ON BELLMAN’S LINEAR SEARCH PROBLEM
FOLKMAR BORNEMANN
Let S be the set of turning point sequences x = ( xk )k∈Z satisfying
· · · < − x2k+1 < − x2k−1 < · · · < 0 < · · · < x2k < x2k+2 < · · · .
The linear search problem (see Alpern and Gal 2003, §8.2.1) amounts for solving
the minimax problem
∑ n +1 x k
V = inf sup k=−∞ .
(1)
xn
x ∈S n∈Z
Gal (1980, Chap. 6) developed a general theory for dealing with that kind of
minimax problems (using advanced concepts such as, e.g., submartingales). However, for the specific problem at hand, Beck and Newman (1970, pp. 421–422)
approached the minimax problem by using elementary methods only. In this note,
we present an even simpler version of their proof.
Theorem. The value of the minimax problem (1) is V = 4, which is attained if and only
if xk = α2k , where α is a positive constant.
Proof. Upon calculating
+1
k
∑nk=−
∞2
=4
( n ∈ Z),
2n
we get V 6 4. To show V > 4, let us assume to the contrary that, for some x ∈ S ,
sup
n ∈Z
+1
∑nk=−
∞ xk
6 γ,
xn
n +1
∑
that is,
k =−∞
xk 6 γxn
( n ∈ Z),
(2)
with γ = 4 − β, β > 0. Set
yn =
1
2n
n +1
∑
k =−∞
xk .
Then yn > 0 and
n
n +2
n
1
2n+2 · (yn−1 + yn+1 ) = 4 ∑ xk + ∑ xk 6 4 ∑ xk + γxn+1
2
k =−∞
k =−∞
k =−∞
n +1
=4
∑
k =−∞
so that
n +1
xk − βxn+1 6 4
∑
k =−∞
x k − γ −1 β
n +2
∑
k =−∞
xk = 2n+2 yn − 2n+1 γ−1 βyn+1 ,
1
1
(yn−1 + yn+1 ) 6 yn − γ−1 βyn+1 6 yn .
2
2
(3)
Date: March 27, 2010. Manuscript written for use in the Proseminar “Search- and Pursuit-Games”.
1
FOLKMAR BORNEMANN
2
Thus, the sequence (yn )n∈Z is positive and concave and therefore, by the Lemma
below, necessarily constant: yn = η. By plugging this constant into (3), we arrive at
1
0 < η 6 η − γ−1 βη,
2
a contradiction.
Finally, if the value V is attained for some x ∈ S , then (2) holds with β = 0. As
the argument that led us to (3) applies here, too, we infer that yn = 4α for some
constant α > 0. Thus,
n
∑
k =−∞
xk = α2n+1
( n ∈ Z),
and, by taking differences,
xn = α(2n+1 − 2n ) = α2n
( n ∈ Z),
which concludes the proof.
Lemma. If the positive sequence (yn )n∈Z is concave, that is, if yn > 0 satisfies
y n −1 + y n +1
6 yn
( n ∈ Z),
2
then yn = const.
(4)
Proof. The concavity condition (4) is equivalent to the monotonicity
y n +1 − y n 6 y n − y n −1
( n ∈ Z).
By induction, we get
y n + k 6 y n + k ( y n +1 − y n )
(n ∈ Z, k > 0),
as there is
y n + k +1 = ( y n + k +1 − y n + k ) + y n + k
6 (yn+1 − yn ) + yn + k(yn+1 − yn ) = yn + (k + 1)(yn+1 − yn ).
If we had yn+1 < yn for some n, we would infer that
y n + k 6 y n + k ( y n +1 − y n ) → − ∞
( k → ∞ ),
contradicting the positivity assumption. Thus, yn+1 > yn for all n ∈ Z: the
sequence (yn ) is increasing. Likewise, since (y−n )n∈Z is positive and concave,
too, we obtain that (y−n ) is increasing or, equivalently, that (yn ) is decreasing: a
sequence, however, that is increasing and decreasing at the same time must be
constant.
References
Alpern, S. and Gal, S.: 2003, The theory of Search Games and Rendezvous, Kluwer Academic Publishers,
Boston.
Beck, A. and Newman, D. J.: 1970, Yet more on the linear search problem, Israel J. Math. 8, 419–429.
Gal, S.: 1980, Search Games, Academic Press, New York.