Kenneth Kuttler

22.6. A MEAN VALUE INEQUALITY 719

Theorem 22.6.4 Suppose f : U → Y where U is an open set in X , a normed linear space.Suppose that f is Gateaux differentiable on U and that the Gateaux derivative is continuouson an open set containing x. Then f is Frechet differentiable at x.

Proof: Denote by G(x) ∈L (X ,Y ) the Gateaux derivative. Thus

G(x)v≡ limλ→0

f(x+λv)− f(x)λ

It is desired to show that G(x) = Df(x). Since G is continuous, one can obtain

f(x+v)− f(x) =∫ 1

0G(x+ tv)vdt

where this is the ordinary Riemann integral.∥∥∥∥ f(x+v)− f(x)−G(x)v∥v∥

∥∥∥∥=∥∥∥∥∥∫ 1

0 G(x+ tv)vdt−G(x)v∥v∥

∥∥∥∥∥=

∥∥∥∥∥∫ 1

0 G(x+ tv)v−G(x)vdt∥v∥

∥∥∥∥∥≤ 1∥v∥

∫ 1

0∥G(x+ tv)−G(x)∥dt ∥v∥

which is small provided ∥v∥ is sufficiently small. Thus G(x) = Df(x) as hoped.Recall the following.

Lemma 22.6.5 Let ∥x∥= sup∥y∗∥X ′≤1 |⟨y∗,x⟩| .

Proof: Let f (kx) = k∥x∥ . Then

sup∥kx∥≤1

|⟨ f ,x⟩|= sup|k|≤1/∥x∥

|k|∥x∥= 1

Then by Hahn Banach theorem, there is y∗ ∈ X ′ which extends f and ∥y∗∥ ≤ 1. Then

∥x∥ ≥ sup∥z∗∥X ′≤1

|⟨z∗,x⟩| ≥ |⟨y∗,x⟩|= ∥x∥

One does not need continuity of G near x. It suffices to have continuity at x. Let y∗ ∈Y ′.Then by the mean value theorem,

⟨y∗, f(x+v)⟩−⟨y∗, f(x)⟩= ⟨y∗,G(x+ tv)v⟩ , t ∈ [0,1]

Then

1∥v∥∥f(x+v)− f(x)−G(x)v∥= 1

∥v∥sup∥y∗∥≤1

|⟨y∗, f(x+v)− f(x)−G(x)v⟩|

=1∥v∥

sup∥y∗∥≤1

|⟨y∗,G(x+ tv)v−G(x)v⟩| ≤ sup|t|≤1∥G(x+ tv)−G(x)∥L (X ,Y )

which converges to 0 as ∥v∥→ 0 thanks to continuity of G at x. This proves the following.

22.6. A MEAN VALUE INEQUALITY 719Theorem 22.6.4 Suppose f: U — Y where U is an open set in X, a normed linear space.Suppose that f is Gateaux differentiable on U and that the Gateaux derivative is continuouson an open set containing x. Then f is Frechet differentiable at x.Proof: Denote by G(x) € (X,Y) the Gateaux derivative. Thus_.. £(x-++Av) —£(x)a eeIt is desired to show that G(x) = Df (x). Since G is continuous, one can obtain1f(x+v) —f(x) = [ G(x-+tv) vat0where this is the ordinary Riemann integral.fo G(x+tv) vdt —G(x)vII[fete —te—ceellyJo. G(x tv) v—G (x) vdtlv1 1< 7p f IGx+1y) -G(x)||ar IvIv Jowhich is small provided ||v|| is sufficiently small. Thus G(x) = Df(x) as hoped. JRecall the following.Lemma 22.6.5 Let ||x|| = supjy+ <1 |(0".4)|-Proof: Let f (kx) = k||x||. Thensup |(f,x)[= sup [al |}x|| = 1|x|) <1 ikl <1/lThen by Hahn Banach theorem, there is y* € X’ which extends f and ||y*|| < 1. ThenIIx] > sup |{z",x)| 2 |0"2)|= lel] 0le lyrOne does not need continuity of G near x. It suffices to have continuity at x. Let y* € Y’.Then by the mean value theorem,(y",f(x+v)) — (y",£(x)) = (y", G(x+rv)v), ¢ € [0,1]Thenihal I|f(x+v) ~ f(x) ~~ G(x) v\|= — sup |(y*,f(x+v)—f(x) —G(x)v)|II¥ll yey1 ok= 7 sup |(y’,G(x+tv)v—G(x)y)| < sup ||G(x+1v) —G(x)|| yy)IVI yi I<which converges to 0 as ||v|| > 0 thanks to continuity of G at x. This proves the following.