Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 325 / 1

Use the extended real numbers $\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty ,+\infty \}$. A proper convex function never takes $-\infty$ and is finite somewhere. Its effective domain isFor a proper extended-real function, convexity meansThe endpoint cases are identities; taking $0<t<1$ avoids the ambiguous product $0\cdot (+\infty )$. Equivalently, its epigraph is a convex set. In particular, the effective domain is a convex set. Properness is an additional condition: it rules out the identically $+\infty$ function and all $-\infty$ values, rather than requiring finiteness everywhere.

For $x\in \mathrm{dom}f$, the subdifferential is the set of supporting slopesEach element is a subgradient. Define $\partial f(x)=\varnothing$ outside $\mathrm{dom}f$. The defining subgradient inequality must hold globally, not merely in a neighborhood of $x$. For a differentiable finite convex function, $\partial f(x)=\{\nabla f(x)\}$.

Write $h=f\square g$ for the infimal convolution. It is proper by the permitted hypothesis. If $h(x_0)$ and $h(x_1)$ are finite, choose $y_0,y_1$ within $\epsilon$ of their respective infima. For $0<t<1$, put $x_t=(1-t)x_0+t{x}_1$ and $y_t=(1-t)y_0+t{y}_1$. Convexity of $f$ and $g$ givesLet $\epsilon \downarrow 0$. If either endpoint value is infinite, the desired inequality is automatic. Thus the infimal convolution is convex, without assuming the infimum is attained.

For its convex conjugate, replace a negative infimum by a supremum and then change variables $u=x-y$:The two suprema separate because $u$ and $y$ are independent. Properness of $f$ and $g$ makes each supremum strictly greater than $-\infty$, so this separation remains valid when one or both are $+\infty$. Therefore $(f\square g{)}^{\ast }=f^{\ast }+g^{\ast }$. This is the conjugate of an infimal convolution. The bounded-domain assumption is unnecessary for these two calculations; the assumed properness and lower semicontinuity of $h$ will be used when applying subgradient inversion under convex conjugacy.

Use the conventionThis proximal map is also the resolvent of a monotone operator $(I+\tau \partial f{)}^{-1}$. A proper lower semicontinuous convex function has an affine minorant, so the quadratic term makes this minimization coercive and strongly convex. A unique minimizer exists for every $x$. Its subgradient optimality condition isThe Moreau–Yosida regularisation is $f_{\tau }=f\square g^{\tau }$. The conjugate of an infimal convolution and the quadratic conjugate giveThe factor $\tau$ here is essential. Apply subgradient inversion under convex conjugacy, followed by the subdifferential sum rule with the everywhere differentiable quadratic:The last equivalence is precisely the unique proximal minimization condition. ThusThis proves both existence and uniqueness of the subgradient, rather than only identifying a possible element. The finite convex function $f_{\tau }$ is therefore differentiable, with $\nabla f_{\tau }=(I-P_{\tau })/\tau$, the gradient of a Moreau envelope.

For completeness, monotonicity of $\partial f$ applied to the two proximal conditions givesHence $P_{\tau }$ is firmly nonexpansive. Expanding the same inequality shows that $I-P_{\tau }$ is firmly nonexpansive too. In particular, $\nabla f_{\tau }$ is $1/\tau$-Lipschitz continuous. None of this requires a bounded effective domain; the result applies to the next example as well.

For the Euclidean norm, rotational symmetry and minimizing the radial objective give the radial soft thresholding formulaIndeed, if $r=\Vert x{\Vert }_2$, an optimal $u$ is a nonnegative scalar multiple of $x$; minimizing $s+(r-s{)}^2/(2\tau )$ over $s\ge 0$ gives $s=(r-\tau {)}_{+}$. The gradient of a Moreau envelope is consequentlyAt $x=0$ this is $\{0\}$, and the two expressions agree when $\Vert x{\Vert }_2=\tau$. The Moreau envelope of the Euclidean norm itself isThus a quadratic core replaces the nondifferentiable tip while the outer gradient remains the normalized radial direction. The Euclidean norm has an unbounded effective domain, unlike the earlier bounded-domain hypothesis; the preceding argument explicitly shows that this restriction is unnecessary here.

For convex optimization, this Moreau–Yosida regularisation permits gradient descent and other methods for functions with a Lipschitz gradient, using a gradient Lipschitz constant $1/\tau$. It approximates the norm uniformly: $0\le \Vert x{\Vert }_2-f_{\tau }(x)\le \tau /2$. Smaller $\tau$ improves approximation but makes the permitted gradient steps smaller. Minimizing $f_{\tau }$ alone preserves the minimizers and minimum value of $f$; replacing one term in a larger objective can shift the minimizer, so the smoothing parameter controls that approximation error.