Appendix A.2. Proof of Lemmas and Propositions
Lemma A1. is right upper semicontinuous in x at .
Proof of Lemma A1. Let
and observe that
at
. Let
and
be such that
for each
n. For each
n, let
be a sequence in
k such that
. For each
n, let
be such that
for all
and choose
for some
. Then note that
Since
, it follows that
. Thus, by the right upper semicontinuity of
in
, there exists an
such that
for all
. Similarly, by the continuity of
in
, there exists an
such that
for all
. Thus, for all
, it must be that
Therefore,
so by definition,
is right upper semicontinuous in
x at
. □
Proof of Lemma 1. Suppose to the contrary that
for each firm
i. Then for any firm
i and any
in support of
, Assumption 6 guarantees that
and
for all
. Observe that
and
since
. Thus, we have
The last inequality follows from the fact that
is a CDF and thus nondecreasing and from Assumption 1 guaranteeing that
for
. This contradicts all
as equilibrium strategies. □
Proof of Lemma 2. Let
be an equilibrium and
x be such that
and
. Suppose that
and
. Consider a sequence of deviations by firm
i to
defined by
where each
is chosen so that
and
. That is,
is the measure created from
by shifting all mass from the price
x to the price
. Then note that
We will show that
for sufficiently large
n, which will guarantee a profitable deviation for firm
i, violating
as an equilibrium strategy. Note that
It follows that the pointwise limit as
is
Thus, since
, then by the Lebesgue dominated convergence theorem,
Since
and
,
is a profitable deviation for firm
i for sufficiently large
n, violating
as an equilibrium. We conclude that either
or
.
From Lemma 1, we know that . Observe that if , it must be that since for each firm i at any price and for some firm i at any price x.
Next, we show that the equilibrium is invariant to the choice of at prices . Let be an equilibrium given the sharing rule with expected profits and consider another sharing rule such that for all . Let denote firm i’s expected payoff when choosing a price x given and the corresponding payoff given . To show that is an equilibrium for the game with sharing rule , it will suffice to show that, for each player i, (i) -almost everywhere and (ii) for all prices x.
(i) Note that the sharing rule does not influence the payoffs at any price x such that , and so at all such prices. Further, at all prices , since . The first part of this lemma demonstrates that for all such that . Since has at most countably many atoms, then . It follows that -almost everywhere.
(ii) As we have shown in part (i), except possibly at prices such that . Since price above is weakly dominated by , it is sufficient to examine prices in . Consider any such price x and let be a sequence such that , for all k, and for all k. Then note that the continuity of and in on from Assumption 7 implies that . Since for all k, then for all k. If , then for sufficiently large k, violating as an equilibrium strategy with the sharing rule . Therefore, for all x.
We conclude that is an equilibrium given the sharing rule . □
Proof of Lemma 3. We first argue that, in any equilibrium, for at least one firm i. Suppose to the contrary that for each firm i. Then Assumption 2 implies that and for all and all . It follows that either player i could choose a price of and receive a payoff of with positive probability since . This contradicts prices in as equilibrium strategies. We conclude that for at least one firm i.
Let be an equilibrium with . Note that for all . If , then Assumptions 2 and 3 imply that , and thus is strictly increasing on , violating prices in as equilibrium strategies for firm i. Thus, and so . Assumption 2 thus implies that for all , so firm i’s equilibrium profit must be zero. From the result proved immediately above, it must be that . Suppose that . Then Assumption 2 implies that for all . Therefore, since , Assumption 3 guarantees that , and thus, is strictly increasing on , violating these prices as equilibrium strategies for firm j. Thus, it must be that . If is nondegenerate, then there is some such that firm j prices strictly higher than x with positive probability. If firm i sets a price of this x, then with positive probability, firm i will receive , which is strictly positive by Assumption 3, contradicting zero as its equilibrium profit. Therefore, is degenerate with . Since there is a positive probability that firm i chooses a price , Assumption 2 implies that there is a positive probability that firm j will obtain a profit if it sets its price at . Consequently, firm j’s equilibrium profit must be positive, and thus .
Suppose that , since by Assumption 4. Assumptions 3 and 4 guarantee that for all and that is strictly increasing on this interval. Thus, it must be that .
Finally, suppose that for some . Then firm j could set any price and still receive with certainty. Since is strictly increasing, this would violate as an equilibrium strategy for firm j. We conclude that for any . □
Proof of Lemma 4. Lemma 3 implies that any nondegenerate mixed strategy equilibrium requires that . Let be an equilibrium with . Recall from the proof of Lemma 3 that at least one firm i must have .
First, we show that . Suppose to the contrary that for some firm i. Then it must be that . By Assumption 2, firm j can set any price and obtain a profit of with certainty. From Assumption 3, is strictly increasing on this interval, contradicting these as equilibrium strategies. Therefore, it must be that .
Second, we show that . Suppose to the contrary that . Then by Assumption 4, for all . Consequently, either firm i could choose any price and earn . Since , Assumption 3 guarantees that is strictly increasing on this interval, violating these prices as equilibrium strategies. Therefore, it must be that .
Third, we show that neither firm i can have an atom at an if . Suppose to the contrary that firm j has an atom at , , noting that since it is in support of each firm’s strategy. From Lemma 2, we know that ; however, as we have just shown, . Thus, for all . We will show that there is some and neighborhood such that for all .
Define
and let
be such that
. If firm
i sets a price
, then its profit will be
with certainty. Since
, Assumption 7 guarantees that
is left continuous, so there exists a
such that
for all
. Note that
at any price
and observe that
at
. Note that
is right upper semicontinuous by Lemma A1. If
, then Assumption 6 guarantees that
for all
, and thus,
is right upper semicontinuous at
. Alternatively, if
, then Assumption 7 guarantees that
is continuous and thus right upper semicontinuous. In either case, the function
is right upper semicontinuous in
x at
, so there exists
such that
for all
. Let
and note that
. Thus,
For all
, we have
and therefore,
Since
, this implies that
for all
. This violates such prices as equilibrium strategies for firm
i. We conclude that
, so neither firm
i can have an atom at
if
. □
Proof of Proposition 3. We will prove the proposition for the two types of equilibria from Proposition 2 separately.
Case 1: Let be a symmetric lower bound equilibrium.
For this case, we first show that
in two parts. First, we argue that
. Suppose to the contrary that
. From Lemma 2, at most one firm can have an atom at
. Without loss of generality, let firm
i be such that that
is in support of
and
. Choose
in support of
such that
; then note that
since
is continuous in
from Assumption 7. Next, since each
is in support of
, it must be that each
. Note that
by definition of
. By definition of
and the fact that
,
for all
for each firm
i. Further, Assumptions 2 and 3 imply that
is nondecreasing, so
for all
. This implies that
, and thus
. This violates
as an equilibrium strategy since firm
i has a profitable deviation to
for sufficiently small
that would guarantee a payoff of
. We conclude that
.
Second, we argue that . Suppose to the contrary that . Let firm i be such that . By definition of and the continuity of from Assumption 7, it must be that . From Lemma 4, , and since , Assumption 2 implies that for all . Thus, . The continuity of and in thus guarantee that and for all prices with , so for all such x. This violates all as equilibrium strategies. We conclude that .
It follows from Lemma 4 that each firm i’s equilibrium expected profit is . The statement of the proposition thus follows from and the facts that is strictly increasing on and that .
Case 2: Let be an asymmetric lower bound equilibrium.
First, consider player i. The lower bound payoff , which implies that . Further, we know from Lemma 3 that . By construction, ; therefore, . Next, consider player From Lemma 3, . This payoff is possible for player j for any pricing by player i. Thus, . By construction, ; therefore, . □
Proof of Lemma 5. We use and to denote the smallest and largest conditional residual maximizer, respectively. That is, and , where the right continuity of ensures that contains a minimal element, while it need not contain a maximal element.
Let
be an equilibrium with the corresponding CDF’s
F. If
, then from Lemmas 3 and 4, either the equilibrium is degenerate or
for some firm
i. If the equilibrium is degenerate, then the statement of Proposition 4 (in
Section 4) guarantees that
for some firm
i. It trivially follows that
. Alternatively,
for some firm
i; then Lemma 3 guarantees that
and
. In order for
to be an equilibrium, firm
i cannot have any profitable deviations, so it must be that
for all
. Thus, by definition,
, so
.
Finally, let
and suppose that either
or
. From Lemma 2, at most one firm may have an atom at
. Let firm
i be such that
and
. Since
is in support of firm
i’s strategy, we may choose
in support of
such that
; then note that
Since each
is a best response for firm
i, this implies that
is also a best response for firm
i. It follows from our supposition that
. Note that, for any price
x,
By definition of
for all
. Thus,
for all
since
for all
. This contradicts
as a best response. We conclude that
. □
Proof of Proposition 4. First, observe that, in any equilibrium with corresponding profits , it must be that . To see why, note that Assumption 7 guarantees that is left continuous at . Thus, firm i can guarantee itself a payoff of by deviating to , with the guarantee that .
We now argue that any pure strategy equilibrium must be symmetric. Suppose to the contrary that there is an asymmetric equilibrium with . This means that firm i obtains with certainty. There are two cases to consider: (i) and (ii) . We may ignore the case in which since firm i would trivially be better off with a price of .
In case (i), if , then Assumption 3 guarantees that some price is strictly better for firm i since its front-side profit is strictly increasing on . Alternatively, if , then Assumption 2 guarantees that and for any . It follows that ; else firm j has a profitable deviation to . Thus, since , firm i’s equilibrium profits must be such that . It follows that , contradicting Assumption 5.
In case (ii), Assumption 3 guarantees that firm j has zero profit. Thus, since Assumption 5 guarantees that , firm j would be better off charging some price in order to obtain a positive profit. We conclude that any equilibrium must be symmetric.
Let be an equilibrium. Now we show that the only possibility is . It follows immediately from Lemma 2 that . Therefore, it remains to show that . Suppose to the contrary that for some firm i. Assumption 4 guarantees that for all . If , then Assumption 3 guarantees that is strictly increasing, thus violating as an equilibrium. Therefore, it must be that , and thus, . Assumption 2 then guarantees that , so it must be that ; else firm j has a profitable deviation. However, Assumption 3 guarantees that , a contradiction. We conclude that for each firm i, so .
Now that we have established that the only possible pure strategy equilibrium is such that , we show that the conditions of the proposition are necessary and sufficient for this to be a pure strategy equilibrium price. We begin with sufficiency by showing that the conditions rule out any profitable defections. For any defection , the condition immediately rules out a profit increase for firm i. The other conditions and imply that for all . Notice that the conditions make it such that for all firms i. Any defection to results in based on the facts that, from Assumption 3, is strictly increasing on the interval , and from Assumption 2, for all .
Next, we prove that the conditions are necessary. Suppose to the contrary that is an equilibrium and neither condition holds. First, we consider the case for all i. If there is a firm i such that , then for any , , a contradiction. Second, consider the case that there is a firm i such that . We will go through a violation of each of four conditions in turn. If , then for any , , a contradiction. If , then , and based on the continuity of for sufficiently close to , . If , then because and . Then , and based on the continuity of for sufficiently close to , . Finally, if and from the previous step, it must be that , then for any , , a contradiction. □
Proof of Proposition 5. Given Proposition 4, we need only show that there is no equilibrium other than . Let be a nondegenerate mixed strategy equilibrium. We consider two cases corresponding to whether or .
Suppose first that . Lemmas 3 and 4 imply that . Using Lemma 2, we may, without loss of generality, assume that and that firm j’s strategy does not have an atom at . Then note that when choosing a price at or near , firm i earns a profit of approximately , where the inequality follows from the fact that is nonincreasing in . Since is the unique maximizer of , it must be that . Further, since , it must be that . This contradicts prices at or near as equilibrium strategies.
Next suppose . From Lemma 3, firm j’s strategy must be degenerate with . If , then Assumption 2 guarantees that is strictly increasing on , and so this interval contains no best responses to . Thus, it must be that and therefore that . This further implies that , so for all . This contradicts as the unique maximizer of . We conclude that is the unique equilibrium. □
Proof of Lemma 6. We go through the verification of each assumption in turn.
Verification of Assumption 1: for all .
First, note that since each is quasiconcave in z, it follows that for all . Observe that , so . Thus, .
Verification of Assumption 2: For each firm i, there exists an such that for all . Further, for all such that .
As defined above, in the special case of the model each . Thus, for any price and any quantity , it follows that , and thus, . It follows immediately that and for all . This further this implies that for all , so . As such, Condition 6 implies that for all . Consequently, , and so it must be that .
Verification of Assumption 3: has a unique maximizer with . is strictly increasing at any price . Further, for all .
From Condition 7,
is strictly quasiconcave and thus has a unique maximizer
on
. We will argue that
satisfies the statement of Assumption 3. To do so, we will argue that
on
. We begin by showing that the correspondence
is nondecreasing.
24 Let
and
, noting that
since
is continuous by Condition 1. Define
and suppose to the contrary that
. Then it must be that
, and thus
. By definition,
, and thus
. Together with the previous inequality, this implies that
, requiring that
, a contradiction.
Next, observe that is convex valued due to the quasiconcavity of in z guaranteed by Condition 2. Let . If , then by definition, , so . Suppose that . Since is nondecreasing, is nonincreasing from Condition 4, and , it follows that, for any price , there is some quantity such that . Further, since is convex valued, it follows that , and so . Thus, and is thus strictly quasiconcave on , and so is the unique maximizer of on .
Next, we argue that is strictly increasing on , which may be empty. Suppose that is nonempty. Let . Since , the definition of implies that there is some quantity such that , and so . This implies that , and so it must be that . By definition of , . Since D is continuous by Condition 4, it follows that for some . Thus, . We conclude that is strictly increasing on . Since is strictly quasiconcave on , it follows that is strictly increasing on , and therefore that is strictly increasing on . We conclude that is the unique maximizer of .
It remains to demonstrate that . From Condition 4, there is some such that . For such a price x, Condition 3 implies that there is some quantity such that , implying that . Thus, . It follows that .
Lastly, we must argue that
for all
. To do so, we will show that
for all
, as this will imply that
and thus
. From Condition 6, it is sufficient to show that
for all
. Suppose to the contrary that
for some
. Since
D is nondecreasing by Condition 4, it follows that
. Observe that
This contradicts
as the maximizer of
. We conclude that
for all
.
Verification of Assumption 4: For each firm i, there exists a price such that Define the set and . We will first show that , implying that . We will then show that if and , then . Lastly, we will show that if , then for all .
First, we have already verified Assumption 3, which guarantees that for all . By Condition 8, , so . Thus, .
We argue that for any quantity , . Suppose to the contrary that there exists a quantity such that . Since D is continuous by Condition 4, it follows that there exists a price such that , and thus , contradicting as the unique maximizer. Thus, . It follows that, for any quantity , .
As we have argued that , it follows that , which requires that . Thus, from Condition 6, . It follows that , and thus . Thus, .
Let
and
. Suppose that
. Then by definition,
, and thus
by Assumption 1. We first argue that
. Suppose to the contrary that
. Then observe that
Thus,
Further, note that
Thus,
This is a contradiction, and so we conclude that
.
Second, we show that . Suppose to the contrary that . Then from Condition 6, it must be that , so . Since Condition 3 implies that for all for sufficiently small z, then it must be that . This further implies that , and since we already verified Assumption 2, this implies that , so . This is a contradiction, so we conclude that .
Observe that since
then
. This further implies that
for any quantity
. The following are equivalent:
Let
and observe that
This can be solved to find
. Thus,
. Since
is quasiconcave by Condition 2, it follows that
for all
. Thus, it must be that
. Since
and
is nonincreasing by Condition 5, it follows that
. Therefore,
, and so
. Thus,
, a contradiction. We conclude that
.
Finally, we argue that if , then for all . Let . Since , it follows by definition of that . Thus, since we already verified Assumption 1, we conclude that . Let and observe that , where . By Condition 5, is nonincreasing in , so for all . Thus, since , it follows that . Further, since by Condition 5, it follows that . Therefore, for all .
Verification of Assumption 5: for each firm i.
This is identical to the statement of Condition 8.
Before proceeding with the verification of Assumptions 6 and 7, we will show that and are continuous in on .
We have already shown that for all . Thus, we need only check continuity on . We will first show that is continuous in on and separately, and finally, that is continuous at .
By the definition of , for all . It follows that . Since is continuous by Condition 1, the theorem of the maximum implies that is continuous in x for all x, and thus, is continuous on .
Define , noting that exists since is continuous by Condition 1. Note further that is nondecreasing, as we showed in the verification of Assumption 3 that is a nondecreasing correspondence. Define and . From Condition 2, is quasiconcave in z, and thus for any . In particular, by choosing and , this implies that . Similarly, by choosing and , this implies that . Note that, for all , . Thus, since D is continuous by Condition 4, it follows immediately that is continuous at all , and thus, is continuous on .
Next, let . Since is continuous in x as noted above, it follows that . We have already demonstrated above that is strictly quasiconcave and that . Thus, is strictly increasing up to the unique maximizer . Therefore, it must be that . We conclude that is continuous at and thus continuous on .
The proof that
is continuous in
is similar to that of
. Define
.
25 For
,
. Thus,
. Condition 5 implies that
is continuous in
, so
is continuous in
at any
given
since
is continuous by Condition 1. For any
,
. As such,
. Thus, the theorem of the maximum implies that
is continuous in
x, and so
is continuous in
at any price
such that
.
Suppose that
. As was the case for
,
. Suppose that
and let
be a sequence of prices such that
for each
k and
. Then from the continuity of
in
at prices
, there exists a
such that
for all
. Let
. If
, then
contradicting
. Thus, it must be that
. It follows from the continuity of
in
by Condition 5 that
. Observe that
is continuous in
since
is continuous by Condition 1, and thus,
. This contradicts the supposition that
. We conclude that
is continuous at
given
.
Lastly, suppose that . Observe that this means that and thus, for all since is nonincreasing in by Condition 5 and is nondecreasing. It follows that for all . The continuity of is thus guaranteed by the fact that is continuous in by Condition 5, while is continuous in since is continuous by Condition 1.
Verification of Assumption 6: There exists a lowest price such that for some price , such that and for all and .
Define the set such that and and ; then define and . It is sufficient to show that for each firm i. We first argue that each is nonempty, and thus, is finite. We will then show that , and lastly that .
To demonstrate that each is nonempty, we will show that . Define and observe that . By the definition of , for all . From Condition 5, for all . As such, since and , it follows that , and thus, . Thus, each is nonempty and is finite.
For each , let be as in the definition of and let and let be a sequence of prices such that for all k, , and . Since , it follows that . Observe that , and since is strictly quasiconcave on , it follows that is strictly decreasing on . Thus, it must be that for all . By definition, for all and all . It follows that for all and all . As shown above, is continuous in , and so . This implies that for all . Therefore, .
Lastly, consider the price and corresponding as in the definition of . By definition, and for all and . In particular, this holds for all , and thus, .
Verification of Assumption 7: Both and are continuous in on . is right upper semicontinuous in , that is, for any sequence such that and .
We have already shown that
and
are continuous in
on
and thus are continuous on
. It remains to show that
is right upper semicontinuous in
. By Condition 5,
is right continuous in
. Let
be a sequence such that
for all
k and
, noting that
. Observe that
. Since
is nonincreasing in
by Condition 5, it follows that
only if
. In this case,
. Note that
Since
and
, it follows that
. Thus, since
is continuous by Condition 1, we may conclude that
, so
is right continuous (and thus right upper semicontinuous) in
. □
Proof of Proposition 6. We begin by showing that under each set of condition is a pure strategy equilibrium.
B.1 Consider any firm i. From Assumption 2, for all . From the assumption of B.1, for all , and so for all . It follows that for all . Thus, neither firm possesses a profitable deviation, so each firm pricing at is an equilibrium.
B.2 Consider firm i with . As noted in the (B.1) case, firm i has no profitable deviations. From the assumptions of B.2, observe that . From Assumptions 2 and 3, is nondecreasing on , and so for all . Lastly, the assumption in B.2 that for all guarantees that there are no profitable deviations for firm j to prices higher than . Thus, neither firm possesses a profitable deviation, so each firm pricing at is an equilibrium.
C. From Condition 6, . Since , it follows that . Observe that if , then by definition . Suppose that . Then from the assumption of C, , so . It follows that for each firm i, and so . As noted above, is nondecreasing on , so there are no profitable deviations to prices . Further, since , there are no profitable deviations to prices . Thus, neither firm possesses a profitable deviation, so each firm pricing at is an equilibrium.
Next, we prove that any pure strategy equilibrium must satisfy either B.1, B.2, or C.
Proposition 4 implies that any pure strategy equilibrium must be symmetric with . Further, by definition, it must be that and for each firm i.
B.1 Suppose that and that for some for some firm i. Then since , from Condition 3, there is some quantity such that , so . Since , it follows that . This contradicts as an equilibrium since .
B.2 Suppose that . If for some , then the preceding argument for the (B.1) case applies and rules out as an equilibrium. If , then by continuity of by Assumption 7, there exists a price such that . This contradicts as an equilibrium. Lastly, suppose that for some . Then since , then x is a profitable deviation from , violating as an equilibrium.
C. Suppose that . If for some firm i, then by definition, any price is a profitable deviation for firm i, violating as an equilibrium. Lastly, suppose that for some firm i with . Then since by Condition 6, it must be that , so . Therefore, since and , it must be that . Then, since is continuous by Assumption 7, there exists a price such that . This violates as an equilibrium. □
Proof of Proposition 7. Without loss of generality, we assume . Suppose to the contrary that there is an equilibrium price with . Then , so Condition 6 guarantees that for all . It follows immediately that for all . From Condition 8, . Therefore, by definition of , we have , contradicting as an equilibrium price. □
Proof of Proposition 8. Since
D and
are unchanged, then
for all
x. Note that
It follows immediately that
. The fact that
and
follows immediately from their definitions. □
Proof of Proposition 9. The fact that and follows directly from Proposition 8 since any change in firm i’s supply has no impact on the front-side profit of firm j, so for all x. It remains to show that and . Observe that for all prices .
Define , with defined analogously for . Let and , with and defined analogously. Then note that . Note that is nondecreasing.
Part 1:
The proof that is conducted in four steps. In Step 1, we argue that for all and then use that fact to argue that we show that . In Step 2, we show that for all . In Step 3, we argue that for all , implying that for all . Finally, in Step 4, we show that if , we can find prices such that , contradicting the assumption that is nondecreasing in q.
Step 1: We first show that for all . Suppose to the contrary that for some . As noted in the proof of Proposition 6, . Thus, . Since is continuous by Assumption 7, there exists a price such that . Observe that . This contradicts as the maximizer of . We conclude that .
We next argue that . Suppose to the contrary that . Since by Assumption 5, we may choose a strictly increasing sequence such that and . By definition of , for all . Thus, since is nonincreasing as noted earlier, this implies that for all k. By continuity of by Assumption 7, , and so . Let and note that by Assumption 1. Observe that . Since is the unique maximizer of , it follows that . As demonstrated in the proof of Proposition 6, and since . By Condition 6, this implies that , so . From the first paragraph of this step, this implies that and thus that . A contradicts to . We conclude that .
Step 2: We show that
for all
. Suppose to the contrary that
for some
. Note that
Since
, it follows that
This contradicts the definition of
as
. We conclude that
for all
.
Step 3: We argue that for all . Let and suppose to the contrary that . Then since , it must be that . It follows that , and so . Since is nondecreasing and is nonincreasing by Condition 5, it follows that for all such that . Thus, for all and such that . Thus, for all , contradicting the definition of . We conclude that for all .
In summary we have now established that, since is nondecreasing, then for any , it follows that .
Step 4: We argue that
. Suppose to the contrary that
. Then for any price
, it must be that
. Let
and
and note that from above,
. Note that
is nonincreasing in
x since
and
is nonincreasing in
x by Condition 5. From above, we may choose the price
such that
. Since
is strictly increasing on
by Assumption 3 and from continuity of
D from Condition 4, we may choose
x so that
and
. Thus, we have
Note that
and
. Putting these together, we have
Next, note that
and
. Putting these together yields
The inequalities (
A1), (
A2), and (
A3) together imply that
which contradicts the assumption that
is nondecreasing in
. We conclude that
.
Part 2:
The proof that is also done by contradiction. Suppose to the contrary that . Recall that . As noted above, is nonincreasing in . Further, by Assumption 1, we can conclude that , and thus . Additionally, since is continuous by Assumption 7, it follows that .
Let
be a sequence such that
. We may without loss of generality choose this sequence such that
for some price
. It follows that
, where
. By definition of
, it must be that
for all
k. Further
, and so
. Therefore,
We briefly argue that . To see this, suppose to the contrary that . Then note that . By definition of , it must be that , and since , it follows that . Thus, Assumption 3 implies that is strictly increasing on . Since for by Assumption 2, it follows that , a contradiction. We conclude that .
We will now argue that for all x in some neighborhood .
We begin by arguing that
for all prices
. Suppose to the contrary that
for some
. Recall that
. Next, since
is quasiconcave in
z by Condition 2, it follows that
, where
. Note that by definition of
and
,
Since
and
, it follows that
, and so
. Thus,
. Therefore,
This is a contradiction. We conclude
for all prices
.
Now, suppose to the contrary that there exists a sequence with and such that for all n. Since , we may without loss of generality assume that for all k and n. Thus, from above, for all k and n. Since , this implies that , a contradiction. We conclude that for all x in some neighborhood . Choose such a .
Now observe that by definition of
,
for any price
. Thus, for any price
, it must be that
since
is strictly increasing by Assumption 3. Recall that
, and as shown above,
, so
is nonempty. Let
with
picked such that
for all
x in some neighborhood
. Note that
Observe that
and
Putting these together, we have
and thus
Recall that from (
A4)
It follows that
which contradictions the assumption that
is nondecreasing in
since
. We conclude that
. □