Finally, we prove two consequences of the Brouwer Fixed Point Theorem that we shall use in the Sections5.4and 6.4in our approximation of reachable sets with cones of needle variations and in establishing the transversality conditions.
C.3 Lemma: (A property of certain maps from compact convex sets)Let K ⊂Rn be com-pact and convex with int(K) 6= ∅, and let f:K → Rn be continuous. If x0 ∈ int(K) has the property that
kf(x)−xk<kx−x0k for each x∈bd(K), then x0∈image(f).
Proof: Without loss of generality suppose thatx0 = 0. Define a mapφ:K→Dnas follows.
For x ∈ K let λx ∈ R>0 have the property that λxx ∈ bd(K). Since K is compact such a λx always exists and since K is convex with 0 ∈ int(K) it follows that λx is uniquely defined (cf. Lemma 1 in the proof of PropositionB.11). We then define φ(x) = λxxkxk. We leave it to the reader to verify thatφis a homeomorphism.
Now suppose that f(x)6= 0 for everyx ∈K. Defineg:Dn→ Rn by g(z) =f◦φ−1(z).
Forx∈bd(K) we have
kf(x)−xk<kxk,
=⇒ 0≤ kf(x)k<2hf(x), xi,
=⇒ hf(x), xi>0.
This means that f(x) has a strictly positive component in the direction of x for each x∈bd(K). This then implies that g(z) has a strictly positive component in the direction of zfor each z∈bd(Dn). We then define h:Dn→Rnby
h(z) =z−g(z) 1− hz, zi 1− hg(z), zi,
and we verify, just as in our proof of the Brouwer Fixed Point Theorem, that h(z) points strictly outwards on bd(Dn). However, as we saw in the proof of the Brouwer Fixed Point Theorem, this leads to a contradiction of the Hairy Ball Theorem. Therefore, we must have
f(x) = 0 for some x∈K.
In Figure C.1 we depict the idea behind the lemma. The gist of the matter is that if the boundary does not deform too much under the continuous map f–specifically, it is deformed sufficiently little that the region containing the image of the boundary does not contain x0–then the image covers x0.
The next result deals with the intersection of transverse planes under the image of a continuous map.
C.4 Lemma: (Intersections of continuous images of transverse planes)Let n, k∈Z>0 with k < n. Define
Cn={(x1, . . . , xn)| max{|x1|, . . . ,|xn|} ≤1}, P1={(x1, . . . , xn)∈Cn| xk+1 =· · ·=xn= 0}, P2={(x1, . . . , xn)∈Cn| x1 =· · ·=xk = 0}.
The Maximum Principle in control and in optimal control 139
x0
Figure C.1. The idea behind LemmaC.3. The solid circle repre-sents the boundary of C and the shaded region is where the boundary gets mapped to.
Suppose that fa:Pa→Rn, a∈ {1,2} are continuous maps such that kfa(xa)−xak< 14, xa∈Pa, a∈ {1,2}.
Then f1(P1)∩f2(P2)6=∅.
Proof: Denote a point in Cn by (x1, x2) ∈ Rk×Rn−k and define a map g:Cn → Rn by g(x1, x2) =f1(x1)−f2(−x2). Then, for every (x1, x2)∈Cn, we have
kg(x1, x2)−(x1, x2)k=k(f1(x1)−x1)−(f(−x2)−(−x2))k
≤ kf1(x1)−x1k+kf(−x2)−(−x2)k< 12. This implies that for every (x1, x2)∈bd(Cn) we have
kg(x1, x2)−(x1, x2)k<k(x1, x2)−(0,0)k,
and so by LemmaC.3we have (0,0)∈image(g). Thus there exists (x1, x2)∈Cn such that
f(x1) =f(−x2), and the lemma thus follows.
In Figure C.2 we depict the idea behind Lemma C.4. The idea is that, provided the planes P1 and P2 do not get deformed too much by f1 and f2, respectively, then they will intersect after the maps are applied provided they intersect before in a sufficiently “robust”
manner.
Figure C.2. The idea behind Lemma C.4. The vertical line rep-resents P1, the horizontal line represents P2, and the shaded regions represent where these sets get mapped to.
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