Let Z be a function whose domain is an open set (a, b), with possibly a number, an open or closed set, or a combination of the above removed from the domain. Let x be any number within that set. Z is considered to be "single-peaked" if at most one x exists such that Z(x) is "peaked". This means that functions Z without a number x within their domain such that Z(x) is peaked will still be called "single-peaked"; this is intentional.
Peakedness is defined thus: for any x, Z(x) is peaked if Z is left-peaked and right-peaked at x.
Z is left-peaked at x if:
1. If Z is leftwise continuous at x, then Z is left-peaked at x if all numbers y within a sufficiently small region (x-delta, x) have the property such that Z(y) is less than or equal to Z(x), and the second leftwise derivative of Z at x is not equal to zero.
2. If Z is not leftwise continuous at x:
a. If the leftwise limit of Z at x exists but is not equal to x, then Z is left-peaked at x if the aforementioned limit is less than Z(x).
b. If the leftwise limit of Z at x does not exist, because there do not exist any y in the domain of Z within a sufficiently small region (x-delta, x), then W is defined as the greatest number w such that w is within the domain of Z, and w is less than x. Z is left-peaked at x if Z(W) is strictly less than Z(x).
Right-peakedness is similarly defined.
If there exists a number x such that no number greater than x is within the domain of Z, then x is not peaked. If there exists a number x such that no number less than x is within the domain of Z, then x is not peaked.
Any function that is monotonic on a set is single-peaked on that set. Let us assume that the slope is always positive or zero within the domain. Then no number x can have the property than Z(x) is rightwise peaked, since every number f such that f is strictly greater than x must have the property that Z(f) is greater than or equal to Z(x).
Any function with a constant slope of 0 within a domain is single-peaked within that domain, since no number x within the domain can have the property that Z(x) is peaked - Z is continuous, and all numbers x have the property that Z(x) has a second derivative of zero.
Lemma A: Let us consider two domains, A (j, k) and B (l, m), such that Z is single-peaked over each. Let A be monotonic. Let every number within a sufficiently small region of l be greater than or equal to every number within a sufficiently small region of k. ((Let l be greater than or equal to k?)) Then Z is single-peaked over the union of A and B. No number x within A can have the property that Z(x) is peaked when considered within the domain of AUB.
Lemma B: Let us consider two domains, A (j, k) and B (l, m), such that Z is monotonic over each. Let the slope of Z over A be always positive or zero, and let the slope of Z of B be always negative or zero. Then Z is single-peaked over the union of A and B, since it cannot be the case that Z(k) is right-peaked and Z(l) is left-peaked - at most one can be strictly greater than the other. Every other Z(x) such that x is not equal to k or l will continue to not be peaked; we have at most one number such that Z(x) is peaked. QED.
=============
Let us consider a single-peaked function P, whose domain is the open set Q (a, b). There is only one number x such that P(x) is peaked. This x is X.
If we modify the domain by removing a number that is not X, then P will continue to be single-peaked, as we will have divided the domain into two sets, one of which is monotonic and thus single-peaked. Let us call the number which is to be removed from the domain R. Let us assume that R is less than x. Every number t such that t is strictly greater than R and strictly less than x will continue to have the property that P(t) is not peaked, as the right-peakedness has not been altered.
If we modify the domain by removing X, then P will have no number x such that P(x) is peaked. Let us consider the open set (a, X). Every number t within that set will continue to have the property that P(t) is not left-peaked. Then there can be no number t such that P(t) is peaked.
If we modify the domain by removing an open set not containing X, then P will continue to be single-peaked. The two resulting sets will satisfy the requirements for Lemma A. QED.
If we modify the domain by removing an open set containing X, then the two remaining sets will satisfy the requirements for Lemma B. QED.
Therefore, removing points or sets from the domain of a single-peaked function over that domain will result in the said function being single-peaked over the new domain. As "single-peaked" was defined to include functions with no peak on their domain, this does not exclude the possibility of there being no peak.
It's semi-rigorous. That's fine. More rigor is good, but not required. (It's not for an assignment, just a project.)