Differentiable Convex Functions

If a function $f$ attains its minimum on a convex set $\mathcal{X}$ at $x$, and $f$ is differentiable at $x$, then $(x-y)'\mathcal{D}f(x)\geq 0$ for all $y\in\mathcal{X}$.

If $f$ attains its minimum on $[a,b]$ at $a$, and $f$ is differentiable at $a$, then $f'(a)\geq 0$. Or, more precisely, if $f$ is differentiable from the right at $a$ and $f'_R(a)\geq 0$.

Suppose $\mathcal{X}=\{x\mid x'x\leq 1\}$ is the unit ball and a differentiable $f$ attains its a minimum at $x$ with $x'x=1.$ Then $(x-y)'\mathcal{D}f(x)\geq 0$ for all $y\in\mathcal{X}.$ This is true if and only if $$\begin{align*} \min_{y\in\mathcal{X}}(x-y)'\mathcal{D}f(x)&= x'\mathcal{D}f(x)-\max_{y\in\mathcal{X}}y'\mathcal{D}F(x)=\\&=x'\mathcal{D}f(x)-\|\mathcal{D}f(x)\|=0. \end{align*}$$ By Cauchy-Schwartz this means that $\mathcal{D}f(x)=\lambda x$, with $\lambda= x'\mathcal{D}f(x)=\|\mathcal{D}f(x)\|.$

As an aside, if a differentiable function $f$ attains its minimum on the unit sphere $\{x\mid x'x=1\}$ at $x$ then $f(x/\|x\||)$ attains is minimum over $\mathbb{R}^n$ at $x$. Setting the derivative equal to zero shows that we must have $\mathcal{D}f(x)(I-xx')=0$, which again translates to $\mathcal{D}f(x)=\lambda x$, with $\lambda= x'\mathcal{D}f(x)=\|\mathcal{D}f(x)\|.$