# Hilbert space &amp; orthogonal projection

1. Feb 2, 2008

### quasar987

[SOLVED] Hilbert space &amp; 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 $\geq$ 0 such that |a(x,y)| $\leq$ M||x|| ||y||
(iii) There exists B>0 such that a(x,x) $\geq$ B||x||^2

(a) Show that the exists a continuous linear operator A-->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 $T(x)=P_C(rz-rAx+x)$ is a contraction, i.e., ||T(x) - T(y)|| $\leq$ 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 $d(z,P_C(z)) = d(z,C)$.

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, $$||P_C(x) - P_C(y)|| \leq ||x - y||$$.

So $$||T(x) - T(y)|| \leq ||x-y-rAx+rAy||$$

And here I've tried using every inequality I know but with no luck. Clearly, we need something stronger than the triangle inequality, because

$$||x-y-rAx+rAy||\leq ||x-y|| + r||A(x-y)||\leq (1+r||A||)||x-y||$$,

which is stricly greater than ||x - y||...

Last edited: Feb 2, 2008
2. Feb 2, 2008

### EnumaElish

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.

3. Feb 2, 2008

### quasar987

They are not necessarily the same constant! I apologize for the confusion and I have edited the OP. Thanks for pointing it out.