advanced32 min studyLesson 278 in Mathematics

Infimal convolution and Moreau envelopes

Split a cost between components and derive smooth regularization from a uniquely attained proximal problem.

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What you’ll understand

  • Distinguish an infimal convolution value from an attained optimal decomposition.
  • Compute proximal maps and Moreau envelopes with consistent parameter scaling.
  • Explain smoothness, gradient Lipschitz bounds and preservation of minimizers.

Ordinary addition charges two functions at the same point. Infimal convolution lets them share a total input by choosing a decomposition and paying one cost for each component. Adding a quadratic through this operation produces the Moreau envelope: a smooth function built from a possibly nonsmooth convex objective. Its value and gradient come from a proximal minimizer. The parameter conventions matter: dividing the distance penalty by twice the parameter gives a threshold proportional to that parameter, not its reciprocal. We will keep that convention explicit throughout the examples.

Choose a decomposition, then take an infimum

For functions f and g on Rd\mathbb R^d, their infimal convolution at x is the infimum over z of f(z)+g(x−z). Convexity of both functions gives convexity of this value function where it is a proper extended-real function, but properness and attainment do not follow automatically. With f(z)=exp(z) and g identically zero, the value is zero at every x, yet no finite z attains it. With f(z)=z and g(w)=−w, the value is negative infinity everywhere. Both input functions in the second example are proper closed convex.

Conjugation exchanges infimal convolution with addition whenever the extended-value expressions are well-defined: the supremum over x and the infimum decomposition can be reparameterized by the independent components z and w. This identity does not itself say that the original infimum is attained or that the infimal convolution is closed. Such details become especially important when attempting to reverse the relation through biconjugation. The quadratic regularization used next supplies stronger structure than a general decomposition problem, so its existence and smoothness conclusions should not be generalized indiscriminately.

(f□g)(x)=inf⁡z{f(z)+g(x−z)},eλf(x)=inf⁡z{f(z)+∥x−z∥22λ},λ>0(f\mathbin{\square}g)(x)=\inf_z\{f(z)+g(x-z)\},\qquad e_\lambda f(x)=\inf_z\left\{f(z)+\frac{\|x-z\|^2}{2\lambda}\right\},\quad\lambda>0
The Moreau envelope is an infimal convolution with a scaled squared norm. The denominator convention determines all later scaling factors.↔ Scroll wide equations horizontally.

Quadratic regularization gives a unique proximal point

Let f be proper lower semicontinuous convex on finite-dimensional Euclidean space. For every x and λ>0, the quadratic-regularized objective has a unique minimizer, denoted prox_{λf}(x). A proper closed convex function admits an affine lower bound. Adding the positive quadratic to that bound prevents escape to infinity, yielding compact sublevels and existence. Strong convexity of the quadratic then gives uniqueness. This reasoning explains why a proximal point exists even when f itself has no minimizer or an unbounded effective domain.

The optimality relation is (x−z)/λ∈∂f(z). It locates the supporting slope at the new point z, unlike an explicit gradient step that uses a slope at the old point x. For f(z)=α|z| with α>0, considering positive, negative and zero candidates yields soft thresholding: keep the sign of x and reduce its magnitude by λα, stopping at zero. For an indicator of a nonempty closed convex set, the proximal map is the nearest-point projection, independent of λ because multiplying a zero-or-infinity indicator by a positive scalar does not alter feasibility.

z=prox⁡λf(x)  ⟺  x−zλ∈∂f(z),prox⁡λα∣⋅∣(x)=sign⁡(x)max⁡{∣x∣−λα,0}z=\operatorname{prox}_{\lambda f}(x)\iff\frac{x-z}{\lambda}\in\partial f(z),\qquad \operatorname{prox}_{\lambda\alpha|\cdot|}(x)=\operatorname{sign}(x)\max\{|x|-\lambda\alpha,0\}
The subgradient is evaluated at the proximal output. The soft-threshold level is λα for the chosen 1/(2λ) quadratic penalty.↔ Scroll wide equations horizontally.

A smooth envelope around a nonsmooth function

The envelope is finite and continuously differentiable everywhere, with gradient (x−prox_{λf}(x))/λ. This gradient is Lipschitz with constant at most 1/λ. A full proof uses the firm nonexpansiveness of the proximal map, developed in the next lesson, together with the minimizing inequalities for nearby input points. The differentiability statement is not obtained by differentiating the original nonsmooth f. It arises because the quadratic forces a unique well-behaved minimizing output. The same results extend to proper lower semicontinuous convex functions on real Hilbert spaces.

The envelope lies below f because choosing z=x is allowed whenever f(x) is finite. Their infimum values agree, and their minimizer sets coincide, including the possibility that both sets are empty. Indeed, a zero envelope gradient means the proximal output equals its input, and the optimality relation then gives zero in ∂f. Smoothing does not manufacture a minimizer when the original problem lacks one. It changes the function values and geometry away from the minimizer, so an envelope value should not be reported as the original objective value without also evaluating f at the relevant point.

A Huber-type envelope with explicit scaling

Take f(z)=2|z|, λ=1/2 and input x=3. Compute the proximal point, envelope value and envelope gradient.

  1. The threshold is λα=1, so the proximal point is z=3−1=2. The output is positive, and its subgradient f'(2)=2 agrees with (x−z)/λ=1/(1/2)=2.

  2. Evaluate the minimized objective at z=2: 2∣2∣+(3−2)2/(2⋅12)=4+1=52|2|+(3-2)^2/(2\cdot\tfrac12)=4+1=5. The original cost at the input is f(3)=6, consistent with the envelope being lower.

  3. The gradient is (3−2)/(1/2)=2. For |x|≤1 the output is zero and the envelope is x²; outside that interval it is 2|x|−1, with matching slopes at both joins.

The result

At x=3 the proximal output is 2, the envelope value is 5, and its gradient is 2. Although f has a corner, its envelope has a continuous first derivative.

Explore the model

Interactive lab

Interactive lab · function

Inspect the quadratic core of an envelope

Pause and Reset, then set a=1,x=1. Predict value 1 and slope 2, exactly the positive join of the envelope of 2|x| with λ=0.5. For a fresh comparison Reset and set a=1,x=−1; predict value 1 and slope −2. The full envelope continues linearly beyond these joins, whereas the instrument continues the quadratic. Its graph therefore displays the core and tangent matching, not the complete piecewise envelope.

Inspect the quadratic core of an envelopeThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA Moreau envelope can smooth a corner by attaching linear tails to a quadratic core with matching first derivatives.

Check your understanding

For proper closed convex f, what is the gradient of e_λf under this lesson’s convention?

Try it yourself

Let f be the indicator of [0,1], λ=2 and x=3. Compute its proximal point, envelope value and gradient. What happens at x=1/2?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The proximal point is projection onto the interval. Then use squared distance divided by 2λ.

Compare with the solution

At x=3 the projection is z=1. The envelope is (3−1)²/4=1 and its gradient is (3−1)/2=1. At x=1/2, projection leaves the input unchanged, the distance is zero, and both envelope value and gradient are zero. Thus every feasible point minimizes this envelope, exactly as every feasible point minimizes the original indicator. The envelope remains finite outside the interval even though the original indicator is infinite there.

Take it with you

  • General infimal convolutions can fail properness or attainment.
  • A proper closed convex function has a unique proximal point for every positive parameter.
  • The Moreau gradient is a scaled proximal residual, and smoothing preserves the minimizer set.

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