Open interval minimum: why the lower bound is never reached

Open interval minimum checks need a feasible point as well as a lower bound. For the invented function f(x) = x² on the exact real domain 1 < x < 2, every value is greater than one. Values can get as close to one as requested, but no allowed point gives that value. In fact, this problem has no minimum at all. Every proposed feasible point admits another with a smaller function value.

Keep the objective and allowed inputs together

The inputs are all real numbers strictly between one and two. Neither endpoint is included. The objective is to make x² as small as possible while keeping x inside that interval. These are stipulated mathematical rules, without a physical unit, measured clearance, product dimension, error tolerance or uncertainty. Replacing the real domain with a printed list of convenient samples would change the allowed inputs.

A minimizing point c must itself be allowed and must satisfy f(c) ≤ f(x) for every allowed x. A lower bound L has a different requirement: L ≤ f(x) for every allowed x. That definition concerns an output number and does not require any input to produce it. A complete answer must distinguish these two roles before treating a bound as a minimum.

For this objective, x > 1 implies x² > 1 because x is positive. Thus one is a lower bound, and every actual function value lies strictly above it. The observation does not yet prove that no minimum exists. An arbitrary unattained lower bound could lie well below an attained minimum in another problem. We will rule out every feasible candidate and then prove that one is the greatest lower bound.

Check the equality point against the domain

The tempting answer uses x = 1 because its square is one. That substitution evaluates the same formula, but it does not supply an allowed input to the original problem. The inequality 1 < x is strict. Writing one beside the answer without a domain check would confuse a formula value available outside the permitted set with an attained value inside it.

The other real solution of x² = 1 is x = −1, which is also outside the stated interval. No allowed input can therefore attain the lower bound one. Keep that equality check beside the domain. It explains why the boundary calculation cannot serve as a witness, while the next argument explains why choosing some larger attained value cannot rescue the proposed minimum.

Improve any proposed feasible point

Take an arbitrary allowed x, so 1 < x < 2. Define y = (1 + x)/2, the midpoint between one and that x. Since x is greater than one, their midpoint is also greater than one and is strictly less than x. Consequently 1 < y < x < 2. The new point is permitted by the same original domain; no endpoint has been added.

Both x and y are positive. Their squared-value difference is x² − y² = (x − y)(x + y). Each factor on the right is positive, so x² − y² > 0. Therefore y² < x². This is a universal improvement rule: whichever feasible x someone proposes, its associated y is another feasible point with a smaller objective value.

For example, x = 3/2 gives value 9/4. Its improving midpoint is y = 5/4, with value 25/16, which is smaller. Starting instead at 5/4 gives midpoint 9/8 and value 81/64. These examples illustrate the same arbitrary-point argument. They are not a finite search from which the general conclusion is guessed.

Separate nonexistence from an unfinished search

Suppose a feasible minimizer c existed. Applying the improvement construction to that same c would produce y with f(y) < f(c), contradicting the required minimum condition. Hence no feasible minimizer exists. The conclusion follows from the declared domain and exact algebra. It is not a report that a search stopped early or that a sufficiently accurate instrument has not yet been found.

Repeated midpoint steps never reach the excluded endpoint. If the initial gap above one is d = x − 1 > 0, one step replaces it with d/2. After k such steps the gap is d/2ᵏ, still positive for every finite nonnegative integer k. Each step stays feasible and improves the value, while none supplies x = 1 as an allowed final point.

Prove that one is the greatest lower bound

We already know all values exceed one. To show that no larger number is a lower bound, let ε be any positive real number. Choose h = min(1/2, ε/4), then use x = 1 + h. The chosen h is positive and at most one half, so this x lies strictly between one and two. Its value is 1 + 2h + h².

Because h ≤ 1/2, its excess above one satisfies 2h + h² ≤ (5/2)h. If ε ≤ 2, the choice is h = ε/4, and the excess is at most 5ε/8, strictly less than ε. If ε > 2, the choice is h = 1/2, giving excess 5/4, again less than ε. Thus an allowed value lies strictly between one and 1 + ε for every positive ε.

Now consider any proposed bound L > 1. Set ε = L − 1. The construction produces an allowed function value less than 1 + ε = L, so L cannot be a lower bound. One is therefore the greatest lower bound, often called the infimum. Its failure to be attained remains explicit: the output bound is one, and the original problem has no minimizing point.

Check the witness with exact fractions

For ε = 1/4, take h = 1/16. Then x = 17/16 and x² = 289/256. Its excess above one is 33/256, less than the requested 1/4 = 64/256. The value remains above one. This gives a concrete instance of approaching the bound without turning an approximation into equality.

An exact arithmetic control checked ninety-nine feasible rational inputs between one and two and confirmed that every constructed midpoint stayed inside the domain and had a smaller square. Separate checks covered twenty-nine finite grids and five positive ε values. Those checks inspect their stated finite inputs. The midpoint and two-case ε arguments establish the conclusions for all the real inputs and positive tolerances in the model.

Include the lower endpoint in a separate problem

Change the domain to 1 ≤ x < 2 while keeping the objective x². Now c = 1 is permitted and has value one. Every other allowed input is greater than one and has a larger square. The changed problem consequently has the unique minimum value one at x = 1. The equality witness now passes the feasibility check that failed before.

This is a new domain, not a numerical correction to the original answer. Keep the two records separate: the open interval has infimum one with no minimum; the lower-endpoint-inclusive interval has an attained minimum one. Replacing a strict sign with a nonstrict sign silently would hide exactly which input rule made the answer change.

Including only the upper endpoint does not help

Consider the separate domain 1 < x ≤ 2. The point two is now available, but its value four does not become a minimum. For every feasible x, including x = 2, the midpoint (1 + x)/2 still lies above one, below x and below two. Its square is smaller. The same universal improvement argument therefore rules out a minimum on this changed domain too.

At x = 2 the improving point is 3/2, with value 9/4 rather than four. This control identifies which endpoint matters for the particular objective. Merely observing that a domain includes an endpoint does not establish that it includes the point needed to attain its lower bound. The objective and the actual membership rule must be checked together.

An open domain can still have a minimum

Return to the original open domain 1 < x < 2, but change the objective to g(x) = (x − 3/2)². A square is nonnegative for every allowed x. At c = 3/2, which lies inside the domain, the value is zero. Thus this separate problem has an attained minimum zero despite using the same open interval.

The minimizing point is unique because the square is zero only when x − 3/2 = 0. This control prevents the false generalization that open domains can never have minima. For the original objective, movement toward the excluded lower endpoint always improves the value. For the changed objective, the equality point is an allowed interior point. The two feasible-set checks reach different valid conclusions.

A finite grid answers its own question

Suppose a different worksheet permits only 5/4, 3/2 and 7/4. Their squared values are 25/16, 9/4 and 49/16. The first is the smallest, so that three-point problem has an attained minimum at 5/4. The continuous problem still permits 9/8, whose square 81/64 is less than 25/16. It was excluded from the grid rather than excluded from the original real interval.

More generally, fix an integer N ≥ 2 and allow x = 1 + k/N only for integers k from one through N − 1. The smallest grid input is 1 + 1/N, and positivity makes its square the grid minimum. That minimum is 1 + 2/N + 1/N², always greater than one. Each finite grid has its own attained answer without supplying a minimum for the original continuum.

A diagram may show a hollow endpoint and filled points approaching it, but a finite picture cannot change membership of the exact interval. Keep the symbolic domain with the illustration. The midpoint rule can produce a valid continuous input missing from a chosen grid, which is precisely why a grid result should not be relabeled as the answer to the unrestricted real-input problem.

Bound the optional native accounts

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Keep neighboring optimization questions distinct

The holiday meeting point minimum finds an attained optimizer for a stated sum of absolute distances over all real locations. Here the equality point needed by the original squared objective is outside its permitted domain. A valid bound must be accompanied by a feasibility check before it can identify an attained answer.

The three-point disk coverage example combines a universal lower bound with a disk that actually attains it. This interval problem has no such feasible witness, and every feasible proposal can improve. It supplies no disk geometry, physical coverage measurement or product-fit conclusion.

For the original record, retain f(x) = x², the strict domain 1 < x < 2, the lower bound one, the midpoint improvement and the ε witness. Report infimum one and no minimum. Keep the endpoint-inclusive, changed-objective and finite-grid controls under their own inputs so none can silently replace the problem that was asked.

An excluded endpoint can remain a bound without becoming an allowed answer.

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