22.2. THE HILBERT SPACE L(U) 581

for u ∈ My and L−1 (y) is therefore characterized by(L−1 (y) ,v

)U = 0 for all v ∈ ker(L)

because ker(L) is a subspace. Also, L(L−1 (y)

)= y. Now from this characterization of

L−1, it is obvious that L−1 is linear. The inner product is well defined because L−1 (y) isuniquely determined. Does it satisfy the axioms? Say 0 = (y,y)L(U) . Then L−1 (y) = 0

and so doing L to both sides, y = 0. It is clear that (y,z)L(U) = (z,y)L(U) because these aredefined in terms of a given inner product on U .

(ay+bŷ,z)L(U) ≡(L−1 (ay+bŷ) ,L−1z

)U

=(aL−1 (y)+bL−1 (ŷ) ,L−1z

)U

= a(L−1 (y) ,L−1z

)U +b

(L−1 (ŷ) ,L−1z

)U

= a(y,z)L(U)+b(ŷ,z)L(U)

Thus this is an inner product as claimed. ■With the above definition, here is the main result.

Theorem 22.2.3 Let U,H be Hilbert spaces and let L ∈L (U,H) . Then Definition22.2.1 makes L(U) into a Hilbert space. Also L : U → L(U) is continuous and L−1 :L(U)→U is continuous. Also,

∥L∥L (U,H) ∥Lx∥L(U) ≥ ∥Lx∥H (22.10)

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.

∥Lu∥2L(U) =

(L−1 (Lu) ,L−1 (Lu)

)U ≤ ∥u∥

2U

(Recall that L−1 (Lu) is the smallest vector in U which maps to Lu. Since u is mapped by Lto Lu, it follows that

∥∥L−1 (L(u))∥∥

U ≤ ∥u∥U .) Hence L is continuous.

Next, why is L−1 continuous? By definition of the norm,∥∥L−1 (y)

∥∥2U ≡ ∥y∥

2L(U). Thus

L−1 is continuous and∥∥L−1

∥∥L (L(U),U)

= 1.Why is L(U) a Hilbert space? Let {yn} be a Cauchy sequence in L(U) .

∥yn− ym∥2L(U) ≡

∥∥L−1 (yn− ym)∥∥2

U

Then from what was just observed, it follows that L−1 (yn) is a Cauchy sequence in U.Hence L−1 (yn)→ x ∈U. Then by continuity of L just shown, yn → Lx. This shows thatL(U) is a Hilbert space. It was already shown that it is an inner product space and this hasshown that 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))∥∥

U

= ∥L∥L (U,H) ∥L(x)∥L(U) .

This verifies 22.10.

22.2. THE HILBERT SPACE L(U) 581for u € My and L~! (y) is therefore characterized by (L~! (y),v),, = 0 for all v € ker(L)because ker (L) is a subspace. Also, L(L~!(y)) = y. Now from this characterization ofL~!, it is obvious that L~! is linear. The inner product is well defined because L~! (y) isuniquely determined. Does it satisfy the axioms? Say 0 = (y,y);(y). Then L(y) =0and so doing L to both sides, y = 0. It is clear that (y,z);(y) = (Z,Y),(y) because these aredefined in terms of a given inner product on U.(ay+b§,z);y) = (L' (ay +b9),L'2)y= (ab '(y) +bL"'(§),L'z)ya(L""(y),Lo'z)y+b(L1' (9), Lb '2)yaly,z)yu) +992.)Thus this is an inner product as claimed.With the above definition, here is the main result.Theorem 22.2.3 Ler U,H be Hilbert spaces and let L€ &(U,H). Then Definition22.2.1 makes L(U) into a Hilbert space. Also L: U + L(U) is continuous and L~! :L(U) > U is continuous. Also,VEll. cow WEAllnwy 2 lla (22.10)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.2 - - 2I|Lu\l7(u) = (L (Lu) ,L (Lu)) y $< |lullo(Recall that L~! (Lu) is the smallest vector in U which maps to Lu. Since u is mapped by Lto Lu, it follows that ||L~! (L(u))||,, < ||u||y .) Hence L is continuous.Next, why is L~! continuous? By definition of the norm, ||L~! (y) Ike = lye): ThusL~! is continuous and Egy v= 1.Why is L(U) a Hilbert space? Let {y,} be a Cauchy sequence in L(U).lyn —ymll2vy = JE" On —ym) IleThen from what was just observed, it follows that L~!(y,) is a Cauchy sequence in U.Hence L~! (y,) + x € U. Then by continuity of L just shown, y, — Lx. This shows thatL(U) is a Hilbert space. It was already shown that it is an inner product space and this hasshown that it is complete.Ifx €U, then ||Lx||_7 < ||LI]_geu,zy Ila lly - It follows thatIL) [ln IE(L* L0))) ly S Ml. gan IE EL OoIZ gu. IE @) In) -This verifies 22.10.