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A feasibility problem asks whether any point satisfies a specified collection of optimization constraints, and seeks such a point when one exists. There need not be a preference among satisfying points. If the feasible set is $F$, the mathematical questions are whether $F$ is empty and, if not, how to produce a feasible point. A feasibility problem can be represented as mathematical optimization with a constant zero objective function.^[ref]

For an original linear example, considerThe pair $(1,0)$ is a directly checkable witness. Replacing the last bound by $x+y\le 0.5$ makes the constraints inconsistent: adding $-x\le -1$, $-y\le 0$ and $x+y\le 0.5$ gives $0\le -0.5$. This short contradiction certifies infeasibility without enumerating candidate points.

More generally, for a real linear system $Ax\le b$, a vector $\lambda \ge 0$ satisfyingis an infeasibility certificate: multiplying and adding the assumed inequalities would force $0\le {\lambda }^{T}b<0$. This is the certificate side of Farkas' lemma. A failed numerical search or a timeout is different from such a proof; it need not imply that the feasible set is empty.

The allowed variable set is part of the question. The equation $2x=1$ is feasible over the real numbers, with $x=1/2$, but infeasible when $x$ must be an integer. Thus replacing integer programming constraints by a real relaxation changes the feasibility problem.

Quasiconvex optimization supplies a useful connection: for each threshold $t$, test whether the original constraints and $f(x)\le t$ admit a feasible point. Such decisions can bracket the optimal value, while a retained witness gives an actual decision attaining the tested bound.^[ref] Numerical tolerances should be stated: a small constraint residual can support approximate feasibility without proving exact satisfaction.

A quasiconvex function is a real-valued function on a convex set whose every sublevel set is convex. If the domain is $D$, this meansis a convex set for every real $\alpha$, including levels for which the set is empty. Equivalently,Boyd and Vandenberghe give these level-set and segment characterizations.^[ref]

To see their equivalence, set $\alpha =\max \{f(x),f(y)\}$. Convexity of that sublevel set puts every convex combination of $x$ and $y$ inside it, proving the inequality. Conversely, if two points lie below the same level, the inequality keeps their connecting segment below that level.

Quasiconvexity is weaker than convexity. An original example is $f(x)=\sqrt{|x|}$ on $\mathbb{R}$. For $\alpha \ge 0$, its sublevel set is $[-{\alpha }^2,{\alpha }^2]$; for $\alpha <0$, it is empty. All are convex, butwhich violates the convex-function inequality. Quasiconvexity limits the height along a segment by the larger endpoint value, rather than by the line joining endpoint values.

A strictly increasing transformation preserves the sublevel ordering and hence quasiconvexity. This can be useful when a complicated objective has simple level sets. Quasiconvex optimization exploits those sets through feasibility tests at trial objective values; it need not treat the objective itself as a convex function.

In one dimension the shape corresponds to the single-minimum convention of unimodality. Flat portions require care: the step function equal to0 for $x\le 0$ and1 for $x>0$ is quasiconvex. Every point strictly to the right of zero is a non-strict local minimizer on a small flat neighborhood, yet its value is not globally minimal. Quasiconvexity alone therefore does not justify replacing every local minimum by a global one.

Quasiconvex optimization minimizes a quasiconvex function over a convex set of feasible points. Its defining advantage is that every sublevel set is convex: for each fixed threshold $t$, asking whether a feasible $x$ satisfies $f(x)\le t$ becomes a convex feasibility problem. The objective function itself need not be a convex function.^[ref]

For an original one-dimensional example, minimizeIts derivative $f^{\prime }(x)=1/(x+2{)}^2$ is positive, so its sublevel sets on this interval are intervals and the function is a quasiconvex function. Yet $f^{\prime \prime }(x)=-2/(x+2{)}^3<0$, so this nonconstant function is strictly concave rather than convex. For fixed $t$, the denominator is positive, and the threshold condition becomes the linear inequalityAt $t=0.4$, no allowed $x$ satisfies it; at $t=0.6$, the allowed interval $0\le x\le 0.5$ does. The true minimum is $f(0)=0.5$.

Interval bisection can search thresholds without directly minimizing the original ratio. Maintain an infeasible lower threshold $L$ and a feasible upper threshold $U$, with a stored feasible witness for $U$. Test $t=(L+U)/2$: on feasibility replace $U$ and its witness; on infeasibility replace $L$. With exact decisions, after $k$ tests the bracket width isStarting with $L=0$, $U=1$ and witness $x=2$, the example's first test finds $x=0$ feasible at $t=0.5$. Later tests below0.5tighten the lower bound.^[ref]

The threshold search is only as reliable as its feasibility problem solver and tolerances. Convexity of the sublevel sets does not supply an inexpensive representation of every such set, and the interval-width guarantee assumes justified feasibility decisions. Returning a feasible witness also matters: an objective bound alone is not an implementable solution.