Kenneth Kuttler

524 CHAPTER 19. HILBERT SPACES

Then from the above characterization,

||w||2 =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∈ker(L)︷︸︸︷w− v + v

∣∣∣∣∣∣∣∣∣∣∣∣∣∣2

= ∥w− v∥2 +∥v∥2

which shows that w= L−1 (y) if and only if w= v just described. From this characterization,it is clear that L−1 is linear. Then it is also obvious that

(y,z)L(U) =(L−1y,L−1z

also specifies an inner product. The algebraic axioms are all obvious because L−1 is linear.If (y,y)L(U) = 0, then

∣∣L−1y∣∣2U = 0 and so L−1y = 0 which requires y = L

(L−1y

)= 0.

With the above definition, here is the main result.

Theorem 19.2.3 Let U,H be Hilbert spaces and let L ∈L (U,H) . Then Definition 19.2.1makes L(U) into a Hilbert space. Also L : U → L(U) is continuous and L−1 : L(U)→Uis continuous. Also,

∥L∥L (U,H) ||Lx||L(U) ≥ ||Lx||H (19.2.11)

If U is separable, so is L(U). Also(L−1 (y) ,x

)= 0 for all x∈ ker(L) , and L−1 : L(U)→U

is linear. Also, in case that L is one to one, both L and L−1 preserve norms.

Proof: First consider the claim that L : U → L(U) is continuous and L−1 : L(U)→Uis also continuous. Why is L continuous? Say un→ 0 in U. Then

∥Lun∥L(U) ≡∥∥L−1 (L(un))

∥∥U .

Now∥∥L−1 (L(un))

∥∥U ≤ ∥un∥U and so it converges to 0. (Recall that L−1 (Lun) is the

smallest vector in U which maps to Lun. Since un is mapped by L to Lun, it follows that∥∥L−1 (L(un))∥∥

U ≤ ∥un∥U .) Hence L is continuous.Next, why is L−1 continuous? Let ∥yn∥L(U) → 0. This requires

∥∥L−1 (yn)∥∥

U → 0 bydefinition of the norm in L(U). Thus L−1 is continuous.

Why is L(U) a Hilbert space? Let {yn} be a Cauchy sequence in L(U) . Then from whatwas just observed, it follows that L−1 (yn) is a Cauchy sequence in U. Hence L−1 (yn)→x ∈U. It follows that yn = L

(L−1 (yn)

)→ Lx in L(U). This is in the norm of L(U). It

was just shown that L is continuous as a map from U to L(U). This shows that L(U) isa Hilbert space. It was already shown that it is an inner product space and this has shownthat it is complete.

If x ∈U, then ∥Lx∥H ≤ ∥L∥L (U,H) ∥x∥U . It follows that

∥L(x)∥H =∥∥L(L−1 (L(x))

)∥∥H ≤ ∥L∥L (U,H)

∥∥L−1 (L(x))∥∥

= ∥L∥L (U,H) ∥L(x)∥L(U) .

This verifies 19.2.11.

524 CHAPTER 19. HILBERT SPACESThen from the above characterization,€ker(L) 2—[wll = }]w—P +v]] = [lw — vl? + [lvlwhich shows that w = L~! (y) if and only if w =v just described. From this characterization,it is clear that L~! is linear. Then it is also obvious that(Y.2)n0) = (L'y,L'z)yalso specifies an inner product. The algebraic axioms are all obvious because L~! is linear.If (%Y)LW) = 0, then JLy|e, = 0 and so L~!y = 0 which requires y = L (L~ly) =0. IWith the above definition, here is the main result.Theorem 19.2.3 Let U,H be Hilbert spaces and let L€ &(U,H). Then Definition 19.2.1makes L(U) into a Hilbert space. Also L: U — L(U) is continuous and L~! : L(U) ++ Uis continuous. Also,ILI guy Ello) 2 lalla (19.2.11)IfU is separable, so is L(U). Also (L~' (y) x) =0 for all x € ker (L), and L~!: L(U) +Uis linear. Also, in case that L is one to one, both L and L~! preserve norms.Proof: First consider the claim that L: U + L(U) is continuous and L~!: L(U) > Uis also continuous. Why is L continuous? Say u, — 0 in U. ThenLinlli(uy = E* (LZ (un) Ip -Now ||L7! (L(un))||y < llunlly and so it converges to 0. (Recall that L~' (Lun) is thesmallest vector in U which maps to Lu,. Since u, is mapped by L to Lu,, it follows that|L~! (L(un))||_y < \lually -) Hence L is continuous.Next, why is L~! continuous? Let ||yn||;(y) > 0. This requires ||L~! (yn)||,, > 0 bydefinition of the norm in L(U). Thus L~! is continuous.Why is L(U) a Hilbert space? Let {y,} be a Cauchy sequence in L(U) . Then from whatwas just observed, it follows that L~! (y,,) is a Cauchy sequence in U. Hence L~! (y,) >x €U. It follows that y, = L(L~! (yn)) + Lx in L(U). This is in the norm of L(U). Itwas just shown that L is continuous as a map from U to L(U). This shows that L(U) isa Hilbert space. It was already shown that it is an inner product space and this has shownthat it is complete.Ifx €U, then ||Lx||_7 < ||LI] geu,zy Ila lly - It follows thatIL) [la E(L* (L0))) ly S Ml. goan IE EOIll gu IE) Ii) -This verifies 19.2.11.