Fréchet Differentiability and the Gradient on an Open Subset of a Real Inner Product Space

In the setting of Real Hilbert Spaces: Standing Notation and Background, let $E$ be a real inner product space, with its inner product $\langle\cdot,\cdot\rangle$, norm $|\cdot|$, distance $d$ and zero vector $0_{E}$ as fixed there, and let $U\subseteq E$ be open in the metric space $(E,d)$.

1. (Differentiability at a point)¶ Let $u:U\to\mathbb{R}$, let $x\in U$ and let $p\in E$. The function $u$ is differentiable at $x$ with gradient $p$ if for every positive $\varepsilon\in\mathbb{R}$ there is a positive $\delta\in\mathbb{R}$ such that every $z\in E$ with $|z|<\delta$ satisfies $x+z\in U$ and

The function $u$ is differentiable at $x$ if it is differentiable at $x$ with gradient $p$ for some $p\in E$.

2. (The gradient)¶ At most one $p\in E$ has the property of clause 1. Indeed, suppose $p$ and $p'$ both have it and let $\varepsilon\in\mathbb{R}$ be positive. Let $\delta$ and $\delta'$ be radii provided by clause 1 for $p$ and for $p'$ with $\tfrac{\varepsilon}{2}$, positive by claim 8 of Elementary Order Arithmetic in an Ordered Field, in place of $\varepsilon$, and let $\delta''$ be the lesser of $\delta$ and $\delta'$ (claim 9 of Elementary Order Arithmetic in an Ordered Field), which is positive because it is one of them. Every $z\in E$ with $|z|<\delta''$ then satisfies

the equality by Elementary Identities in a Real Inner Product Space §bilinear and the inequality by claims 2 and 5 of Properties of the Absolute Value in an Ordered Field. Since $\varepsilon$ was an arbitrary positive real number, Vanishing of Uniformly Small Linear, Quadratic and Bilinear Terms in a Real Inner Product Space §linear gives $p-p'=0_{E}$, whence $p=(p-p')+p'=0_{E}+p'=p'$ by the associativity, inverse and identity axioms of the vector space $E$. When $u$ is differentiable at $x$ we write $Du(x)$ for the unique $p$ with the property of clause 1 and call it the gradient of $u$ at $x$.

3. (Differentiability on an open set)¶ The function $u$ is differentiable on $U$ if it is differentiable at every point of $U$. Its gradient map is then the map $Du:U\to E$ sending $x\in U$ to $Du(x)$.