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An equality-constrained convex optimization problem has the formwhere $f$ is convex and the constraint is affine. Its stationarity and feasibility equations can be combined into a monotone primal-dual operator.

Suppose $|Dv|\le M$. The coefficient matrices in part (ii) then have uniform ellipticity constants depending only on $M$. Applying the Harnack inequality for uniformly elliptic divergence-form equations to the nonnegative solutions $M+D_k{v}_R$ and $M-D_k{v}_R$ gives a scale-independent oscillation contractionScaling back,For fixed $r$, iterate this estimate with $R=4^{m}r$ and use the global bound $|D_{k}v|\le M$ to obtain ${\mathrm{osc}}_{B_r}{D}_{k}v\le 2M{\theta }^m\to 0$. Every partial derivative is therefore constant, so $v(x)=a\cdot x+b$ is an affine function. This is a bounded-gradient Bernstein theorem for entire minimal graphs.

Each map $x\mapsto ⟨a_i,x⟩+b_i$ is an affine function. For $0\le t\le 1$, the pointwise maximum of convex functions satisfiesso $f$ is convex.

A vector $g$ is a subgradient of a convex function $f$ at $x$ whenfor every $y$. Choose any active index $j\in I(x):=\{i:f(x)=⟨a_i,x⟩+b_i\}$. Thenand hence $\boxed{a_j\in \partial f(x)}$. More generally, every convex combination of the active vectors is a subgradient, and in fact

Write $z_i(x)=⟨a_i,x⟩+b_i$. The log-sum-exp function is convex and composition with the affine functions $z_i$ preserves convexity, so $f_{\beta }$ is convex. Directly, its Hessian matrix will also be shown positive semidefinite in part d.

Let $M=f(x)={\max }_i{z}_i(x)$. Factoring $e^{\beta M}$ out of the sum givesAt least one term in the sum is $1$, while every term is at most $1$. Thereforeand henceThus $f_{\beta }$ is a uniform smooth maximum of the affine pieces of $f$.

Composing the log-sum-exp function with finitely many affine functions gives a differentiable approximation to their pointwise maximum. Increasing the inverse-temperature parameter $\beta$ reduces the uniform approximation error while increasing the Lipschitz gradient constant.