The local projection x→s(x) will be discontinuous in general, so solving (3) raises the question of a differential equation involving a discontinuous state function. As x crosses v, s(x) suddenly moves to the vicinity of point C. All strings acting on x are omitted from the figure.).
Exercise
The first part of the following result follows easily from the cone property of NSP(s) and the characterization (d) of Theorem 1.3; the second part. 1 Nearest points and proximal normals 25 shows that P-normality is essentially a local property: the proximal normal cones NSP1(s) and NSP2(s) are the same as the two sets S1 and S2.
Proposition
We then say that f is Fr´echet-differentiable on x, and in this case write f(x) (the Fr´echet derivative) instead of fG(x). Moreover, if the above limit holds uniformly for bounded sets of X1, then F is Fr´echet-differentiable and we write F(x) instead of FG(x).
Exercise
Put in purely heuristic terms, the content of Theorem 2.5 is that there exists such a parabola that "locally fits" the epigraph on. The proximal subdifferential is a “one-sided” object suitable for the analysis of semicontinuous functions below.
Exercise
For a theory applicable to upper semi-continuous functions f, the proximal superdifferential ∂Pf(x) is the appropriate object, and can be simply defined as −∂P(−f)(x). Using the notation of the previous paragraph, suppose there exists ζ ∈ ∂Pf(x), and write ζ = (ζ1, ζ2) according to the direct sum decomposition.
Exercise
In particular, dom(∂Pf) is dense in domf. 1) We first give a simple motivational proof in the case X =Rn. In this section, we investigate the proximal subgradients of the distance functions associated with a non-empty closed subset S of X. The results of the analysis will enable us to derive geometric analogues of the minimization principles of §4. s∈S so that the following holds:.
Exercise
But the argument just given clearly implies that every such x0 belongs to int(domUf), so that cl(domUf) is seen to be open (as well as closed) with respect to U; hence cl(domUf)∩U = domUf =U. There is a technique, sometimes called "decoupling", to approximate the first (paired) situation above with the second (unpaired). We shall give an exact result of this kind, the relevance of which to the Sum Rule will be apparent shortly.
Exercise
We now turn to the case of hypothesis (ii), letting Kbe a Lipschitz constant for one of the functions onC, sayf1. We are now ready to prove a result known as the "fuzzy sum rule". i) f1 and f2 are weakly lower semicontinuous (automatically the case if X is finite-dimensional); or. ii) one of the functions is Lipschitz nearx0. We can take η small enough that both f1 and f2 are bounded below on C and so in case (ii) one of the functions is Lipschitz on that set.
Exercise
Again, the assumption that one of the functions (say f1) is locally Lipschitz would be fine. 11 Problems in Chapter 1 65 (b) Show that the conclusion in the form is false. c) Obtain the limiting form of the theorem in terms of ∂Lf when ε→0. For each i, by the upper bound definition, there exists yi in X and ti>0 such that.
Proposition
1 Definition and Basic Properties 73 is chosen on X, it follows that ∂f(x) is also independent of the particular norm on X. Some immediate intuition about ∂f is available from the following exercise, where we see that the relation between f◦ and ∂f generalizes the classical formula(x;v).
Exercise
Let ε >0 be given; then we want to show that for everything you are close to, we have. We can therefore separateζi from the compact convex set in question: for some vi= 0 we have. Since we are in finite dimensions, we can extract convergent subsequences from {ζi} and {vi} (we do not relabel): ζi → ζ, vi → v, where v = 1.
Exercise
The two closed convex sets that appear in the equation are actually intervals in R, so it suffices to prove that we have for v=±1. A real-valued function f defined on an open convex subset U of X is called convex provided that for any two points x we have y∈U. If K is the Lipschitz constant for f nearx, then we have for allx in B(x;εδB), for allε sufficiently small,.
Exercise
To show that NS(x) is present in the set on the right-hand side of the formula asserted there, it suffices to show that ∂dS(x) is present in the right-hand side, in light of Theorem 5.4. In light of (a), this would in turn follow from proving that every ζ is of the form w-limζi, where ζi. To complete the proof of the theorem, it now suffices to prove that every point ζ of the form w-limζi, where ζi ∈ NSP(xi) and xi → x, belongs to NS(x).
Exercise
The meaning of the formula is the following: consider any sequence {xi} that converges to x while avoiding both Ω and points where f is not differentiable, and such that the sequence. Since ∂f(x) is convex, we deduce that the left side of the formula asserted by the theorem contains the right side. Note that the limits of these three sets form a set Ω of measure 0, and that if (x, y) does not lie in Ω, then f is differentiable and.
Exercise
The question now becomes that of going to the extreme in the context of the sequences provided by the statement. In the "completely convex" case of the problem, the distinction disappears, and the formula becomes exact; see the problems at the end of the chapter. Here are the counterparts of the necessary conditions (Exercise 1.8) and the solvability consequences (Theorem 1.9) for the inequality case under consideration.
Theorem
For a continuous Gˆateaux-differentiable function f, the new Mean Value Inequality claims that this is the case for somez∈B. this is where we exploit the uniformity of the conclusion). Accordingly, we will start by proving a version of the theorem for a differentiable function on Rn. We will see that the conclusions of the theorem hold for these choices of ζ and ¯z, provided that η is chosen sufficiently small.
Exercise
Clearly, there is no loss of generality if we assume that the left-hand side of (27) is finite, since otherwise the conclusion of the theorem is immediate. Since hC(ζ) ≤ 0 by assumption and since ε is arbitrary, we derive the existence of y∈x+tC such that f(y)≤f(x); i.e. f falls slightly. Letti↓0 and extract a subsequence of corresponding points ci (without relabeling) so that they converge to c∈C.
Exercise
Of course, part of the interest of (4) is that it implies the non-emptiness of Φ(α) (since d. We note that the conclusion of the theorem subordinates as special cases stability results with respect to either a parameter change or to a change in the variable x. 3 Solving Equations 129 under the hypotheses of Lemma 3.3, and especially in the presence of the constraint qualification (6).
Exercise
Theorem
Systems of Mixed Equations and Inequalities We now consider the solutions of a parameterized system of the form. Armed with Lemma, applying Solvability Theorem 3.1 is immediate and yields the required result. 3 Solving Equations 133 If there is no inequality in (8), it is natural to look for non-smooth analogues of the assertions of the implicit function from Theorem 3.6.
Exercise
We note that, in accordance with the previous formula, an arbitrary set of measures 0 can be avoided when computing ∂G(x) without affecting the results below. We identify ∂G(x) with a convex set of n ×m matrices, and we say that ∂G(x) is of maximal rank given that every matrix in ∂G(x) is of maximal rank. Then the restriction qualification(9) holds, and for someη >0, for some quarter Ωofx0, we have The set P of pointsx in x0+rB, where G(x) does not exist, is measure 0, and therefore by Fubini's theorem, for almost every x in π, the ray.
Exercise
Exercise
We note that in light of (1), the ˆrof the theorem satisfies ˆr≥ρt+t2, so ¯r=ρ is a suitable choice in the application of the theorem. Show that the statement is false if the convexity of E is deleted from the hypotheses, by considering X=R2,f(u) :=u, x= 0,E= unit circle. We denote the set of all such ζ by ∂Df(x) and refer to it as the D-subdifferential.
Exercise
Motivated by this, as well as by the duality between f◦(x;·) and∂f(x) observed in Chapter 2, we proceed to define a subdifferential corresponding to the subderivative Df(x;·). We saw in the previous theorem that the graph of the multifunction ∂Pf is dense in that of ∂Df. This fact can be used to quite easily derive the basic calculation of ∂Df by appealing to the known results for ∂Pf.
Exercise
On the contrary, notice that the equality of the three sets in question implies that Df(x;v) =f◦(x;v) for all v since. We can extract a subsequence (without relabeling) such that the vectors vi :=ui/ui converge to the limit v. We note that the set of subgradients defined by (5) is sometimes called the Fr´echet subdifinential ∂Ff(x), an object that can differ from.
Exercise
The proof of what follows will assume known the fact that a locally Lipschitz function of a single variable is differentiable almost everywhere in the sense of Lebesgue measure on the line. The left-hand side of (6) is always contained in the right-hand side, for any y ∈ Y, of Exercise 4.15(a), so it is sufficient to prove that the opposite inclusion holds for almost ally∈Y. In other words, Sis contained in the countable union of the sets C(vi, wi, ri, εi), and it is therefore sufficient to prove that every such set has measure 0.
Exercise
Then from the theory of integration on a straight line, we know that for almost all t g(t) exists and coincides with χC(t). To begin with, note that the gradient of f, ∇f(x), exists for almost all x when f is a Lipschitz function on Rn. Apply Corollary 4.17 to tof and −f; we derive that for almost all x both ∂Df(x) and ∂Df(x) are nonempty.
Exercise
Γ is measurable if Γ−1(V) is measurable for any closed setV (or compact setV, or (open or closed) ballV). Prove that when Γ has a closed value, Γ is measurable as the function u → d. assigning Rm to [0,∞] is measurable for eachx∈Rn. Prove that Γ is measurable if gr Γ is closed. f) A function γ:Rm → Rn is measurable as the multifunction Γ(u) :=. is measurable asγ:Rm →Ris semi-continuous bearing. h) If Γ1 and Γ2 are two measurable closed-valued multifunctions, then Γ(u) := Γ1(u)∩Γ2(u) defines another measurable closed-valued multifunction.
Exercise
So (5) can be appealed to yield b. 0 a.e. Since ˆϕ is measurable and continuous in y, it now follows from Corollary 5.5 that forta.e. a) Let ∂Pσϕ(x) denote the “global proximal subgradients of rank σ”; i.e. those that are satisfactory. To understand this property, we must consider the natural presumption that if ζ ∈∂Lf(x), then ζ(t) ∈∂Lϕ. We call this “natural” given that we established this fact when ∂L is replaced by ∂P.) It turns out that this conjecture is incorrect, and the following example will provide insight into why. The last part we need is the next version of a celebrated achievement from the nineteenth century.
Exercise
Furthermore, the property intTSC(x)=∅ will serve to identify a class of strings with important properties that play a role, for example, in equilibrium theory and in constructing feedback controls. The first of several immediate consequences of the theorem given below explains our use of the term "stuck" for the condition 0∈intTSC(x) and reveals that it is local and unitary in nature.
Exercise
Exercise
The sufficiency part of the proof is no more difficult than necessity; it constitutes one of the final problems of the chapter. If NSC(x) is pointed, then NSC(·) is graph-closed at x, TSC(·) is atx lower semicontinuous, and S is fixed at x. a) Kurn= 1,NSC(·) is always a closed graph at every point of S. b). Between any two points of the form 2−n−1 and 2−n, the graph of f describes an isosceles triangle whose vertex is located at.
Exercise
If each of the functions gi accepts directional derivatives at x0, then every vectorv∈TSB(x0) satisfies. Show that any absolutely continuous function x that satisfies the Euler inclusion and transversality condition of Theorem 5.22 is a solution of the variational problem when φandare is convex. It is known that there exists a subset S e R with the following property: for everya, b∈Rwitha < b, the LebesgueL measure of the set S∩[a, b] satisfies.
Exercise
Many of us have already seen methods of calculating solutions to ordinary differential equations; how to concretely study the calculation of differential inclusion trajectories (1). Thus, by the definition of x, the Euler solution of the initial value problem (4) is on [a, b] and assertion (a) of the theorem is proved. For every point that is not one of the finite number of points where xπj(t) is a node, we have ˙xπj(t) =f.
Exercise
We are certain from part (c) of the theorem that when f is continuous (which is the minimal assumption under which the classical study of differential equations operates), the Euler arcs satisfy the usual point-by-point definition of a solution. The following important consequence of the theorem justifies our attempt to compute trajectories of F by means of selections and proves that trajectories exist, given any x0 and [a, b]. Now it is a consequence of the theorem thatx, the uniform limit ixπj, is a trajectory of F.
Exercise
