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The convex conjugate records a function using slopes rather than input positions. Let $f:{\mathbb{R}}^n\to \mathbb{R}\cup \{+\infty \}$ be a proper extended-real function: it is finite somewhere and never takes $-\infty$. Its convex conjugate iswhere $p\cdot x$ is the dot product and $\mathrm{dom}f=\{x:f(x)<+\infty \}$ is the effective domain. An infinite value of $f$ excludes that input from the supremum. The value $f^{\ast }(p)$ can be $+\infty$, but cannot be $-\infty$ because an input with finite $f(x)$ supplies a finite lower bound.

The convex conjugate is automatically a convex function, even when $f$ is not convex: each fixed input gives an affine function $p\mapsto p\cdot x-f(x)$, and their supremum is convex. It is also lower semicontinuous. It need not be proper: for $f(x)=-x^2$ on $\mathbb{R}$, $px-f(x)=px+x^2$ is unbounded above for every $p$. The definition and the automatic convexity are discussed in Boyd and Vandenberghe, Convex Optimization (2004), §3.3.1.

The best lower bound with a prescribed slope. Whenever $f^{\ast }(p)$ is finite, the affine functionis an affine minorant of $f$. Indeed, an affine function $p\cdot x+b$ lies below $f$ everywhere exactly when $b\le {\inf }_x(f(x)-p\cdot x)=-f^{\ast }(p)$. Thus $-f^{\ast }(p)$ is the largest possible intercept at slope $p$. The convex conjugate trades positions for slopes: $x$ specifies where to inspect $f$, whereas $p$ specifies which global affine minorant to use. Taking the supremum over all these affine minorants produces the biconjugate; the Fenchel-Moreau theorem says this recovers a proper convex function when it is lower semicontinuous.

A quadratic, computed by completing the square. For the convex quadratic function $f(x)=x^2/2$,Consequently $f^{\ast }(p)=p^2/2$, and the unique maximizing input is $x=p$. The affine minorant ${\ell }_p(x)=px-p^2/2$ is the tangent line at $(p,p^2/2)$: its slope is $p$ and its intercept is $-f^{\ast }(p)=-p^2/2$. For example, $p=2$ gives $f^{\ast }(2)=2$ and the tangent line $2x-2$, touching at $(2,2)$. This is the same curve described either by its point values or by the intercepts of its tangent lines.

A corner turns into a constraint on slopes. For the absolute value $f(x)=|x|$, if $|p|\le 1$ then $px-|x|\le 0$, with value zero at $x=0$. If $p>1$, taking $x\to +\infty$ makes $(p-1)x\to +\infty$; if $p<-1$, taking $x\to -\infty$ does the same. Therefore its convex conjugate is zero on $[-1,1]$ and $+\infty$ outside: it is the indicator functional of a constraint set for this interval. For $|p|<1$ the unique maximizing input is zero; for $p=1$ every $x\ge 0$ maximizes, and for $p=-1$ every $x\le 0$ maximizes. The corner's many supporting slopes are exactly the subdifferential $\partial f(0)=[-1,1]$.

A supremum need not be attained. For a differentiable convex function finite on all of ${\mathbb{R}}^n$, an input maximizes $p\cdot x-f(x)$ exactly when $p=\nabla f(x)$. If $f$ is strictly convex, there is at most one such input; existence is a separate issue. For the exponential function $f(x)=e^x$, the derivative is positive. At $p>0$, solving $p=e^x$ gives $x=\log p$ and $f^{\ast }(p)=p\log p-p$. At $p=0$, the supremum of $-e^x$ is zero as $x\to -\infty$, with no maximizing input. At $p<0$, $px-e^x\to +\infty$ as $x\to -\infty$. Thus a strictly convex differentiable function can have both unattained finite conjugate values and infinite conjugate values.

Exact fitting in the small-parameter limit requires data that can be approximated by forward images of elements in the regulariser's effective domain. Membership merely in the operator range is insufficient when an extended-real regulariser imposes a hard constraint.

The effective domain of an extended-real function is $\mathrm{dom}E=\{u:E(u)<+\infty \}$. An infinite value can impose a hard constraint through an indicator functional of a constraint set.

If two nonnegative finite-valued convex functions are defined on all of a finite-dimensional space, their infimal convolution is finite, convex and continuous everywhere. Convex perturbation duality then gives strong duality and dual attainment at every argument. It does not require or imply attainment of the primal decomposition infimum.

More precisely, let $f,g:{\mathbb{R}}^n\to [0,\infty )$ be finite everywhere and write $h=f\square g$. Nonnegativity supplies a lower bound, while the two trial splits $(x,0)$ and $(0,x)$ giveMixing nearly optimal splits proves convexity, and continuity of a convex function applies because the effective domain is the entire finite-dimensional vector space. These statements concern the optimized value, rather than existence of an optimal split. For the continuity theorem and dual attainment under the equality constraint $u+v=x$, see Bertsekas, Convex Optimization Theory: A Summary, Propositions 1.3.11 and 5.3.3(a). The equality has a feasible point in the whole space for every $x$, and the displayed bound makes its optimal value finite.

A finite value without any optimal split. On $\mathbb{R}$, take the convex functions $f(u)=e^u$ and $g(v)=0$. Both are nonnegative and finite everywhere, butfor every $x$. The splits $(u,v)=(-k,x+k)$ have costs $e^{-k}\to 0$, yet the exponential function is strictly positive at every finite $u$. No split attains the infimum. The value function is even constant and differentiable; regularity of the value function does not prevent optimizing splits from escaping to infinity. For example, $k=1,3,6$ gives costs approximately $0.367879$, $0.049787$ and $0.002479$.

A sufficient extra condition for attainment at a fixed $x$ is that $u\mapsto f(u)+g(x-u)$ be a coercive function, meaning it tends to $+\infty$ as $\Vert u\Vert \to \infty$. Under the finite-valued assumptions it is a continuous function, so a sufficiently large closed ball contains a global minimum by the extreme value theorem. If this cost is also a strictly convex function, the optimal split is unique. The quadratic example above satisfies both conditions; the exponential function example does not satisfy coercivity.

Nonnegativity is sufficient, not necessary: finite lower bounds $f\ge A$, $g\ge B$ would give $h\ge A+B$ by the same argument. Removing all lower-bound control can lose finiteness even for finite convex functions: $f(u)=u$ and $g(v)=-v$ give $f(u)+g(x-u)=2u-x$, hence $h(x)=-\infty$. Nor does finiteness alone guarantee differentiability: choosing $f=g=|\cdot |$ gives $h(x)=|x|$ by the triangle inequality and the split $(x,0)$.

A global minimizer is a point attaining the infimum of an objective. For an extended-real functional, it is usual to require a finite objective value; this excludes points outside its effective domain. A convention allowing $+\infty$ as the attained value gives every point as a formal minimizer of the identically infinite function.

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.

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, $\partial f(x)=\{\nabla f(x)\}$.

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.

A finite-valued global minimizer of the functional $E$ is an element $\bar{u}\in \mathrm{dom}E$ satisfyingHere $\mathrm{dom}E$ is its effective domain. The finite-value convention avoids treating an infeasible problem as solved.

The functional is a proper extended-real function if its effective domain is nonempty; the specified codomain already excludes $-\infty$. It has coercivity if $\Vert u_n\Vert \to \infty$ always implies $E(u_n)\to +\infty$, equivalently every finite sublevel set is bounded. It is $\tau$-sequentially lower semicontinuous ifThe topology in this definition matters: norm sequential lower semicontinuity and weak sequential lower semicontinuity need not coincide for a nonconvex functional.

On $\mathcal{U}=\mathbb{R}$ with its usual topology, the following examples isolate the three failures.

A nonproper functional is $E(x)=+\infty$ for every $x$. Its effective domain is empty, so it has no finite-valued global minimizer. It nevertheless has coercivity and is sequentially lower semicontinuous. If a minimizer is instead defined only by $E(x)=\inf E$ without requiring finiteness, every point formally minimizes this function; under that convention the requested nonproper counterexample is impossible with the given codomain.

A noncoercive functional is $E(x)=e^x$. It is a proper extended-real function and is continuous, but its infimum zero is approached as $x\to -\infty$ and is never attained. Hence it has no global minimizer.

A failure of sequential lower semicontinuity isIt is a proper extended-real function with coercivity, but $E(1/n)\to 0<E(0)$. Its infimum is zero, while every function value is positive. Thus it has no global minimizer.

The small-parameter limit depends on data compatibility with the regularizer domain. SetFor every such $v$, optimality givesTaking the upper limit as $\alpha \downarrow 0$ and then the infimum over $v$ proves ${\Phi }_{\alpha ,f}(u_{\alpha })\to m$. Its two nonnegative contributions above $m$ must vanish, yielding the general resultIn particular, the requested zero-misfit limit holds when $f\in \overline{K(\mathrm{dom}J)}$. A sufficient condition is an exact solution $v$ with $Kv=f$ and $J(v)<\infty$; thenMembership of $f$ in $\mathcal{R}(K)$ alone is insufficient. Take $\mathcal{U}=\mathcal{V}=\mathbb{R}$, $K=I$, $f=1$, and $J$ the indicator functional of a constraint set $\{0\}$. All positive-parameter objectives are proper extended-real functions with coercivity and sequential lower semicontinuity, with the unique minimizer $u_{\alpha }=0$, yet $D(u_{\alpha })=1/2$ for every $\alpha$. Thus the first printed limit needs compatibility with the regulariser's effective domain; the second limit remains true under the given assumptions.

For $p\in {\mathcal{U}}^{\ast }$, the definition of the convex conjugate giveswhere the substitution $w=u-v$ is a bijection of the independent pairs. ThereforeThis infimal convolution identity does not need convexity of $E,F$ or attainment of the inner infimum. The assumptions ensure that both functions have nonempty effective domains; their conjugates never take $-\infty$, so the separated sum is well defined, allowing $+\infty$.