Kenneth Kuttler

19.2. THE HILBERT SPACE L(U) 525

If U is separable, then letting D be a countable dense subset, it follows from the con-tinuity of the operators L,L−1 discussed above that L(D) is separable in. To see this, notethat

∥Lxn−Lx∥L(U) =∥∥L(L−1 (Lxn−Lx)

)∥∥≤ ∥L∥L (U,H)

∥∥L−1 (L(xn− x))∥∥

≤ ∥L∥L (U,H) ∥xn− x∥U

As before, L−1 (L(xn− x)) is the smallest vector which maps onto L(xn− x) and so itsnorm is no larger than ∥xn− x∥U .

Consider the last claim. If L is one to one, then for y ∈ L(U) , there is only one vectorwhich maps to y. Therefore,

L−1 (L(x)) = x.

Hence for y ∈ L(U) ,∥y∥L(U) ≡

∥∥L−1 (y)∥∥

Also,∥Lu∥L(U) ≡

∥∥L−1 (L(u))∥∥

U ≡ ∥u∥UNow here is another argument for various continuity claims.

∥Lx∥L(U) ≡∥∥L−1 (Lx)

∥∥U ≤ ∥x∥U

because L−1 (Lx) is the smallest thing in U which maps to Lx and x is something whichmaps to Lx so it follows that the inequality holds. Hence L ∈ L (U,L(U)) and in fact,∥L∥L (U,L(U)) = 1. Next, letting y ∈ L(U) ,∥∥L−1y

∥∥U ≡ ∥y∥L(U)

and so∥∥L−1

∥∥L (L(U),U)

= 1 and this shows that L∈L (U,L(U)) while L−1 ∈L (L(U) ,U)

and both have norm equal to 1.Now

∥Lx∥H =∥∥L(L−1 (Lx)

)∥∥H ≤ ∥L∥L (U,H)

∥∥L−1 (Lx)∥∥

U ≡ ∥L∥L (U,H) ∥Lx∥L(U)

Now here are some other very interesting results. I am following [108].

Lemma 19.2.4 Let L ∈L (U,H) . Then L(

B(0,r))

is closed and convex.

Proof: It is clear this is convex since L is linear. Why is it closed? B(0,r) is compact inthe weak topology by the Banach Alaoglu theorem, Theorem 17.5.4 on Page 461. Further-more, L is continuous with respect to the weak topologies on U and H. Here is why this isso. Suppose un→ u weakly in U. Then if h ∈ H,

(Lun,h) = (un,L∗h)→ (u,L∗h) = (Lu,h)

19.2. THE HILBERT SPACE L(U) 525If U is separable, then letting D be a countable dense subset, it follows from the con-tinuity of the operators L,L~! discussed above that L(D) is separable in. To see this, notethat[Ln —Lxl| yy = ||L(L* (Lt — Lx) |LI (un) IIE (L (xn —x)) llu<< Ll gH) Ilxn — llyAs before, L~! (L(x, —x)) is the smallest vector which maps onto L(x, —x) and so itsnorm is no larger than |x, —x||yy.Consider the last claim. If L is one to one, then for y € L(U), there is only one vectorwhich maps to y. Therefore,L“'(L(x)) =x.Hence for y€ L(U),—|J7-1IIy lew) = IIL (lyAlso,Lelie) = WE" Ly = lelNow here is another argument for various continuity claims.= |\7-1IZx\|;(u) = |Z (Lx)||,, < |lallybecause L~! (Lx) is the smallest thing in U which maps to Lx and x is something whichmaps to Lx so it follows that the inequality holds. Hence L © &(U,L(U)) and in fact,ILIl gu.w)) = 1. Next, letting ye L(U) 5eyo = Ibvllewyand so oa ll eau) = | and this shows that L € Y (U,L(U)) while L~! € & (L(U) ,U)and both have norm equal to 1.NowLl) = EL Le) Il S Mell. gom le" lo = Wellgo lili) 0Now here are some other very interesting results. I am following [108].Lemma 19.2.4 Let L€ &(U,H). Then L (3 (0, r)) is closed and convex.Proof: It is clear this is convex since L is linear. Why is it closed? B(0,r) is compact inthe weak topology by the Banach Alaoglu theorem, Theorem 17.5.4 on Page 461. Further-more, L is continuous with respect to the weak topologies on U and H. Here is why this isso. Suppose uy, — u weakly in U. Then if h € H,(Lun, h) = (un, L*h) > (u,L*h) = (Lu,h)