Kenneth Kuttler

548 CHAPTER 19. HILBERT SPACES

L(y(a) ,y′ (a)

)= 0,

L̂(y(b) ,y′ (b)

)= 0. (19.8.56)

andlimn→∞|λ n|= ∞ (19.8.57)

such that for all f ∈C ([a,b]), whenever w satisfies 19.8.54 and the boundary conditions,

w(x) =∞

∑n=1

1λ n

( f ,yn)yn. (19.8.58)

Also the functions, {yn} form a dense set in L2 (a,b,q) which satisfy the orthogonalitycondition, 19.8.44.

Proof: Let Ay(x) ≡∫ b

a G(t,x)q(t)y(t)dt where G is defined above in 19.8.53. Thenfrom symmetry and Fubini’s theorem,

(Ay,z)L2(a,b,q) =

∫ b

aG(t,x)y(t)z(x)q(x)q(t)dtdx =

∫ b

aG(x, t)y(x)z(t)q(t)q(x)dxdt

=∫ b

∫ b

aG(t,x)z(t)q(t)y(x)q(x)dxdt

= (Az,y)L2(a,b,q)

This shows that A is self adjoint. For y ∈ L2 (a,b,q) ,

Ay(x) =∫ x

(−c−1 (v(t)u(x))

)y(t)q(t)dt +

∫ b

(−c−1 (v(x)u(t))

)q(t)y(t)dt

If you have yn → y weakly in L2 (a,b,q) , then it is clear that Ayn (x)→ Ay(x) for each x,this from the above formula. Consider now ∥Ayn−Ay∥L2(a,b,q). Look at the first term in theabove. Is it true that the following converges to 0?∫ b

∣∣∣∣∫ x

(−c−1 (v(t)u(x))

)(yn (t)− y(t))q(t)dt

∣∣∣∣2 q(x)dx (**)

We know that the integrand converges to 0 for each x. Is there a dominating function? Ifso, then the dominated convergence theorem gives the result.∣∣∣∣∫ x

(−c−1 (v(t)u(x))

)q(t)(yn (t)− y(t))dt

∣∣∣∣≤ |u(x)|C (c)

∫ x

a|v(t)(yn (t)− y(t))|dt

≤ |u(x)|C∥v∥L2(a,b,q) ∥yn− y∥L2(a,b,q)

548 CHAPTER 19. HILBERT SPACESL (y (a) ’ ‘ (a = ’L(y(b),y(b)) = 0. (19.8.56)andlim |An| = 0 (19.8.57)such that for all f € C ([a,b]), whenever w satisfies 19.8.54 and the boundary conditions,- 1> 7, Peyn)on- (19.8.58)Also the functions, {y,} form a dense set in L* (a,b,q) which satisfy the orthogonalitycondition, 19.8.44.Proof: Let Ay (x) = f? G(t,x) q(t) y(t) dt where G is defined above in 19.8.53. Thenfrom symmetry and Fubini’s theorem,(AY, 2) 72(a,bq) =b pb b pb[ [ eexvOz@awa@aax = [ [Gly @)zWa(a(saratb rb= [| f Gi.x)zaWy wae) axar= (Az, y)2 (a,b,q)This shows that A is self adjoint. For y € L’ (a,b,q),Xx bays) = [ (-e vu) y@aWars [| (-e v@)u))) a yOarIf you have y, + y weakly in L? (a,b,q), then it is clear that Ay, (x) —> Ay (x) for each x,this from the above formula. Consider now ||Ay, — Ay||;2 (a,b,q)’ Look at the first term in theabove. Is it true that the following converges to 0?bIWe know that the integrand converges to 0 for each x. Is there a dominating function? Ifso, then the dominated convergence theorem gives the result.2q(x) dx (8)[ eu) On vO) alaexX(6 ula) a0 Onl) -yO)at< ulgic [WO Onl) —¥)\ae< |u(x)|C IIV Il 22(a,b,g) lyn —yllz2(a,b,)