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An infimal convolution is the cheapest way to split a prescribed total between two costs. Suppose two activities contribute $u$ and $v$ to a required total $x$, costing $f(u)$ and $g(v)$. Their combined cost is optimized over all splits $u+v=x$, rather than charging both activities for the entire total.

For proper extended-real functions $f,g:X\to \mathbb{R}\cup \{+\infty \}$ on a real vector space $X$, defineThe definition does not require convexity. An infinite cost forbids that contribution, as with an indicator functional of a constraint set. If no split has two finite costs, set $h(x)=+\infty$; if the costs of admissible splits are unbounded below, $h(x)=-\infty$. Thus proper inputs need not produce a proper extended-real function. The notation $E\square F$ denotes the same operation. See Parikh and Boyd, Proximal Algorithms, §3.1 for the standard convex analysis definition and its connection to the Moreau envelope.

An infimum is a best lower bound, whereas a minimum is a value realized by some split. Writing $\min$ in place of $\inf$ requires an attaining pair $(u_{\ast },v_{\ast })$. For convex functions, mixing two nearly optimal splits and applying convexity to each cost proves that their infimal convolution is convex whenever it is proper; one must use nearly optimal splits because exact minimizers need not exist. Its convex conjugate is $f^{\ast }+g^{\ast }$ whenever the extended real number arithmetic is well defined; see conjugate of an infimal convolution for the precise identity. This turns the optimization over splits into addition in the conjugate variables.

Two quadratic costs: share the load according to its price. Let $X=\mathbb{R}$ and let the convex quadratic functions beFor fixed $x$, eliminate $v=x-u$. The derivative of the total cost is $au-b(x-u)$, so equal marginal costs giveThe cheaper activity receives more of a positive total: when $a<b$ and $x>0$, $u_{\ast }>v_{\ast }$. Completing the square gives an exact certificate for every competing split:Consequently the infimum is a unique minimum, andThe effective quadratic coefficient is $c=ab/(a+b)$, with $1/c=1/a+1/b$. It is smaller than either $a$ or $b$: allowing both activities reduces the cost compared with assigning everything to either one. If $a=b$, each receives half the total and $c=a/2$. This is a simple quadratic optimization problem whose answer preserves the form of the original costs.

Smoothing a corner by splitting its cost. Take the absolute value function $f(u)=|u|$ and the convex quadratic function $g(v)=v^2/(2t)$, with $t>0$. This infimal convolution is the Moreau envelope of the absolute value function:Here $\mathrm{sgn}$ is the sign function, with $\mathrm{sgn}(0)=0$. The optimal $u_{\ast }$ is soft thresholding. In the central region, the quadratic function carries the whole total; outside it, $v_{\ast }$ stays at $t\mathrm{sgn}(x)$ and the absolute value function carries the excess. At $t=1$ the result is the Huber loss in the normalization $x^2/2$ near zero and $|x|-1/2$ outside. Its derivative is $x$ for $|x|\le 1$ and $\mathrm{sgn}(x)$ outside, so the corner becomes a continuous change of slope. Parikh and Boyd's ^[ref] uses twice this normalization for the Huber loss; the explicit formula here fixes the convention. This is also the one-dimensional case of the existing Moreau envelope of the Euclidean norm.

Compute the convex conjugate of $f_z$ at a general pair $(v,y)$. With $w=z+u-x$, the independent variables become $(x,w)$ and $u=w+x-z$, soTherefore the Lagrange dual function isUsing strong duality from the preceding solution givesEquivalently the conjugate of an infimal convolution is $F^{\ast }=k^{\ast }+h^{\ast }$, and the continuous convex $F$ equals its biconjugate.

For practical subgradient computation, minimize the known convex dual objectiveEvery optimizer, and only an optimizer, belongs to $\partial F(z)$ by Fenchel–Young inequality. The full characterization isThis is infimal-convolution dual subgradients; the set is nonempty because $F$ is finite convex everywhere. If both conjugates are differentiable at the optimizer, solve $\nabla k^{\ast }(y)+\nabla h^{\ast }(y)=z$. For nonsmooth conjugates, use the displayed aggregate subdifferential or a convex minimization algorithm. Replacing it by $\partial k^{\ast }(y)+\partial h^{\ast }(y)$ requires the usual subdifferential sum rule qualification; it is not automatically justified solely by knowing the two conjugates.

The conjugate of an infimal convolution is the sum of the conjugates:Here $g^{\ast }(p)=\frac{1}{2}\Vert p{\Vert }_2^2$, and $h^{\ast }(p)={\delta }_{[-1,1{]}^n}(p)$, the indicator functional of the unit infinity-norm ball. ThusThe infimal convolution is finite convex and continuous, so the Fenchel-Moreau theorem applies without a closure defect. By equality in the Fenchel–Young inequality and the subdifferential sum rule,This is precisely the variational characterization of projecting $x$ onto the cube. ConsequentlyFor the primal split, $y_i=\mathrm{clip}(x_i,-1,1)$ and $z_i=\mathrm{sign}(x_i)(|x_i|-1{)}_{+}$. These directly minimize the two scalar terms. The Huber loss is continuously differentiable, including at $x_i=\pm 1$, but its second derivative changes there. The one-dimensional sketch shows a quadratic center joined tangentially to linear tails:

Applied to a discrete gradient, the Huber gradient regularizer penalizes small slopes quadratically and large slopes linearly. Compared with pure squared-gradient smoothing it preserves large edges better; compared with pure total variation denoising it encourages small smooth variations and reduces the strong preference for piecewise-constant plateaus. It can therefore be useful for denoising signals or images containing both smooth regions and sharp transitions. It still penalizes edges and can bias their amplitude, and it does not guarantee complete elimination of staircasing in total variation denoising. The unit threshold must be scaled appropriately for data units and grid spacing.

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.

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.