22.4. ORTHONORMAL SETS 585

This equals ∥x∥2 +∑nk=1 |ck− (x,xk)|2−∑

nk=1 |(x,xk)|2and therefore, this minimum is ach-

ieved when ck = (x,xk) and equals ∥x∥2−∑nk=1 |(x,xk)|2. Now since span(S) is dense, if

n large enough then for some choice of constants ck, ∥x−∑nk=1 ckxk∥2 < ε. However, from

what was just shown, ∥∥∥∥∥x−n

∑i=1

(x,xi)xi

∥∥∥∥∥2

≤

∥∥∥∥∥x−n

∑k=1

ckxk

∥∥∥∥∥2

< ε

showing that limn→∞ ∑ni=1 (x,xi)xi = x as claimed.

For the last claim, from what was just shown and the observation that if xn → x andyn→ y, then (xn,yn)→ (x,y) ,

(x,y) = limn→∞

(n

∑i=1

(x,xi)xi,n

∑j=1

(y,x j)x j

)= lim

n→∞∑

i, j≤n(x,xi)(y,x j)(xi,x j)

= limn→∞

n

∑i=1

(x,xi)(y,xi) =∞

∑i=1

(x,xi)(y,yi) ■

The proof of this theorem contains the following corollary.

Corollary 22.4.3 Let S be any orthonormal set of vectors and let

{x1, · · · ,xn} ⊆ S.

Then if x ∈ H ∥∥∥∥∥x−n

∑k=1

ckxk

∥∥∥∥∥2

≥

∥∥∥∥∥x−n

∑i=1

(x,xi)xi

∥∥∥∥∥2

for all choices of constants, ck. In addition to this, Bessel’s inequality follows,

∥x∥2 ≥n

∑k=1|(x,xk)|2 .

If S is countable and span(S) is dense, then letting {xi}∞

i=1 = S, 22.14 follows.

Corollary 22.4.4 If V is a closed subspace of a Hilbert space H and if

V = span({uk}N

k=1

)where the uk are an orthonormal set and N ≤ ∞, then for P the projection map onto V, itfollows that Py = ∑

Nk=1 (y,uk)uk and when N = ∞ the series converges to Py. In particular,

when y ∈V,y = Py = ∑Nk=1 (y,uk)uk.

Proof: The case where N < ∞ was done above. For y ∈ H,(y−

∞

∑k=1

(y,uk)uk,um

)= (y,um)− (y,um) = 0

assuming the series makes sense. Therefore, if this happens, then Py=∑∞k=1 (y,uk)uk. Now

note that (y,uk) = (y−Py,uk)+(Py,uk) = (Py,uk). Now, by assumption, there are scalars