2.3 Proof of Proposition 1.3
Before we prove anything we would like to recall the fact that we can write see Remark ii after Theorem 1.2
f
+
J
1
, J
2
, h
1
= ψJ
2
, τ
+
J
1
, J
2
, h
1
J
1
, J
2
0, h
1
∈ R,
where τ
+
J
1
, J
2
, h
1
= t
+
J
1
, h
1
+ |J
1
− J
2
| and the map t 7→
ψJ
2
, t is continuous see Figure 1.2 for a picture. In the rest of the proof, we will use this fact without further notification. For example, the above immediately gives that
h
1
7→ f
+
J
1
, J
2
, h
1
is continuous at a point h
1
∈ R if h
1
7→ t
+
J
1
, h
1
is so. Proof of Proposition 1.3. We will only prove part a and c. The proof of part b follows the same
type of argument as the proof of part a. To prove part a, we start to argue that for given J
1
0 the map h
1
7→ t
+
J
1
, h
1
is right-continuous at every point h
1
∈ R. To see that, take a sequence of reals {h
n
} such that h
n
↓ h
1
as n → ∞ and note that since the map h
1
7→ t
+
J
1
, h
1
is increasing, the sequence {t
+
J
1
, h
n
} converges to a limit ˜t with ˜t ≥ t
+
J
1
, h
1
. Moreover, by taking the limit in the fixed point equation we see that ˜t = h
1
+ dφ
J
1
˜t 2.8
and since t
+
J
1
, h
1
is the largest number satisfying 2.8 we get ˜t = t
+
J
1
, h
1
. Next, assume h
1
6= −h
∗
J
1
and h
n
↑ h
1
as n → ∞. As before, the limit of {t
+
J
1
, h
n
} exists, denote it by T . The number T will again satisfy 2.8. By considering the different cases described in Figure
2.3, we easily conclude that T = t
+
J
1
, h
1
. Hence, the function h
1
7→ t
+
J
1
, h
1
is continuous for all h
1
6= −h
∗
J and so we get that h
1
7→ f
+
J
1
, J
2
, h
1
is also continuous for all h
1
6= −h
∗
J
1
. Now assume h
1
= −h
∗
J
1
. By considering sequences h
n
↓ −h
∗
J
1
and h
n
↑ −h
∗
J
1
we can similarly as above see that
τ
+
J
1
, J
2
, −h
∗
J
1
+ : = lim
h↓−h
∗
J
1
τ
+
J
1
, J
2
, h = t
+
J
1
, −h
∗
J
1
+ |J
1
− J
2
| τ
+
J
1
, J
2
, −h
∗
J
1
− : = lim
h↑−h
∗
J
1
τ
+
J
1
, J
2
, h = t
−
J
1
, −h
∗
J
1
+ |J
1
− J
2
| and so
τ
+
J
1
, J
2
, −h
∗
J
1
+ = τ
+
J
1
, J
2
, −h
∗
J
1
− ⇐⇒
h
∗
J
1
= 0. Since h
∗
J
1
= 0 if and only if 0 J
1
≤ J
c
the continuity of h
1
7→ f
+
J
1
, J
2
, h
1
at −h
∗
J
1
follows at once in that case. If J
1
= J
2
, then τ
+
J
1
, J
2
, −h
∗
J
1
+ = t
+
J
2
, −h
∗
J
2
τ
+
J
1
, J
2
, −h
∗
J
1
− = t
−
J
2
, −h
∗
J
2
and since ψJ
2
, t
+
J
2
, −h
∗
J
2
= ψJ
2
, t
−
J
2
, −h
∗
J
2
, the continuity is clear also in that case. If J
1
J
c
and 0 J
2
≤ J
c
, then τ
+
J
1
, J
2
, −h
∗
J
1
+ 6= τ
+
J
1
, J
2
, −h
∗
J
1
− 1815
t |h
1
| h
∗
J
1
, h
∗
J
1
t h
1
= h
∗
J
1
t |h
1
| h
∗
J
1
-30 -20
-10 10
20 30
-30 -20
-10 10
20 30
-30 -20
-10 10
20 30
-20 20
-20 20
-20 20
Figure 2.3: A picture of the different cases in the fixed point equation that can occur when h
1
6= −h
∗
J
1
. Here, d = 4 and J
1
= 3.
and the map t 7→ ψJ
2
, t becomes strictly increasing, hence h
1
7→ f
+
J
1
, J
2
, h
1
is discontinuous at −h
∗
J
1
. For the case when J
1
J
c
, J
2
J
c
, J
1
6= J
2
just note that h
1
7→ f
+
J
1
, J
2
, h
1
is continuous at −h
∗
J
1
if and only if a and b defined in the statement of the proposition are in the flat region in the upper graph of Figure 1.2.
To prove part c we take a closer look at the map J
2
, t 7→ ψJ
2
, t. By definition, this map is ψJ
2
, t = −h
∗
J
2
if t
−
J
2
, −h
∗
J
2
≤ t t
∗
J
2
t − d φ
J
2
t if
t ≥ t
∗
J
2
or t t
−
J
2
, −h
∗
J
2
. From the continuity of t 7→
ψJ
2
, t for fixed J
2
and the facts that J
2
7→ t
∗
J
2
, J
2
7→ t
−
J
2
, −h
∗
J
2
, J
2
7→ −h
∗
J
2
and J
2
, t 7→ t − d φ
J
2
t are all continuous, we get that ψ is jointly continuous and so the result follows.
1816
2.4 Proof of Proposition 1.4
Kategori