 Subgradient optimality condition

For a proper convex function $f:{\mathbb{R}}^n\to \mathbb{R}\cup \{+\infty \}$ and $x\in \mathrm{dom}f$,This is exactly the subgradient inequality with supporting vector zero: $f(y)\ge f(x)$ for every $y$.

If $f=g+h$, where $g:{\mathbb{R}}^n\to \mathbb{R}$ is differentiable and convex on all of ${\mathbb{R}}^n$ and $h$ is a proper convex function, the subdifferential sum rule yieldsThe full-domain smooth term supplies the regularity needed for this sum rule. For $f(x)=\frac{1}{2}(x-v{)}^2+\lambda |x|$, with $\lambda >0$, the subdifferential of absolute value at zero is $[-1,1]$. Thus $x=0$ minimizes $f$ exactly when $|v|\le \lambda$. Ordinary zero-derivative tests would miss this nondifferentiable minimizer.