12.1. THEOREMS BASED ON BAIRE CATEGORY 301

≤ limn→∞

n

∑i=1

2−(i−1) ∥xi∥< limn→∞

n

∑i=1

2−(i−1)b = 2b.

This proves the lemma. ■Proof of Theorem 12.1.8: Y = ∪∞

n=1L(B(0,n)). Thus Y is a countable union of closedsets. By Corollary 12.1.6, 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. Since L is linear B(−y,r) ⊆ L(B(0,n0)) also. Here is why.If z ∈ B(−y,r), then −z ∈ B(y,r) and so there 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 followsthat

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 and so (y1 + y2) ∈ L(B(0,2n0)).By Lemma 12.1.9, L(B(0,2n0))⊆ L(B(0,4n0)) which shows

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). (12.2)

Now let U be open in X and let x+B(0,r) = B(x,r)⊆U . Using 12.2,

L(U)⊇ L(x+B(0,r))

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

Hence Lx ∈ B(Lx,ar)⊆ L(U) which shows that every point, Lx ∈ LU , is an interior pointof LU and so LU is open. ■

This theorem is surprising because it implies that if |·| and ∥·∥ are two norms withrespect to which a vector space X is a Banach space such that |·| ≤ K ∥·∥, then there existsa constant k, such that ∥·∥ ≤ k |·| . This can be useful because sometimes it is not clear howto compute k when all that is needed is its existence. To see the open mapping theoremimplies this, consider the identity map idx = x. Then id : (X ,∥·∥)→ (X , |·|) is continuousand onto. Hence id is an open map which implies id−1 is continuous. Theorem 2.9.1 givesthe existence of the constant k.