Gateaux and Frechet derivatives of a functional

Let $X$ be a real Banach space, $U\subseteq X$ open and $F:U\to \mathbb{R}$. Frechet differentiability. $F$ is Frechet differentiable at $u\in U$ if there is a bounded linear functional $DF(u)\in X^{\ast }$ (Fréchet derivative between Banach spaces, A bounded linear operator between normed spaces, The dual space X^* of a normed space and its dual norm) with $F(u+h)=F(u)+DF(u)h+o(\mathrm{\parallel }h\mathrm{\parallel })(h\to 0),$ that is, ${\lim }_{h\to 0}\mathrm{\mid }F(u+h)-F(u)-DF(u)h\mathrm{\mid }\mathrm{/}\mathrm{\parallel }h\mathrm{\parallel }=0$; such $DF(u)$ is unique. Gateaux differentiability. $F$ is Gateaux differentiable at $u$ in the direction $v\in X$ if the limit $\delta F(u;v):={\lim }_{\epsilon \to 0,\epsilon \mathrm{\ne }0}\frac{F(u+\epsilon v)-F(u)}{\epsilon }\in \mathbb{R}$ exists; $F$ is Gateaux differentiable at $u$ if that limit exists for every $v\in X$ and the resulting map $v\mapsto \delta F(u;v)$ is a bounded linear functional, written $\delta F(u)\in X^{\ast }$ and called the Gateaux (variational) derivative of $F$ at $u$. Relation and caveat. Frechet differentiability at $u$ implies Gateaux differentiability at $u$ with $\delta F(u)=DF(u)$, because $F(u+\epsilon v)-F(u)=DF(u)(\epsilon v)+o(\mathrm{\mid }\epsilon \mathrm{\mid })$; the converse fails, and the directional limits $\delta F(u;v)$ need not be linear or bounded in $v$ when only the one-dimensional limits exist. For each fixed $v$ the function $\phi (\epsilon ):=F(u+\epsilon v)$ satisfies ${\phi }^{\mathrm{\prime }}(0)=\delta F(u;v)$.