advanced27 min studyLesson 273 in Mathematics

Extended-real convex functions

Encode feasibility with infinity while keeping properness, domain geometry, and objective arithmetic precise.

On this page

What you’ll understand

  • Use extended-value indicators to encode convex constraints.
  • Check properness and convexity without differentiating outside the domain.
  • Separate domain restrictions from the existence and uniqueness of minimizers.

A function that assigns an ordinary number everywhere cannot directly forbid a candidate. Assigning positive infinity to infeasible points solves that modeling problem while allowing constrained and unconstrained optimization to share one notation. This is not the probability indicator that equals one on an event. It is an extended-value indicator that equals zero on an allowed set and infinity elsewhere. The convention is powerful, but it must be accompanied by precise domain and arithmetic rules; otherwise a compact expression can conceal an empty feasible problem or an undefined subtraction.

Infinity is a constraint marker

Here a proper function maps Rd\mathbb R^d into R∪{+∞}\mathbb R\cup\{+\infty\}, is finite somewhere, and never takes negative infinity. Its effective domain is the set where its value is finite. Calling a function proper does not mean that it is bounded below. The affine function f(x)=x is proper but has infimum negative infinity. Likewise, properness alone says nothing about continuity, closedness, or whether an infimum is attained. These are additional properties that must be checked rather than inferred from a familiar-looking formula.

For a nonempty set C, define its extended-value indicator to be zero on C and positive infinity outside. Minimizing h over C becomes minimizing h plus this indicator over the whole space, provided h is finite on C or their common domain is explicitly identified. The sum of two proper functions can fail to be proper when their domains do not meet. With the present convention there is no negative-infinity value to combine with positive infinity, but expressions involving subtraction of two infinite objective values remain undefined.

δC(x)={0x∈C,+∞x∉C,dom⁡f={x:f(x)<+∞}\delta_C(x)=\begin{cases}0&x\in C,\\+\infty&x\notin C,\end{cases}\qquad \operatorname{dom}f=\{x:f(x)<+\infty\}
This indicator encodes feasibility. It differs from the ordinary 0–1 indicator used in probability and integration.↔ Scroll wide equations horizontally.

The convexity inequality includes the domain

An extended-real function is convex when its epigraph is convex, equivalently when the convexity inequality holds for interior mixture weights and finite-domain endpoints. Checking finite endpoints is sufficient because a positive multiple of infinity makes the right side infinite otherwise. Endpoint weights zero and one can be handled separately, avoiding any need to interpret zero times infinity. The inequality forces the effective domain to be convex. For an indicator function, that condition is also sufficient: indicators are convex exactly when their allowed sets are convex.

A Hessian test addresses only the smooth interior of a domain. It does not independently prove that the entire extended-value model is convex. For instance, assigning x² to two disjoint intervals and infinity elsewhere produces positive curvature wherever differentiable, but the midpoint of two finite points can be forbidden. Conversely, a convex indicator has no useful ordinary derivative across its boundary and can still be a perfectly valid convex objective. Domain geometry and finite-value inequalities should therefore come before derivative-based recognition rules.

f((1−t)x+ty)≤(1−t)f(x)+tf(y),x,y∈dom⁡f,0<t<1f((1-t)x+ty)\le(1-t)f(x)+tf(y),\qquad x,y\in\operatorname{dom}f,\quad0<t<1
A proper convex function has a convex effective domain. Testing this inequality at finite endpoints avoids ambiguous endpoint infinity arithmetic.↔ Scroll wide equations horizontally.

Reformulation preserves a problem, not every guarantee

Adding a convex indicator to a convex loss keeps the combined objective convex whenever their common domain is nonempty. Composing a convex extended-real function with an affine map also preserves the inequality, although the new domain can be empty. A resource constraint Ax=b becomes the indicator of an affine inverse image. A nonnegative-variable restriction becomes the indicator of an orthant. Such notation makes later subgradient and conjugate calculations uniform, but the modeler must still verify that at least one candidate has finite total cost.

Convexity implies that every local minimizer relative to the domain is global. It does not imply a minimizer exists: exp(x) has infimum zero on the real line but never reaches it. Strict convexity gives at most one minimizer, still without guaranteeing existence. Finally, an open domain can prevent attainment even for a simple linear loss. The next lesson adds lower semicontinuity and compactness conditions that control these issues. Keeping the concepts separate makes it possible to diagnose a modeling failure instead of blaming an optimization routine for a mathematically nonexistent answer.

A box-constrained quadratic as one function

Minimize h(x)=(x−3)² on C=[−1,1] by using F=h+δ_C. Identify the domain and minimizer.

  1. The effective domain of F is [−1,1], since h is finite everywhere. The interval is nonempty and convex, so F is proper and convex.

  2. The unconstrained quadratic decreases while x<3. On the entire feasible interval its derivative 2(x−3)2(x-3) is negative, so the smallest value occurs at the right endpoint x=1.

  3. At x=1 the cost is (1−3)²=4. An outside point such as x=3 has h=0 but F=+∞, so it cannot beat the feasible solution.

The result

The constrained and extended-value problems have the same unique minimizer x=1 and minimum 4. Uniqueness follows from the strictly convex finite quadratic, not merely from the indicator notation.

Explore the model

Interactive lab

Interactive lab · function

Curvature is only one part of convexity

Pause and Reset, then set coefficient a=1 and position x=1. Predict tangent slope 2 and an upward convex graph. For a fresh comparison Reset and set a=−1,x=1; predict slope −2 and downward curvature. The instrument plots finite ax² over its display window, without an indicator or excluded domain. Use it to check the curvature of a finite piece, then separately explain why a disconnected allowed domain would invalidate convexity.

Curvature is only one part of convexityThis 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 forPositive interior curvature cannot repair a nonconvex effective domain.

Check your understanding

Which assertion about proper extended-real convex functions is correct?

Try it yourself

Let f=δ_[0,1] and g=δ_[2,3]. Are f and g proper convex? Is f+g proper? Then replace g by δ_[1,3] and describe the sum.

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

Give me a hint

Write the effective domain of the sum as an intersection before considering numerical values.

Compare with the solution

Each original indicator is proper and convex because its interval is nonempty and convex. Their sum is infinite everywhere: no point belongs to both intervals, so it is not proper. After replacing g, the common domain is the singleton {1}. The sum is zero at 1 and infinite elsewhere, exactly δ_{1}; it is proper and convex. This example distinguishes closure under the convexity inequality from preservation of properness.

Take it with you

  • Extended-value indicators turn constraints into objective terms.
  • Properness requires a nonempty finite domain and excludes negative-infinity values.
  • Convexity, closedness, existence, and uniqueness are different assertions.

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