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 $\lVert x-y\rVert$ 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

• 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:

$\|y_{n}-y_{m}\|^{2}=\|y_{n}-x\|^{2}+\|y_{m}-x\|^{2}-2\langle y_{n}-x\,,\,y_{m}-x\rangle$

and

$4\|{\frac {y_{n}+y_{m}}{2}}-x\|^{2}=\|y_{n}-x\|^{2}+\|y_{m}-x\|^{2}+2\langle y_{n}-x\,,\,y_{m}-x\rangle$

We have therefore:

$\|y_{n}-y_{m}\|^{2}=2\|y_{n}-x\|^{2}+2\|y_{m}-x\|^{2}-4\|{\frac {y_{n}+y_{m}}{2}}-x\|^{2}$

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 :

$\|y_{n}-y_{m}\|^{2}\;\leq \;2\left(\delta ^{2}+{\frac {1}{n}}\right)+2\left(\delta ^{2}+{\frac {1}{m}}\right)-4\delta ^{2}=2\left({\frac {1}{n}}+{\frac {1}{m}}\right)$

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:

$\|y_{2}-y_{1}\|^{2}=2\|y_{1}-x\|^{2}+2\|y_{2}-x\|^{2}-4\|{\frac {y_{1}+y_{2}}{2}}-x\|^{2}$

Since ${\frac {y_{1}+y_{2}}{2}}$ belongs to C, we have $\|{\frac {y_{1}+y_{2}}{2}}-x\|^{2}\geq \delta ^{2}$ 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 $t\in \mathbb {R}$.

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

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

QED

References

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

See also

• Orthogonality principle