528 CHAPTER 19. HILBERT SPACES

19.3 Approximations In Hilbert SpaceThe Gram Schmidt process applies in any Hilbert space.

Theorem 19.3.1 Let {x1, · · · ,xn} be a basis for M a subspace of H a Hilbert space. Thenthere exists an orthonormal basis for M, {u1, · · · ,un} which has the property that for eachk ≤ n, span(x1, · · · ,xk) = span(u1, · · · ,uk) . Also if {x1, · · · ,xn} ⊆ H, then

span(x1, · · · ,xn)

is a closed subspace.

Proof: Let {x1, · · · ,xn} be a basis for M. Let u1 ≡ x1/ |x1| . Thus for k = 1, span(u1) =span(x1) and {u1} is an orthonormal set. Now suppose for some k < n, u1, · · · , uk havebeen chosen such that (u j ·ul) = δ jl and span(x1, · · · ,xk) = span(u1, · · · ,uk). Then define

uk+1 ≡xk+1−∑

kj=1 (xk+1 ·u j)u j∣∣∣xk+1−∑kj=1 (xk+1 ·u j)u j

∣∣∣ , (19.3.16)

where the denominator is not equal to zero because the x j form a basis and so

xk+1 /∈ span(x1, · · · ,xk) = span(u1, · · · ,uk)

Thus by induction,

uk+1 ∈ span(u1, · · · ,uk,xk+1) = span(x1, · · · ,xk,xk+1) .

Also, xk+1 ∈ span(u1, · · · ,uk,uk+1) which is seen easily by solving 19.3.16 for xk+1 and itfollows

span(x1, · · · ,xk,xk+1) = span(u1, · · · ,uk,uk+1) .

If l ≤ k,

(uk+1 ·ul) = C

((xk+1 ·ul)−

k

∑j=1

(xk+1 ·u j)(u j ·ul)

)

= C

((xk+1 ·ul)−

k

∑j=1

(xk+1 ·u j)δ l j

)= C ((xk+1 ·ul)− (xk+1 ·ul)) = 0.

The vectors,{

u j}n

j=1 , generated in this way are therefore an orthonormal basis becauseeach vector has unit length.

Consider the second claim about finite dimensional subspaces. Without loss of gener-ality, assume {x1, · · · ,xn} is linearly independent. If it is not, delete vectors until a linearlyindependent set is obtained. Then by the first part, span(x1, · · · ,xn)= span(u1, · · · ,un)≡Mwhere the ui are an orthonormal set of vectors. Suppose {yk} ⊆ M and yk → y ∈ H. Isy ∈M? Let

yk ≡n

∑j=1

ckju j

528 CHAPTER 19. HILBERT SPACES19.3. Approximations In Hilbert SpaceThe Gram Schmidt process applies in any Hilbert space.Theorem 19.3.1 Let {x,-+- ,X)} be a basis for M a subspace of H a Hilbert space. Thenthere exists an orthonormal basis for M, {u,,-+- ,Un} which has the property that for eachk <n, span(x,,-++ ,xX¢) = span (uj,--- , ug). Also if {x1,+-+ ,xn} CH, thenspan (x1,--+ ,Xn)is a closed subspace.Proof: Let {x),--- ,x,} be a basis for M. Let uy =x, /|x;|. Thus for k = 1, span (wu) =span (x,) and {w,} is an orthonormal set. Now suppose for some k <n, u1, +++, uz havebeen chosen such that (u;-u;) = 6 jy and span (x1,--+ ,x,) = span (u,--- , uz). Then definekXe — Loy eg uy) uyWest = Zuei\ ket i) aj (19.3.16)Meet — Djat ep Uy) ujwhere the denominator is not equal to zero because the x; form a basis and soXe € span (x1,-+- ,X~) = span (uy,--- , uz)Thus by induction,Uk+1 € span (uy,° * Uk, Xk+1) = span (x1,°°° Xk X41) :Also, x441 € span (u1,+++ ,ux,Wx4+1) which is seen easily by solving 19.3.16 for x4, and itfollowsspan (x1,- “ Xk, X41) = span (uy, ute Uk, Ue+1) :If <k,k(Uppi-ul) = e(ima ul) hI Xk+1°Uj) ra)k= © (ona) uj) hI Xk41 Uj) sa)= C(ep1-u) — on ‘uj)) =0.The vectors, {uj} , generated in this way are therefore an orthonormal basis becauseeach vector has unit length.Consider the second claim about finite dimensional subspaces. Without loss of gener-ality, assume {x1,--- ,x,} is linearly independent. If it is not, delete vectors until a linearlyindependent set is obtained. Then by the first part, span (x1,--- ,X,) =span(u1,--+ ,Uun) =Mwhere the u; are an orthonormal set of vectors. Suppose {y,} C M and y, > y € H. Isy€M? Letn_ kVe = Ye ju;jal