ADVANCES IN MATHEMATICS 26, 206-222 (1977)

A Variational Problem for Random Young Tableaux

B. F. LOGAN AND L. A. SHEPP*

Bell Laboratories, Murray Hill, Nm Jersey 07974

Stanley posed the problem of minimizing the functional

f(X)

dy lodf (4 - Y + f-‘(y) - x) (1)

over nonincreasing nonnegative f on (0, to) of integral unity. We show that the minimum is unique and has the value - 4 , as was conjectured by Stanley. The minimizing function f . , H( f,,) = - 4 , is given parametrically by

fo(x) = (2/?r)(sin B - B cos 6), x =focx) + 2 cos 8, 0 G e G T (2)

for 0 < x < 2; and fo(x) = 0 for x > 2. Closely related unpublished results have been obtained by Hammersley. We also find the minimum of H(f) subject to the constraints f(0) < a and f-i(O) = inf(x: f(x) = 0) < b where a and b are given. Proofs of the results for the case of constraints are com- plicated and will be given elsewhere. Let X, be the shape of the random Young tableau with n unit squares obtained from sampling from the Schensted distribu- tion where

W”) = 4eAJ, (3)

where rr(r\,J is the product of the n hook lengths of An. Consider the stochastic processes

X,(t) = (l/?V) &(t &a), n>l, t>o, (4)

where h,(t) is the height of the tableau h, at a horizontal distance t from the corner. We show

‘x" -fo (5)

in the sense of weak convergence in a certain metric, where f,, is the deter- ministic function in (2).

Let Z(U,,) denote the length of the longest increasing subsequence of a random permutation 0, of 1, 2,..., n. Hammersley showed that

l(O”)/?w + c in probability, n -+ co. 05)

Schensted showed that Z(uJ has the same distribution as X,(O) under the distribution (3) on X, . It has long been conjectured (apparently first by Baer and Brock) that c = 2. We show here that c > 2 as a by-product of (5).

* This paper was presented as a Special Invited Paper at the Annual Meeting of the Institute of Mathematical Statistics August 1975.

206 Copyright ~$3 1977 by Academic Press, Inc. All rights of reproduction in any form reserved. ISSN 0001-8708

RANDOM YOUNG TABLEAUX 207

1. INTRODUCTION

Baer and Brock [l] conjectured c = 2 on the basis of extensive computer calculations (see also [3]). Using a clever imbedding of the problem into the Poisson process, and results of Kingman [5], Hammersley [3] showed the exis- tence of the limit (5). Hammersley [3] made several attacks on determining the actual value of c, including an unpublished one similar to the present attack (but apparently not obtaining uniqueness).

The present attack relies on a combinatorial identity of Schensted [7], expressing the probability distribution of Z(a,) in terms of Young tableaux. A Young shape [7] of size n is an array of n unit squares, left and bottom justified (note that the array is usually top justified, but we prefer to think of the shape as a nonnegative function) whose column lengths are nonincreasing from left to right, as in Fig. 1. With each square S E X = An , let L(S; h) denote the hook length of S in A, i.e., the number of squares in /\ directly to the right and directly above S, counting S itself exactly once. The hook lengths are written in each square of Fig. 1 as an illustration. Let T = p(h) denote the hook product, i.e., the product of all the integersL(S; h), S E h. Then, Schensted [A

m4 = 4 = ,(& $i 3 k = 1, 2 ,..., n, (l-1)

where the sum is taken over all shapes h = X, with n squares where the first (leftmost) column has length X(0) = K. From (1.1) we see that if we randomly choose a shape h, of size n according to the distribution

y=*(xl

4 n=lo 1 r(X)= 8.6.4.3*.2*

3

3 1

2.

4 2

1

8 6 3 2 1

0 1 2 3 4 5 - II * s-?Y)

FIG. 1. A Young tableau for n = 10 with hook lengths P(h) = lO!/ti(X).

(1.2)

208 LOGAN AND SHEPP

(note [7] that these probabilities actually sum to unity) then h,(O), the length of the first column of h, , is a random variable with the same distribution as Z(U~). A direct mapping of a, + X, is given by Schensted [7], but we make no

use of this. Since T@,) varies greatly with h, it is expected and we shall prove that P&J is concentrated very sharply around one specific shape X,, foreach n, where

7r&,) = In” ?T(h,). (1.3)

We would then further expect that Z(an) would be concentrated around As,(O). The problem (1.3) of minimizing I thus arises in finding the most likely

Young tableaux under the distribution (1.2). The problem (1.3) also arises in finding the maximal degree of an irreducible representation of the symmetric group S, . Frobenius [6] gave a one-one correspondence between a shape h (whose column lengths correspond to a partition of n) and an irreducible repre- sentation of the symmetric group as a group of matrices (with complex entries) of common size f h x f A. The size fA is called the degree of the representation corresponding to h in the Frobenius correspondence, and using a formula of Frobenius, it was shown [9], that the degree f A is given by f * = n!/r(h). The question of finding the shape of maximum degree fA thus again reduces to the problem (1.3) of minimizing n(X), X = h, , and was apparently first mentioned in [lo, Footnote 91, as was brought to our attention by Stein. In [ll], the com- puter calculations of h,, in [l] are extended from n = 36 to n = 75. For n = 75, [ll] shows that X,, has column lengths 14, 11, IO, 8, 7, 6, 5, 4, 3, 2, 2, 1, 1, 1 which does begin to have the general shape predicted by our asymptotic result (2), (4), and (5) and shown in Fig. 3.

Thus it is of interest to find shapes h = h, minimizing rr(h,) in (1.3). We express this problem as a calculus of variations minimization of a functional. Note

log Tr(X) = 1 logL(S; h) SEA

= g* h(Wxs) - Ys + WYS) - 4, (1.4)

where X(x) is the height of the shape X at x, when X is placed at the origin in the first quadrant of the x-y plane, (x~ , ys) is the center of the square SE X, and h-l is the inverse function to X. Since log is a convex function, and S has unit area,

&$(x,) - ys + A-l(ys) - xs) 3 JJs log(+) - y + qy> - x) A &. (1.5)

Normalizing X to have unit area by setting

f*(x) = (l/n’/“) /\(nrl%), x30 (1.6)

where n is the number of squares in X = X, , and defining for any f E 9, where .F

RANDOM YOUNG TABLEAUX 209

is the class of nonnegative nonincreasing functions on (0, co) of integral unity,

H(f) = la /‘@) log(f(x) -Y +f-YY) - 4 dY dx (1.7) 0 0

we have from (1.2), (1.4)-(1.7),

P(h) < n! exp [-2 C JJ log(W - Y + WY) - XI dY h] SEA s

=n!exp[-2~~~‘)10g(A(r)-y+~-‘(Y)-~)dy~]

= n! exp [-28 S, 6”“’

(l-8)

log n”“(fn(4 - Y +f?(Y) - 4 dY q

= (n!/n”) exp[-2nH(fJ] N (2?m)l’* e-an(H’A)+*).

Since the inequality in (1.8) is reasonably tight it is natural to expect (since P(A) < 1) that min Hcf) = -3. Further if A, is any set of A,, for which X = h, E A, implies that

WA) > -4 + 6, s > 0, (1.9)

where 6 is independent of n, then since the number p(n) of shapes X of order n is the number of partitions of n, [4],

p(n) = exp O(n1/2), (1.10)

we see from (1.8)-(1.10) that

P(A,) < p(n)(2?7n)%-s”s -+ 0, exponentially. (1.11)

In particular if A, is the set of h = A, for which X,(O) < (2 - E) nl/s, or fA(0) < 2 - E by (1.6) we will show that (1.9) holds. It then follows using (1.11) that P(2(u,)/n1P > 2 - 6) = P(A~(0)/rz1/2 2 2 - l ) -+ 1 for every E > 0, i.e., that c > 2 in (6). The reason that the present method fails to yield c = 2 is that there are f E 9 with f (0) arbitrarily large but with Hcf) arbitrarily close to -4. Of course such f (x) fall quickly to 2 in the neighborhood of x = 0. Thus the continuous function problem of minimizing H(f) no longer is a good approx- imation to the discrete problem of minimizing n(h,J.1

We show that in a somewhat complicated metric d on F’, the stochastic processes,

X,(x) = (l/n’/“) h,(xnlb), x>o (1.12)

* Vedtik and Kerov (cf. LbkL Acad. Nauk 233 (1977), 1024-1028) show c = 2 by a separate combinatorial argument based on the notion of Young diagrams.

60712612-g

210 LOGAN AND SHEPP

converge as n -+ co weakly in the sense of [2] to the deterministic (i.e., constant) processf,(x) in (2). It is probably also true as a consequence of weak convergence in the metric d defined below, but it is not proved, that

fYl L&(x> - fowl 2 4 - 0, R-+03 (1.13)

for every x > 0. Of course if (1.13) could be proved for x = 0 then c = 2 would follow immediately sincef,(O) = 2 by (2).

To define the metric d, let for f~ 9, Tf denote the function on (-co, W)

defined by

Tf (6) = mi@, f (4) if [=x-f(x), x20 (1.14)

setting Tf (6) = 0 if -co < 5 < -f (0). It will be seen in Section 2 that g = Tf is nonnegative and integrates to unity. We define

where d(f, 3 fi) = Q(Tfl - TW2 (1.15)

Q(g) = t j-1 I !(w)12 I w I .dw (1.16)

and 8 is the Fourier transform of g. Since Q1J2 is a norm, it is easy to check that d is a metric on 9.

We will obtain in Section 2 the lower bound for H(f), f E 9

H(f) B Hcfo) + SW- Tf,)

= -3 + d2(f,fo) (1.17)

by (1.15). Let A, be the set of h = A, for which d(& , fo) > 6. By (1.17) we see that (1.9) holds and (1.11) follows, fA = A, -+ f. weakly in (3, d), i.e.,

f’(d(k ,h) 3 6) -+ 0 exponentially. (1.18)

We will show (2.35)-(2.41) that if f (0) < 2 - E then Q(Tf - Tfo) > 8 > 0 which, using (1.17) gives (1.9) and shows c > 2.

We state in Section 3 the more precise result determining explicitly the value of the infimum H,,(a, b) of H(f) over f E 9 subject to the constraints f(0) < a and f-l(O) = inf{x:f(x) = 0} < b. Note that by Schensted’s correspondence Ho(a, b) gives an asymptotic index via (1 .S) of the probability that the longest increasing subsequence and the longest decreasing subsequence have lengths less than un112 and &l/2 9 respectively. The unique minimizing functions fo(x; a, b) are given explicitly. Graphs of some representativef,(x; a, b) are seen in Fig. 3. The graph labeled 1 in Fig. 3 is that of the unconstrained minimizing f,, of (2). Proofs of the results stated in Section 3 will be given elsewhere.

RANDOM YOUNG TABLEAUX 211

It is of course natural to expect that for appropriate narmalizing constants c, + co (perhaps c, = tP would do) the stochastic processes

Aa@) = 4&) -h(t)), t 2 0 (1.19)

would tend weakly to a nonzero limiting process W(t), t > 0, as lt + co. It would be of interest to know what the process W is. It is clear only that W integrates pathwise to zero and that W(t) > 0 for t > 2. Perhaps W(t) = 0 for t > 2 and is the Wiener process in 0 < t < 2 conditioned to integrate to zero over [0,2] and to vanish at 0 and 2, but this is just a guess.

We mention that Greene [12, Theorem 3.11 has given an interpretation of the length of the ~5th cohunn, for K > 1, of the Young tableau, A, in terms of increasing subsequences of the corresponding permutation, a, under the Schensted correspondence. Namely, he shows that the number of entries in the first k columns of X is the maximum length of the union of K increasing sub- sequences of a.

2. THE VARIATIONAL PROBLEM

Let 9 be the class of nonincreasing nonnegative functionsf(x) on (0, 00) with

s omf(x)dx = 1. (2.1)

Consider the region R in the x-y plane defined by 0 < y < f(x), 0 < x < a illustrated in Fig. 2.

FIG. 2. The hook length for a continuous Young tableauf.

212 LOGAN AND SHEPP

The hook distance of a point (x, y) in R is defined as 4 + d, where

4 = f(x) - Y, d, = f-‘(y) - x.

We are interested in minimizing the functional

= ff 1ogW) - Y + f-‘(r) - 4 dx dye

R

It is important to make the right change of variables in (2.2). It isinstructive to do this in stages. Assume for the moment that f(0) = Q <’ co, f-l(O) = b < co, and that f is absolutely continuous. Setting y = f(t) in (2.2) and using the change of variables

x --f(x) = -5, x = h(5); t-f(t) =‘I, t = h(7))

we obtain after a short calculation

H(f) = - /‘f’(t) dt 1” log(f(x) -f(t) + t - x) dx 0 0

= f+ (1 - r(71)) 4 jn logh - E) Wd d5 -a

= s s bdq ’ log(rl - e-1 h’(5) d5 --la -a

The further transformation

(2.3)

(2.4)

(2.5)

gives a great simplification. It is easy to verify that there is a one-one corre- spondence between f E 9 and g given by

f(x) = km - BE + 4 I f I, (2.6)

RANDOM YOUNG TABLEAUX 213

where the inverse g = 2”’ is given by (1.14). The function g has the properties that

0 < g’(6) < 1 for f < 0,

-1 <g’(5) d 0 for 4 > 0, (2.7)

j-1 g(t) d5 = So”f(x) dx = 1. (2.8)

Even functions g correspond to functions f with f = f-l as follows from (2.6) by replacing [ by -5. By (2.7), g is absolutely continuous whether or not f is, and from (2.4), (2.5), after a calculation

H(f) = I(g) A Q(g) +W

where (since g’ vanishes outside of (--a, b))

(2.9)

(2.10)

is a quadratic functional of g’ and

L(g) = - jm g’(x)(x log 1 x 1 - x) dx -m

(2.11)

is a linear functional of g’. Note that since there is a one-one correspondence between f E s and g we may regard the problem (1) of minimizing H(f) as that of minimizing the much simpler functional I(g) = Q(g) + L(g) in (2.10H2.11) over g E 3 where c!? is the class of functions satisfying (2.7) and (2.8), i.e., g E 9 has 1 g’ / < 1, and g is unimodal with mode at zero, nonnegative, and of integral unity. Showing uniqueness of the solution to the problem of minimizing H(f) is of course the same as that for minimizing I(g).

We next express the functional I(g) of the Fourier g and Hilbert g” transforms

of g,

k(x) = /g(t) dxt dt (2.12)

g’(x) = ; j $ dt. (2.13)

Formally integrating by parts, which can be justified easily using the fact that g is unimodal and integrable,

B(x)=+jg’(t)logIx-tldt (2.14)

so that from (2.10)

Q(g) = - + j g'(x)&) dx. . (2.15)

607/26/z-9

214 LOGAN AND SHEPP

Since the Fourier transform ofg” is i(sgn w) d(w) and that of g’ is iw&w) we obtain from Parseval’s identity,

Q(g) = Q j Id(~ I w I &J (2.16)

so that Q is a positive definite quadratic form. Thus integrating by parts in (2.11) gives from (2.9)

U) = jR(x)logIxI~~+tjI$(w)l2lwl~~. (2.17)

We can obtain a crude lower bound for I(g) using (2.7),

I(g) s(g) = -1 -J’g~(x)sgnxjxIiog~x~dx

I ’ (2.18)

3 -l- 1 x 1 log 1 x / dx = -; -1

since Q(n) > 0. To get the best lower bound (-4) introduce a comparison .- .-. function g, E G!? and write for an arbitrary g E 8, from (2.15)

Qk - go) = - $ j k’(x) - go’(xNW) - &o(xN dx

= Q(g) + QW + T j g'(x) &&4 dx.

Thus,

where

W = Q(g) + %) = Qk - go) - Qko) - %, go)

R(g, go) = j g'(x)(x log I x I - x + 444 dx.

We will exhibit a g, E 3 for which

p(x) A x log I x / - x + 7&(x) = 0 for lx]<2

>o for x > 2

<o for x < -2

go(x) = 0 for / x 1 > 2.

(2.19)

(2.20)

(2.21)

(2.22)

(2.23)

Assuming (2.22) and (2.23) hold for some g,, E 9, note that the integrand in (2.21) is identically zero in x when g = g, , i.e., R(gO , g,,) = 0 so that from (2.20)

%J = Q(O) - Qko) - Wo 9 go) = -Q(gd (2.24)

RANDOM YOUNG TABLEAUX 215

Further, since g E Q satisfies (2.7) the integrand of (2.21) for any g E B is non- positive because of (2.22) and so R(g, g,) < 0. Thus from (2.24), for any g E 9

U) 2 %J + Q(g - go) (2.25)

which gives (1.17) where g,, = 2”‘s and f0 is the function in (2.6) corresponding to g,, . Since Q(g - g,,) 2 0 and Q(g - g,) = 0 only for g - g, = 0 from (2.16), the uniqueness of the minimizing g,, (or f,,) follows.

It remains to produce a g,, E B satisfying (2.22) and (2.23). The function g, was found by one of us (BFL) and although we considered giving a motivation for the choice of g,, it seems best to simply give g, and verify its properties.

Consider the function

G,(z) = 1 (( 1 - $)“’ + $) - $ log (:)

-“log((1-$)l’2+$), (2.26)

?T

which is analytic in the upper half-plane Im z > 0. On the real axis z = x,

Ix1 <2

- 1)1’2, (2*27) 1 x 1 > 2

as can be seen by indenting around z = +2. Thus on the real axis z = x

log ((1 - $,,‘, + i;) =

i sin-l x ,

gsgfx + l;;; ,;; - _ (f - f2), (2*28)

IxI>2.

We then verify that on the real axis z = x,

(2.29)

216 LOGAN AND SHEPP

It is easy to check from (2.29) that G,(z) = 0(1/l a 1) as / z / 4 co in Im z 3 0 and so [8, p. 1251, the real and imaginary parts of G,,(X) are conjugate Hilbert transforms, i.e.,

go(x) = Re G,,(X) =

3(1 -$)1;2+~sin1~---!$, 1x1 <2 (230)

0, [xl>2 *

has Hilbert transform

&,(x) = Im G,,(x) = - 1)1’2) - G log 1 x I (2.31)

-$log(!$ (f- 1)1’2), 1x1 > 2.

It is easy to check from (2.30) and (2.31) that ga satisfies (2.23) and that in

(2.22),

p(x) = x [log (F + ($ - 1)1’2) - & (f - 1)1’2], I x I > ‘;z 32)

It is easy to check that p(x)/x has a positive derivative for / x 1 > 2 and since it vanishes for x = 2, (2.22) follows easily. We must check also that g, E 9, i.e., (2.7) and (2.8) hold. Since

&l’(x) = ; ($ 2 sin-l ” - sgn x ), 1x1 G-2, (2.33)

we verify directly that (2.7) holds and

- j

xgo’(x) = go(x) dx = 1, s

(2.34)

i.e., g, E 9. Since we have exhibited g, E 3 satisfying (2.22) and (2.23), the argument below (2.23) shows that (2.25) h o Id f s or every g E 3. By the one-one correspondence between g = Tf and f we have proved (1.17). Note that (2) follows from (2.30) by setting E = 2 cos 8, 0 < 0 < T.

We next prove that if f(0) < 2 - E then Q(Tf - Tfo) > S > 0 which by (1.17) gives c 3 2 as pointed out at the end of Section 1. Note first that g = Tf satisfies

g(c3 = 0, 5 < 42 - 6) (2.35)

RANDOM YOUNG TABLEAUX 217

because from (1.14), g(5) = T’(5) = 0 for [ < -f(O). Since g,,‘(t) < 0 for -2 < 6 < -(2 - E) by (2.33) we see that K(t) defined by

K(t) = jm (g’(t - u) - g,,‘(t - u)) e-u du 0

(2.36)

satisfies, for -2 < t < -(2 - E),

K(t) = -I”” gi(t - u) e-” du > e-(2+t)go(t), 0

(2.37)

whence for any g = Tf with f (0) < 2 - E, there is a fixed 6, > 0 with

I K2(t) dt > 8. (2.38)

From (2.16) and the inequality 2 1 w j < 1 + w2,

Q(Tf - Tfo) = Q(g - go)

t(w) - Jo( I w I dw

&JJ> - ~okJ)12 -gg lwldw

iwj(w) - iwto(w)12 / & 1’ dw

= ,rr J K2(t) dt > 6,

(2.39)

where we have used (2.36) and Parseval’s identity in the last line. This shows that f (0) < 2 - E implies that

H(f) < -4 + 6 WY

so that (1.9) holds, which yields c > 2. We remark that the linear part in (2.20) g ives an inequality similar to (2.25)

which may be useful in estimating H(f) = I(g),

&T) 2 %o) - a!, go). (2.41)

This immediately follows from (2.20) and (2.24) since Q(g - go) > 0.

218 LOGAN AND SHEPP

3. MINIMIZATION UNDER CONSTRAINTS

Let .9& denote the set offE 9 for which

f(o) d a, f-l(O) = inf[x: f (x) = 0] < 6. (3.1)

If 0 < a < co, 0 < b < co, and ab > 1,9& is not empty and we denote

fb(a, 4 = &‘, Wf ), (3.2)

H,,(a) = i:f H,,(a, b), (3.3)

H,, = i:f H,(a). (3.4)

It follows from the results of Section 2 that

Ho = -4 = Ho(a, b), a 3 2, b > 2, (3.5)

since f0 E Fa,b for a > 2, b > 2 and achieves the minimum of H(f) with no constraints.

To state the explicit formula for Ho(a, b) and H,(a) it is enough to consider the case a < b by symmetry and by (3.5) we may also assume that a < 2. Let

6, = b,(a) = f - : + (2 + $-)“‘, a < 2.

Then, for a < 2,

(3.6)

f&W = f&h 4J

= f&da, b), b 3 6,. (3.7)

For the case a < 2, a < b < b,,(a) the values of H,,(a, 6) and the unique minimizing functions f,,(x; a, b) E 9& are given as follows:

Starting with new fundamental parameters 01 and /3, 0 < 01 < 1, 0 < /3 < 1, define -

1 +(Y2 c=-, dz1+p2 1 - /3”

2a --Tp y= 2p 1 - LX2 ~, (3.8a)

2a -.

A = 2 + (1 - a2)(1 - B”> a2 + P” 4 4@

-4’

a = 1 - B” + 243 2j3A”2 ’

b = 1 - a2 + 243 2orA1r2 . (3.9)

For a and b satisfying

O<a<2, a < b < h,(a) (3.10)

(3.8b)

BANDOM YOUNG TABLEAUX 219

there is a unique pair 0 < a < /3 < 1 satisfying (3.8) and (3.9). For (a, 6) satisfying (3.10), the unique minimizing function f&; u, b) E SaSa satisfies the constraints (3.1) with equality, i.e.,

MO; 4 4 = 4 &‘(O; a, b) = b, (3.11)

whereas if, e.g., a < 2, b > b,(a) then f$(O; a, b) = b, . Thus for (a, 6) satisfying (3.10) we have

H,(u, b) = -& - i - f log(4ajL4) - t (b2 - 9) log t

+ (a+og

(3.12)

4 + 5 (ub - 1) log ap

and

f,(x; a, b) = a, (3.13)

foP-; 4 6) = $-$: , (3.14)

fo(x; 4 b) = 0, x 3 b, (3.15)

f&G a, 4 = d- cos8 1

Al/Z + ,~1/2 (sin 0 + e(a - B + cos 4)

+ 2(cos B - d) tan-l /3 sin e

i - p ~0~ 8

- ~(COS e + C) tan-l a sin e

i + a cos e ’ (3.16)

where 0 < 0 < n and

x = f&; a, b) + cos e - y

All2 . (3.17)

The formulas simplify somewhat in the extreme cases b = a and b = &(a). Forb = awehavea =/3and

Ho(U, U) = 1 + ,9” _ a4 - 2 - i log( 1 + a2 - a”)

u2 - 1 +-T-

2 log a2 + 2a2 log 1 + a2 ,

where a is determined by setting a = /? in (3.9) as

(1 + a212 a2 = 1 + a2 _ a4 ’ 1 <u<2.

(3.18)

(3.19)

220 LOGAN AND SHEPP

In the limiting case a = 1, we have n = 0 and

E&)(1, 1) = -8 + 2 log2 (3.20)

which corresponds to fs(x; 1, 1) = 1, 0 < x < 1, which is of course the sole member of Sq,i . In general, fs(x; a, u) does not greatly simplify.

For the case a < 2, b = b,(u), we have /3 = 1 and

H,(u) = H&z, 6,) = -1 + f + log ; - (1 + f) log -& (3.21)

l f j&; UI, &I) = ~ n-Al/” 1 cd + sin @ - Z(c + cos 0) tan-r

01 sin 6 1 + olsin 8 ’

cos 6 - y (1 - Lx)” (3.22)

x = f&Y; a, 6,) + ~- Al'2 ' 2aA1J2 <x<b,

where 0 < 8 < rr, and

u’ f&; 4 44 = I

o<s<;+~-(2+q2 (323)

0, x 3 6,

and from (3.8) and (3.9),

L&z2 a2 -

4+u2' 6, = ; - ; + (2 + q2 (3.24)

Thus when b >, b,(a), a < 2 the optimal f,,(x; a, b) is continuous at 3~ = b, where it vanishes. However, under the constraintf(0) < 2 the optimal function is constant over an interval including the origin. This interval is relatively small even for a = 2ri2 having a length (5(2’/*)/4) 31f2 = 0.0357. For comparison, the triangular function f(x) = 2112 - x, 0 < x < 2rj2 is not optimal for the case a = b = 21j2. We have by direct calculation

H(f) = $(log 2 - 1) = -0.4602792, (3.25)

whereas from (3.18) and (3.21)

H,,(21/2, 21/2) = $ log 3 + $ log 2 - 5 log(1 + 31/2) + 2(3lp)

-Q = -0.4701211, (3.26)

Ho(21f2) = -$ - 2 log 2 + Q log 3 = -0.4883759, (3.27)

bo(21i2) = a21j2 + 3112 = 2.085604. (3.28)

RANDOM YOUNG TABLEAUX 221

It is perhaps worth remarking that the overall optimal function, fO E 9,

given in (2) satisfies

f&c; 2,2) = 2 - (3~rx/2)~/~ + O(x), X-+0 (3.29)

and fO is symmetric about the line y = x. Some graphs off&x; a, b) are given in Fig. 3.

f (xi

1. 012, bz2; asp=1

2. o=b= 1.1470... a=@= 0.5

3. 0=0.90476 b=1.7142...; . . . . a=0.5, 890.75

4. ~80.75592 bL2.6457...; . . . .

a’0.5, B=l

0 0.4 a6 1.2 1.6

FIG. 3. fs(+; a, b) for various values of (I, b.

The proofs of the results in this section are omitted because they are cumber- some. The basic method is the same as the proof for the unconstrained case, i.e., to find a function g&v; a, b) with conditions analogous to those in (2.22) and (2.23). We hope to give full proofs elsewhere.

Note added in proof. The result c = 2 has now been proved by A. M. Vershik and C. V. Kerov.

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222 LOGAN AND SHEPP

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