542 CHAPTER 21. BANACH SPACES

21.2.3 The Complex Version of the Hahn Banach TheoremFirst is a lemma which is quite interesting for its own sake.

Lemma 21.2.5 Let h : V → R where V is a complex normed linear space. Then h islinear with respect to real scalars if and only if F (x)≡ h(x)− ih(ix) is linear with respectto complex scalars.

Proof:⇐ By assumption, F is linear with respect to real scalars. Let c ∈ R. Then

cF (x) = ch(x)− cih(ix) = F (cx) = h(cx)− ih(cix) .

Equating real parts, it follows ch(x) = h(cx). Also

F (x+ y) = h(x+ y)− ih(i(x+ y)) = F (x)+F (y)

= h(x)− ih(ix)+h(y)− ih(iy)

Equating real parts, h(x+ y) = h(x)+h(y).⇒ I need to show that F is linear with respect to complex scalars.

F (ix) ≡ h(ix)− ih(−x) = h(ix)+ ih(x)

= i(h(x)− ih(ix)) = iF (ix)

It is fairly obvious that F (x+ y) = F (x)+F (y) . Also, if c is real, it is clear that F (cx) =cF (x). Therefore,

F ((a+ ib)x) = F (ax)+F (ibx)

= aF (x)+ ibF (x) = (a+ ib)F (x) ■

Corollary 21.2.6 (Hahn Banach) Let M be a subspace of a complex normed linearspace X, and suppose f : M→ C is linear and satisfies | f (x)| ≤ K ∥x∥ for all x ∈M. Thenthere exists a linear function F, defined on all of X such that F(x) = f (x) for all x ∈M and|F(x)| ≤ K ∥x∥ for all x ∈ X.

Proof: First note f (x) = Re f (x)+ i Im f (x) and so

Re f (ix)+ i Im f (ix) = f (ix) = i f (x) = iRe f (x)− Im f (x).

Therefore, Im f (x) =−Re f (ix), and

f (x) = Re f (x)− iRe f (ix).

This is important because it shows it is only necessary to consider Re f in understanding f .From Lemma 21.2.5 Re f is linear with respect to real scalars.

Consider X as a real vector space and let ρ(x)≡ K ∥x∥. Then for all x ∈M,

|Re f (x)| ≤ | f (x)| ≤ K ∥x∥ ≡ ρ(x).

From Theorem 21.2.4, Re f may be extended to a function h which satisfies

h(ax+by) = ah(x)+bh(y) if a,b ∈ Rh(x) ≤ K ∥x∥ for all x ∈ X .