\bot\), which itself implies that \(\inner{y, z} \geq 0\) and we also get \(y \in \cone{K}^*\). The converse is easy to check and we obtain:
\[-N_x(\cone{K}) = \cone{K}^* \cap x^\bot\]Putting everything together, this means that the optimality conditions for minimizing \(f\) over a closed convex cone \(\cone{K}\) are the following:
\[\cone{K} \ni x\ \bot\ \nabla f(x) \in \cone{K}^*\]We consider the problem of minimizing a quadratic function over some closed convex cone \(\cone{K}\):
\[\min_{x \in \cone{K}} \ \frac{1}{2}x^TMx + q^Tx\]The optimality conditions are the following (linear) Cone Complementarity Problem:
\[\begin{align} \quad Mx + q &= \lambda \\ \cone{K} \ni x&\ \bot\ \lambda \in \cone{K}^* \end{align}\]When \(\cone{K}=\RR^n_+\) is the positive orthant (self-dual), these conditions are known as a Linear Complementarity Problem (LCP):
\[\begin{align} \quad Mx + q &= \lambda \\ 0 \leq x&\ \bot\ \lambda \geq 0 \end{align}\]If we wish to impose constraints of the form \(Ax \in \cone{K}^*\) for some dual cone \(\cone{K}^*\), we’ll need to compute the preimage of cone by a linear map. One can immediately check that the preimage \(A^{-1}\cone{K}^*\) is a closed cone. Furthermore:
\[\begin{align} x \in \inv{A}\cone{K}^* &\iff Ax \in \cone{K}^* \\ &\iff \forall y \in \cone{K}\quad \inner{Ax, y} \geq 0 \\ &\iff \forall y \in \cone{K}\quad \inner{x, A^Ty} \geq 0 \\ &\iff \forall z \in A^T\cone{K}\quad \inner{x, z} \geq 0 \\ &\iff x \in \block{A^T\cone{K}}^* \end{align}\]And we obtain \(\inv{A}\cone{K}^* = \block{A^T\cone{K}}^*\), a result known as the (generalized) Farkas lemma. Note that we’ve been carefully avoiding adherence issues, in particular the image of a closed cone by a linear map in not necessarily closed2, which explains why the Farkas Lemma is sometimes given as:
\[\block{\inv{A}\cone{K}}^* = \bar{\block{A^T\cone{K}^*}}\]The version above sidesteps the issue by taking duals, which are always closed, but this is something to keep in mind.
We now consider the following optimization problem:
\[\min_{x \in E} \ f(x)\ \st \ c(x) \in \cone{K}\]for some convex closed cone \(\cone{K}\). The admissible directions at \(x\) must satisfy:
\[\dd c(x).\dd x \in T_{c(x)}(\cone{K})\]In other words, tangent vectors outputted by \(c\) must be admissible in \(\cone{K}\), that is: belong to \(T_{c(x)}\cone{K}\). From Farkas’ lemma, we get:
\[\dd c(x)^{-1}\block{T_{c(x)}(\cone{K})} = \block{\dd c(x)^T \block{T_{c(x)}\cone{K}}^*}^*\]where the inverse is understood as a preimage. One generally asks that \(\dd c(x)^T \cone{K}^*\) be closed (see the discussion above) via a constraint qualification condition, so that the dual cone is:
\[\begin{align} \block{\dd c(x)^{-1} \block{T_{c(x)}(\cone{K})}}^* &= \dd c(x)^T \block{T_{c(x)}\cone{K}}^*\\ &= \dd c(x)^T \block{\cone{K}^* \cap c(x)^\bot} \end{align}\](see the discussion above on normal cones to convex closed cones). We are now ready to state the optimality conditions for the constrained problem:
\[\nabla f(x) \in \dd c(x)^T \block{\cone{K}^* \cap c(x)^\bot}\]which expands to:
\[\begin{align} \nabla f(x) &= \dd c(x)^T \lambda\\ \cone{K} \ni c(x) &\ \bot\ \lambda \in \cone{K}^* \\ \end{align}\]These are known as the Karush, Kuhn & Tucker (KKT) conditions.
Here we minimize a quadratic function:
\[f(x) = \frac{1}{2}x^T Q x + c^T x\]subject to constraints:
\[g(x) = Ax - b \geq 0\]The positive orthant cone \(\RR^n_+\) is self-dual, and the KKT conditions are:
\[\begin{align} Qx + c &= A^T \lambda\\ 0 \leq Ax - b &\ \bot\ \lambda \geq 0 \\ \end{align}\]Let \(C\) be a closed convex set. the projection of a point \(x\) onto \(C\) minimizes the Euclidean norm (squared):
\[\pi_C(x) = \argmin{y \in C}\ \norm{x - y}^2\]TODO (weierstrass)
From the convexity of \(C\) one can derive a characterization of the projection similar to the stationary conditions in optimization as follows. Let us call \(c = \pi_C(x)\), for any \(y \in C\) the point \((1 - \lambda) c + \lambda y\) belongs to \(C\) for \(\lambda \in [0, 1]\), therefore:
\[\begin{aligned} \norm{x - c}^2 &\leq \norm{x - (1 - \lambda) c + \lambda y}^2 \\ &= \norm{x - c + \lambda\block{c - y}}^2 \\ &= \norm{x - c}^2 + \lambda\block{x - c}^T\block{c - y} + \lambda^2 \norm{c - y}^2 \\ \end{aligned}\]This means that for any \(\lambda \in [0, 1]\) we have:
\[\lambda \block{x - c}^T{y - c} \leq \lambda^2 \norm{y - c}\]which for all \(\lambda \in ]0, 1]\) gives:
\[\block{x - c}^T{y - c} \leq \lambda \norm{y - c}\]Therefore \(\block{x - c}^T{y - c} \leq 0\). Conversely, if \(\block{x - c}^T{y - c} \leq 0\) we get
\[\begin{aligned} \norm{x - y}^2 &= \norm{x - c + c - y}^2 \\ &= \norm{x - c}^2 + \underbrace{2 \block{x - c}^T\block{c - y}}_{\geq 0} + \underbrace{\norm{c - y}^2}_{\geq 0} \\ \end{aligned}\]therefore \(\norm{x - y}^2 \geq \norm{x - c}^2\) and \(c = \pi_C(x)\).
Actually, the notion of dual cone can be expressed using only the canonical pairing between \(E\) and its dual \(E^*\) ↩
Consider the Lorentz cone projected on the plane of normal \((1, 1, \ldots, 1)^T\): the projection is the union of slices of the cone by planes parallel to the projection plane. Each slice is delimited by a parabola passing through the origin of the slicing plane, and whose shape flattens as the slicing plane gets further from the origin. The union of all these slices is the open half-plane plus the origin, which is not closed. ↩
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