De Vaucouleurs' law

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 closed convex $C\subset H$, there exists a unique point $y\in C$ for which $\Vert x-y\Vert$ 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

• Let us show the existence of y:

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

$\Vert y_n-y_m{\Vert }^2=\Vert y_n-x{\Vert }^2+\Vert y_m-x{\Vert }^2-2⟨y_n-x,y_m-x⟩$

and

$4\Vert \frac{y_n+y_m}{2}-x{\Vert }^2=\Vert y_n-x{\Vert }^2+\Vert y_m-x{\Vert }^2+2⟨y_n-x,y_m-x⟩$

We have therefore:

$\Vert y_n-y_m{\Vert }^2=2\Vert y_n-x{\Vert }^2+2\Vert y_m-x{\Vert }^2-4\Vert \frac{y_n+y_m}{2}-x{\Vert }^2$

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

$\Vert y_n-y_m{\Vert }^2\le 2({\delta }^2+\frac{1}{n})+2({\delta }^2+\frac{1}{m})-4{\delta }^2=2(\frac{1}{n}+\frac{1}{m})$

The last inequality proves that (y_n) 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 minimizer. Then:

$\Vert y_2-y_1{\Vert }^2=2\Vert y_1-x{\Vert }^2+2\Vert y_2-x{\Vert }^2-4\Vert \frac{y_1+y_2}{2}-x{\Vert }^2$

Since $\frac{y_1+y_2}{2}$ belongs to C, we have $\Vert \frac{y_1+y_2}{2}-x{\Vert }^2\ge {\delta }^2$ and therefore

$\Vert y_2-y_1{\Vert }^2\le 2{\delta }^2+2{\delta }^2-4{\delta }^2=0$

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 $⟨z-x,a⟩=0$ for all $a\in M$. $\Vert x-a{\Vert }^2=\Vert z-x{\Vert }^2+\Vert a-z{\Vert }^2+2⟨z-x,a-z⟩=\Vert z-x{\Vert }^2+\Vert a-z{\Vert }^2$ which proves that $z$ is a minimizer.

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

$\Vert (y+ta)-x{\Vert }^2-\Vert y-x{\Vert }^2=2t⟨y-x,a⟩+t^2\Vert a{\Vert }^2=2t⟨y-x,a⟩+O(t^2)$

is always non-negative. Therefore, $⟨y-x,a⟩=0.$

QED

References

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

See also

• Orthogonality principle