Theorem of projection on convex closed

In mathematics, the theorem of orthogonal projection on convex closed is a result of minimization of the distance which generalizes the orthogonal Projection on a vector space. It can be stated for a convex part closed of a Euclidean Espace, or more generally of a Espace hilbertien.

Statement of the theorem

Are $\ mathbb H$ a Espace of Hilbert on $\ mathbb R$ or $\ mathbb C$ and M a closed convex whole (not vacuum) of $\ mathbb H$ . If $\ quad v$ indicates an unspecified vector of $\ mathbb H$ , the problem $\ min \ nolimits_ {W \ in M} \|VW \|$ admits a single solution $\ quad w^*$ . One notes then

\ exists! \, w^* \ in M \ quad \|v-w^* \|= \ min_ {W \ in M} \|VW \|

This solution is characterized by the variational inequation :

\ forall W \ in M \ quad \ Re (w-w^*|v-w^*) \ the 0 \,

Moreover projection : $p_M: \ quad v \ longrightarrow w^*$ is 1-lipschitzienne and consequently uniformly continues .

Like shows it the following demonstration, it is enough to suppose that $\ mathbb H$ is a Espace préhilbertien and that M is a convex and complete part (not vacuum) of $\ mathbb H$ . Indeed only the complétude of M intervenes in the proof.

When $\ mathbb H$ is a space of Hilbert, very closed part is complete, which gives the result stated above.

This theorem is often stated as here within the framework of spaces of Hilbert, because it is the case in the majority of the practical situations.

Demonstration

first of all Let us show the equivalence of the initial definition with the variational inequation.

w^*

is a solution of the problem, let us suppose that the variational inequation is false: there thus exists

w_1 \ in M

such as

\ Re (w_1-w^*|v-w^*) >0

. M being convex, for all

\ lambda \ in

the vector

\ lambda w_1+ (1 - \ lambda) w^*

belongs to Mr. But then

\|v (\ lambda w_1+ (1 - \ lambda) w^*) \|^2= \|v-w^* \|^2+ \ lambda^2 \|w_1-w^* \|^2-2 \ lambda \ Re (v-w^*|w_1-w^*)

The coefficient of $\ lambda$ is strictly positive under the terms of our assumption. For $\ lambda$ sufficiently small not no one, the term in $\ lambda^2$ is strictly dominated by the term in $\ lambda$ and consequently the algebraic sum of the last two terms is strictly negative. Consequently in this case $\|v (\ lambda w_1+ (1 - \ lambda) w^*) \|^2< \|v-w^* \|^2$ , which is contradictory.

Reciprocally let us suppose that the variational inequation is true. Then for all $W \ in M \ quad \|VW \|^2= \|v-w^* \|^2 + \|w-w^* \|^2-2 \ Re (v-w^*|w-w^*) \ Ge \|v-w^* \|^2$ , which is the announced result.

the whole of realities $\ {\|VW \|\ mid W \ in M \}$ admits a lower limit D (it is undervalued by 0). There thus exists at least a continuation minimizing $(w_n) \ quad$ of elements of M such as $\ lim_ {N \ to \ infty} \|v-w_n \|=d$ . We will show that it is a Suite of Cauchy.

Are thus

w_n

and

w_m

two elements of the continuation. It results from the Théorème of the median that, if I indicates the medium of

\ quad

one has

\|w_n-w_m \|^2=2 (\|w_n-v \|^2+
\|w_m-v \|^2) - 4 \|VI \|^2

. Like

I \ in M

(convexity)

\|VI \|^2 \ Ge d^2

. In addition since

\|v-v_n \|

and

\|v-v_m \|

tends towards D , whatever the

\ epsilon >0

one can find NR such as ( N > NR and m > NR ) involves

2 (\|v-w_n \|^2+ \|v-w_m \|^2) <4d^2+ \ epsilon^2

. And thus

\|w_n-w_m \|^2< \ epsilon^2

, is

\|w_n-w_m \|< \ epsilon

. This shows that one has well a Suite of Cauchy. Maintaining as M is Fermé in

\ mathbb H

and thus Complet , the continuation

(w_n)

converges towards an element

w^*

of M . This shows the existence of a solution of the problem since by the preceding definition

\ quad w^*

checks

\|v-w^* \|= \ lim_ {N \ to \ infty} \|v-w_n \| = D

Prouvons that $w^*$ is well single solution initial problem. So indeed, $w^ {**}$ was a different solution, one would have $\|v-w^ {**} \|^2= \|v-w^* \|^2+ \|w^*-w^ {**} \|^2-2 \ Re (v-w^*|w^ {**} - w^*) >d^2$ since the second term is strictly positive and the positive last (because of the variational inequation checked by $w^ {**}$ ). This would contradict the fact that $w^ {**}$ is solution.
Is now two vectors $\ quad v_1$ and $\ quad v_2$ and is $w^*_1$ and $w^*_2$ their projections respective. As one can write

v_1-v_2=w^*_1-w^*_2+ (v_1-w^*_1) - (v_2-w^*_2)

there is immediately

\|v_1-v_2 \|^2= \|w^*_1-w^*_2 \|^2+ \|(v_1-w^*_1) - (v_2-w^*_2) \|^2+2 \ Re (w^*_1-w^*_2|(v_1-w^*_1) - (v_2-w^*_2))

But the last two terms of the second member are positive (for the last, that comes from the variational inequations implying:

\ Re (w^*_2-w^*_1|v_1-w^*_1) \ the 0

and

\ Re (w^*_1-w^*_2|v_2-w^*_2) \ the 0

).
Consequently

\|v_1-v_2 \|^2 \ Ge \|w^*_1-w^*_2 \|^2

, which proves well that projection is 1-lipschitzienne .

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