21.1. THEOREMS BASED ON BAIRE CATEGORY 537

To aid in the proof, here is a lemma.

Lemma 21.1.11 Let a and b be positive constants and suppose B(0,a) ⊆ L(B(0,b)).Then L(B(0,b))⊆ L(B(0,2b)).

Proof: Let z ∈ L(B(0,b)). Then let x1 ∈ B(0,b) with ∥z−Lx1∥ < a21 . Then it follows

that on multiplying by 2, ∥2z−2Lx1∥< a and so there is x2 ∈ B(0,b) with

∥(2z−2Lx1)−Lx2∥<a2

which implies∥∥∥z−

(Lx1 +

Lx22

)∥∥∥< a22 . If xi ∈ B(0,b) and

∥∥∥z−∑ni=0

Lxi2i

∥∥∥< a2n , then∥∥∥∥∥2n

(z−

n

∑i=0

Lxi

2i

)∥∥∥∥∥< a

Continuing this way, there is xn+1 ∈ B(0,b) such that∥∥∥∥∥2n

(z−

n

∑i=0

Lxi

2i

)−L(xn+1)

∥∥∥∥∥< a2

and so∥∥∥z−∑

n+1i=0

Lxi2i

∥∥∥ < a2n+1 . Let x ≡ ∑

∞i=0

xi2i . Then from the triangle inequality, ∥x∥ <

∑∞i=0

b2i = 2b and by continuity of L,

∥z−Lx∥= limn→∞

∥∥∥∥∥z−L

(n

∑i=0

xi

2i

)∥∥∥∥∥≤ limn→∞

a2n = 0 ■

Proof of Theorem 21.1.10: Y = ∪∞n=1L(B(0,n)). By Corollary 21.1.3, L(B(0,n0)) has

nonempty interior for some n0. Thus B(y,r)⊆ L(B(0,n0)) for some y and some r > 0. SinceL is linear B(−y,r)⊆ L(B(0,n0)) also. Here is why. If z∈B(−y,r), then−z∈B(y,r) and sothere exists xn ∈ B(0,n0) such that Lxn→−z. Therefore, L(−xn)→ z and −xn ∈ B(0,n0)also. Therefore z ∈ L(B(0,n0)). Then it follows that

B(0,r) ⊆ B(y,r)+B(−y,r)≡ {y1 + y2 : y1 ∈ B(y,r) and y2 ∈ B(−y,r)}⊆ L(B(0,2n0))

The reason for the last inclusion is that from the above, if y1 ∈ B(y,r) and y2 ∈ B(−y,r),there exists xn,zn ∈ B(0,n0) such that Lxn→ y1, Lzn→ y2. Therefore, ∥xn + zn∥ ≤ 2n0 andso (y1 + y2) ∈ L(B(0,2n0)).

By Lemma 21.1.11, L(B(0,2n0))⊆ L(B(0,4n0)) so B(0,r)⊆ L(B(0,4n0)). Letting a =r(4n0)

−1, it follows, since L is linear, that B(0,a)⊆ L(B(0,1)). It follows since L is linear,

L(B(0,r))⊇ B(0,ar). (21.2)

Now let U be open in X and let x+B(0,r) = B(x,r)⊆U . Using 21.2,

L(U)⊇ L(x+B(0,r))

= Lx+L(B(0,r))⊇ Lx+B(0,ar) = B(Lx,ar).