|Feb2-08, 09:56 AM||#1|
[SOLVED] Hilbert space & orthogonal projection
1. The problem statement, all variables and given/known data
Let H be a real Hilbert space, C a closed convex non void subset of H, and a: H x H-->R be a continuous coercive bilinear form (i.e.
(i) a is linear in both arguments
(ii) There exists M [itex]\geq[/itex] 0 such that |a(x,y)| [itex]\leq[/itex] M||x|| ||y||
(iii) There exists B>0 such that a(x,x) [itex]\geq[/itex] B||x||^2
(a) Show that the exists a continuous linear operator A:H-->H such that a(x,y)=(Ax,y) for all x,y in H.
(b) Let z be in H. Show that the exists a constant r>0 such that the operator T:C-->C defined by [itex]T(x)=P_C(rz-rAx+x)[/itex] is a contraction, i.e., ||T(x) - T(y)|| [itex]\leq[/itex] k||x - y|| for some constant k < 1. P_C is the operator "orthogonal projection on C", i.e. is the (unique) element of C such that [itex]d(z,P_C(z)) = d(z,C)[/itex].
3. The attempt at a solution
(a) is done. The operator in question is obtained via Riesz's representation theorem: we choose Ax to be the unique element of H such that a(x,y) = (Ax,y) for all y. Linearity and continuity of A follow from the corresponding properties of a.
(b) I already know that the orthogonal projection map is non expensive: for all x,y in H, [tex]||P_C(x) - P_C(y)|| \leq ||x - y||[/tex].
So [tex]||T(x) - T(y)|| \leq ||x-y-rAx+rAy||[/tex]
And here I've tried using every inequality I know but with no luck. Clearly, we need something stronger than the triangle inequality, because
[tex]||x-y-rAx+rAy||\leq ||x-y|| + r||A(x-y)||\leq (1+r||A||)||x-y||[/tex],
which is stricly greater than ||x - y||...
|Feb2-08, 10:18 PM||#2|
Do you know whether the r used in the definition of T is the same r used in the defintion of a contraction mapping? E.g., I could have written ||T(x) - T(y)|| < k||x - y|| for some constant k < 1.
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