How to prove that Frechet derivative exists and coincides with the Gateau derivative?

Question

Let $X$ and $Y$ be Banach spaces and $U \subseteq X$ be open.Let $F:U \to Y$ be Gateaux differentiable and let the mapping $x \to F′(x)$ be continuous from $U \in L(X,Y)$. How can I prove that the Frechet derivative exists at the point $x \in U $and coincides with the Gateaux derivative?

Answer 1

You can use the weak mean-value theorem: $$ \|F(y)-F(x)\| \le \sup_{t\in(0,1)} \|F'(x+t(y-x)\| \cdot \|x-y\| $$ and apply it to the mapping $y\mapsto F(y)-F'(x)y$. Then $$ \|F(y)-F'(x)y - (F(x)-F'(x)x)\| = \|F(y)-F(x)-F'(x)(y-x)\|\\ \le \sup_{t\in(0,1)} \|F'(x+t(y-x)-F'(x)\| \cdot \|x-y\|. $$ Due to the continuity of $F'$, the right-hand side is $o(\|y-x\|)$.