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A TURNPIKE THEOREM FOR A FAMILY OF FUNCTIONS* - Ivie

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and<br />

u(k u ;k u ) ¸u(x;y) +q u (y ¡x) for all(x;y) 2X; and<br />

u(k u ;k u )>u(x;y) +q u (y ¡x) for all(x;y) 2X;<br />

such that(x;y) 6=(k u ;k u ):<br />

(21)<br />

Proof: The existence of a subsequence fu ns g of fu n g and a concave function<br />

u 2 U ¹ such that limu ns =uis the content of i) in Lemma 3. Note that, without<br />

s!1<br />

loss of generality, we can supposeu n !u:<br />

Now, fk u n<br />

g ½ X 1 by A4, andq u n<br />

= D 1 u n (k u n<br />

;k u n<br />

) for anyn¸1by (16).<br />

Therefore, fk un g and fq un g are bounded becauseX 1 is compact and because of<br />

the uniform bound over functions and derivatives. Hence, there exists a point<br />

(k u ;q u ) 2X 1 £ < n + such that, without loss of generality,k un !k u andq un !q u<br />

(taking subsequences, if necessary): Note that(k u ;k u ) 2X:<br />

To prove that u(x;y)+ 1 2 B(x2 +y 2 ) is concave at (k u ;k u ) we will proceed<br />

by contradiction. Hence, we suppose this is not the case, that is, there exists<br />

(x;y) 2 X with (x;y) 6= (k u ;k u ) and ¸ 2 (0;1) such that if v¸ = (x¸;y¸) =<br />

¸(x;y)+(1 ¡¸)(k u ;k u ),<br />

1<br />

©¸u(x;y)+¸B<br />

2 (x2 +y 2 )+(1 ¡¸)u(k u ;k u )+B(1 ¡¸)(k u ) 2ª ><br />

u(v¸)+ 1B(x2¸+y2¸)<br />

(22)<br />

2<br />

Take v ņ := ¸(x;y)+(1 ¡¸)(k u n<br />

;k u n<br />

). Then (22) implies that for all n large<br />

enough we have<br />

©¸un (x;y)+¸B 1 2 (x2 +y 2 )+(1<br />

n<br />

¡¸)u n (k un ;k un )+B(1<br />

o<br />

¡¸)(k un ) 2ª ><br />

u n (v ņ )+ 1 2 B ((x (ku n;k u n)<br />

¸<br />

) 2 +(y (ku n;k u n)<br />

¸<br />

) 2 (23)<br />

)<br />

where(x (kun ;k un )<br />

¸<br />

;y (kun ;k un )<br />

¸<br />

)=v ņ ; because from Lemma 3 we have that<br />

u n (k un ;k un ) ! u(k u ;k u ); u n (x;y) ! u(x;y) and lim u n (v ņ ) = u(v¸) as well.<br />

n!1<br />

But (23) is in contradiction with assumption A6, thenu(x;y)+ 1 2 B(x2 +y 2 ) is<br />

concave at(k u ;k u ):<br />

This proves the …rst part of the Lemma.<br />

To show thatu(k u ;k u ) ¸u(x;y) for all(x;y) 2D such thaty¸x; we will<br />

suppose this is not the case to get a contradiction; …rst note thatuis indeed de-<br />

…ned at(x;y) 2Dfory ¸xbecause we have that(x;y) 2X by A4; now suppose<br />

17

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