42.3. SUBGRADIENTS 1391

This shows this set and the set of 42.3.29 are equal. It is also clear the set of 42.3.29 isclosed and convex. It only remains to show this set is nonempty.

Define f : Rx→ R by f (αx) = α ||x||2. Then the norm of f on Rx is ||x|| and f (x) =||x||2. By the Hahn Banach theorem, f has an extension to all of X x∗, and this extension isin the set of 42.3.29, showing this set is nonempty as required.

The next theorem shows this duality map is the subgradient of 12 ||x||

2.

Theorem 42.3.17 For X a real Banach space, let φ (x)≡ 12 ||x||

2. Then F (x) = δφ (x).

Proof: Let x∗ ∈ F (x). Then

⟨x∗,y− x⟩ = ⟨x∗,y⟩−⟨x∗,x⟩

≤ ||x|| ||y||− ||x||2 ≤ 12||y||2− 1

2||x||2.

This shows F (x)⊆ δφ (x).Now let x∗ ∈ δφ (x). Then for all t ∈ R,

⟨x∗, ty⟩= ⟨x∗,(ty+ x)− x⟩ ≤ 12

(||x+ ty||2−||x||2

). (42.3.30)

Now if t > 0, divide both sides by t. This yields

⟨x∗,y⟩ ≤ 12t

((∥x∥+ t ∥y∥)2−∥x∥2

)=

12t

(2t ||x|| ||y||+ t2 ||y||2

)Letting t→ 0,

⟨x∗,y⟩ ≤ ||x|| ||y|| . (42.3.31)

Next suppose t =−s, where s > 0 in 25.7.66. Then, since when you divide by a negative,you reverse the inequality, for s > 0

⟨x∗,y⟩ ≥ 12s

[||x||2−||x− sy||2

]≥

12s

[||x− sy||2−2 ||x− sy|| ||sy|| + ||sy||2−||x− sy||

]2. (42.3.32)

=12s

[−2 ||x− sy|| ||sy||+ ||sy||2

](42.3.33)

Taking a limit as s→ 0 yields⟨x∗,y⟩ ≥ −||x|| ||y||. (42.3.34)

It follows from 42.3.34 and 42.3.31 that

|⟨x∗,y⟩| ≤ ||x|| ||y||