Hilbert projection theorem

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In mathematics, the Hilbert projection theorem is a famous result of convex analysis that says that for every point x in a Hilbert space H and every nonempty closed convex C \subset H, there exists a unique point y \in C for which \lVert x - y \rVert is minimized over C.

This is, in particular, true for any closed subspace M of H. In that case, a necessary and sufficient condition for y is that the vector  x-y be orthogonal to M.

Proof[edit]

  • Let us show the existence of y:

Let δ be the distance between x and C, (yn) a sequence in C such that the distance squared between x and yn is below or equal to δ2 + 1/n. Let n and m be two integers, then the following equalities are true:

\| y_n - y_m \|^2 = \|y_n -x\|^2 + \|y_m -x\|^2 - 2 \langle y_n - x \, , \, y_m - x\rangle

and

4 \left\| \frac{y_n + y_m}2 -x \right\|^2 = \|y_n -x\|^2 + \|y_m -x\|^2 + 2 \langle y_n - x \, , \, y_m - x\rangle

We have therefore:

\| y_n - y_m \|^2 = 2\|y_n -x\|^2 + 2\|y_m -x\|^2 - 4\left\| \frac{y_n + y_m}2 -x \right\|^2

By giving an upper bound to the first two terms of the equality and by noticing that the middle of yn and ym belong to C and has therefore a distance greater than or equal to δ from x, one gets :

\| y_n - y_m \|^2 \; \le \; 2\left(\delta^2 + \frac 1n\right) + 2\left(\delta^2 + \frac 1m\right) - 4\delta^2=2\left( \frac 1n + \frac 1m\right)

The last inequality proves that (yn) is a Cauchy sequence. Since C is complete, the sequence is therefore convergent to a point y in C, whose distance from x is minimal.

  • Let us show the uniqueness of y :

Let y1 and y2 be two minimizers. Then:

\| y_2 - y_1 \|^2 = 2\|y_1 -x\|^2 + 2\|y_2 -x\|^2 - 4\left\| \frac{y_1 + y_2}2 -x \right\|^2

Since \frac{y_1 + y_2}2 belongs to C, we have \left\| \frac{y_1 + y_2}2 -x \right\|^2\geq \delta^2 and therefore

\| y_2 - y_1 \|^2 \leq 2\delta^2 + 2\delta^2 - 4\delta^2=0 \,

Hence y_1=y_2, which proves unicity.

  • Let us show the equivalent condition on y when C = M is a closed subspace.

The condition is sufficient: Let z\in M such that \langle z-x, a \rangle=0 for all a\in M. \|x-a\|^2=\|z-x\|^2+\|a-z\|^2+2\langle z-x, a-z \rangle=\|z-x\|^2+\|a-z\|^2 which proves that z is a minimizer.

The condition is necessary: Let y\in M be the minimizer. Let a\in M and t\in\mathbb R.

\|(y+t a)-x\|^2-\|y-x\|^2=2t\langle y-x,a\rangle+t^2 \|a\|^2=2t\langle y-x,a\rangle+O(t^2)

is always non-negative. Therefore, \langle y-x,a\rangle=0.

QED

References[edit]

  • Walter Rudin, Real and Complex Analysis. Third Edition, 1987.

See also[edit]