Proof of At a Point of Differentiability, the Subdifferential of a Convex Function on an Open Convex Subset of $\mathbb{R}^n$ Is the Gradient Alone

By Differentiability at a Point for Functions Between Euclidean Spaces (iii), $f$ is differentiable at $x$ with derivative vector $w$ for some $w\in\mathbb{R}^{n}$, and $w=\nabla f(x)$ by The Derivative Is the Jacobian Matrix, and Is Unique (i). Thus for every real $\varepsilon>0$ there is a real $\delta>0$ such that every $h\in\mathbb{R}^{n}$ with $\lVert h\rVert<\delta$ satisfies $x+h\in U$ and

$$\bigl|f(x+h)-f(x)-w\cdot h\bigr|\le\varepsilon\lVert h\rVert. \tag{1}$$

We first show that $w\in\partial f(x)$. Let $y\in U$ with $y\neq x$ and put $h=y-x$. Let $\varepsilon>0$ be given, and choose $\delta$ for this $\varepsilon$ as in (1); then put $s=\min\{1,\delta/(2\lVert h\rVert)\}$, so that $0<s\le1$ and $\lVert sh\rVert<\delta$ (Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (v)). Since $x+sh=(1-s)x+sy$, the convexity inequality of Convex Function on a Convex Subset of $\mathbb{R}^n$ (i) and (1) for $sh$ give

$$s\,w\cdot h-\varepsilon s\lVert h\rVert\le f(x+sh)-f(x)\le s\bigl(f(y)-f(x)\bigr).$$

Dividing by $s>0$ yields $w\cdot h-\varepsilon\lVert h\rVert\le f(y)-f(x)$, and since $\varepsilon>0$ was arbitrary,

$$f(y)\ge f(x)+w\cdot(y-x).$$

For $y=x$ this holds trivially. So $w$ is a subgradient of $f$ at $x$, and $w\in\partial f(x)$ by Subgradients and the Subdifferential of a Real-Valued Function on a Subset of $\mathbb{R}^n$ (ii).

It remains to show that there is no other subgradient. Let $a\in\partial f(x)$ and $v=a-w$, and suppose that $v\neq0$, so $\lVert v\rVert>0$. Take $\varepsilon=\lVert v\rVert/2$ in (1), choose its $\delta$, and then a real $s>0$ with $s\lVert v\rVert<\delta$. Then $x+sv\in U$, and the subgradient inequality for $a$ at $x+sv$, together with (1) for $sv$, gives

$$s\,a\cdot v\le f(x+sv)-f(x)\le s\,w\cdot v+\varepsilon s\lVert v\rVert .$$

Dividing by $s$ and using bilinearity of the dot product,

$$\lVert v\rVert^{2}=v\cdot v=(a-w)\cdot v\le\varepsilon\lVert v\rVert=\lVert v\rVert^{2}/2,$$

which is impossible. Hence $a=w$, and $\partial f(x)=\{\nabla f(x)\}$.