u i log v i + 1 T (u v).

Transcription

1 where f(v) = n v log v s the negatve entropy of v. Soluton. The negatve entropy s strctly convex and dfferentable on R n ++, hence f(u) > f(v) + f(v) T (u v) for all u, v R n ++ wth u v. Evaluatng both sdes of the nequalty, we obtan u log u > v log v + (log v + 1)(u v ) = u log v + 1 T (u v). Re-arrangng ths nequalty gves the desred result Convex-concave functons and saddle-ponts. We say the functon f : R n R m R s convex-concave f f(x, z) s a concave functon of z, for each fxed x, and a convex functon of x, for each fxed z. We also requre ts doman to have the product form dom f = A B, where A R n and B R m are convex. (a) Gve a second-order condton for a twce dfferentable functon f : R n R m R to be convex-concave, n terms of ts Hessan 2 f(x, z). (b) Suppose that f : R n R m R s convex-concave and dfferentable, wth f( x, z) = 0. Show that the saddle-pont property holds: for all x, z, we have f( x, z) f( x, z) f(x, z). Show that ths mples that f satsfes the strong max-mn property: sup z nf x f(x, z) = nf x sup z f(x, z) (and ther common value s f( x, z)). (c) Now suppose that f : R n R m R s dfferentable, but not necessarly convexconcave, and the saddle-pont property holds at x, z: Soluton. for all x, z. Show that f( x, z) = 0. f( x, z) f( x, z) f(x, z) (a) The condton follows drectly from the second-order condtons for convexty and concavty: t s 2 xxf(x, z) 0, 2 zzf(x, z) 0, for all x, z. In terms of 2 f, ths means that ts 1, 1 block s postve semdefnte, and ts 2, 2 block s negatve semdefnte. (b) Let us fx z. Snce xf( x, z) = 0 and f(x, z) s convex n x, we conclude that x mnmzes f(x, z) over x,.e., for all z, we have f( x, z) f(x, z). Ths s one of the nequaltes n the saddle-pont condton. We can argue n the same way about z. Fx x, and note that zf( x, z) = 0, together wth concavty of ths functon n z, means that z maxmzes the functon,.e., for any x we have f( x, z) f( x, z). (c) To establsh ths we argue the same way. If the saddle-pont condton holds, then x mnmzes f(x, z) over all x. Therefore we have f x( x, z) = 0. Smlarly, snce z maxmzes f( x, z) over all z, we have f z( x, z) = 0.

2 Exercses Examples 3.15 A famly of concave utlty functons. For 0 < α 1 let u α(x) = xα 1 α, wth dom u α = R +. We also defne u 0(x) = log x (wth dom u 0 = R ++). (a) Show that for x > 0, u 0(x) = lm α 0 u α(x). (b) Show that u α are concave, monotone ncreasng, and all satsfy u α(1) = 0. These functons are often used n economcs to model the beneft or utlty of some quantty of goods or money. Concavty of u α means that the margnal utlty (.e., the ncrease n utlty obtaned for a fxed ncrease n the goods) decreases as the amount of goods ncreases. In other words, concavty models the effect of sataton. Soluton. (a) In ths lmt, both the numerator and denomnator go to zero, so we use l Hoptal s rule: (d/dα)(x α 1) x α log x lm uα(x) = lm = lm = log x. α 0 α 0 (d/dα)α α 0 1 (b) By nspecton we have The dervatve s gven by u α(1) = 1α 1 α u α(x) = x α 1, = 0. whch s postve for all x (snce 0 < α < 1), so these functons are ncreasng. To show concavty, we examne the second dervatve: u α(x) = (α 1)x α 2. Snce ths s negatve for all x, we conclude that u α s strctly concave For each of the followng functons determne whether t s convex, concave, quasconvex, or quasconcave. (a) f(x) = e x 1 on R. Soluton. Strctly convex, and therefore quasconvex. Also quasconcave but not concave. (b) f(x 1, x 2) = x 1x 2 on R Soluton. The Hessan of f s 2 f(x) = whch s nether postve semdefnte nor negatve semdefnte. Therefore, f s nether convex nor concave. It s quasconcave, snce ts superlevel sets are convex. It s not quasconvex. (c) f(x 1, x 2) = 1/(x 1x 2) on R Soluton. The Hessan of f s {(x 1, x 2) R 2 ++ x 1x 2 α} 2 f(x) = 1 x 1x 2 2/(x 2 1 ) 1/(x 1x 2) 1/(x 1x 2) 2/x 2 2 Therefore, f s convex and quasconvex. It s not quasconcave or concave., 0

3 (d) f(x 1, x 2) = x 1/x 2 on R Soluton. The Hessan of f s 2 f(x) = 0 1/x 2 2 1/x 2 2 2x 1/x 3 2 whch s not postve or negatve semdefnte. Therefore, f s not convex or concave. It s quasconvex and quasconcave (.e., quaslnear), snce the sublevel and superlevel sets are halfspaces. (e) f(x 1, x 2) = x 2 1/x 2 on R R ++. Soluton. f s convex, as mentoned on page 72. (See also fgure 3.3). Ths s easly verfed by workng out the Hessan: 2 f(x) = 2/x 2 2x 1/x 2 2 2x 1/x 2 2 2x 2 1/x 3 2 = (2/x 2) 1 2x 1/x 2 1 2x1/x 2 0. Therefore, f s convex and quasconvex. It s not concave or quasconcave (see the fgure). (f) f(x 1, x 2) = x α 1 x 1 α 2, where 0 α 1, on R Soluton. Concave and quasconcave. The Hessan s 2 α(α 1)x α 2 1 x 1 α 2 α(1 α)x α 1 1 x α 2 f(x) = α(1 α)x α 1 1 x α 2 (1 α)( α)x α 1 x α 1 2 = α(1 α)x α 1 x 1 α 1/x 2 1 1/x 1x 2 2 1/x 1x 2 1/x 2 2 = α(1 α)x α 1 x 1 α 1/x1 1/x1 2 1/x 2 1/x 2 0. f s not convex or quasconvex Suppose p < 1, p 0. Show that the functon ( f(x) = x p ) 1/p wth dom f = R n ++ s concave. Ths ncludes as specal cases f(x) = ( n x1/2 ) 2 and the harmonc mean f(x) = ( n 1/x) 1. Hnt. Adapt the proofs for the log-sum-exp functon and the geometrc mean n Soluton. The frst dervatves of f are gven by f(x) ( ) 1 p f(x) = ( x p x )(1 p)/p x p 1 =. The second dervatves are 2 f(x) x x j = 1 p x ( ) p ( f(x) f(x) x x j ) 1 p x T ( ) = 1 p f(x) 2 1 p f(x) x x j for j, and 2 f(x) x 2 = 1 p f(x) ( ) f(x) 2 1 p ( 1 p f(x) x 2 x x ) 1 p.

4 whch s a nonnegatve sum of the convex functons x 1, x 1 + x 2, x 1 + x 2 + x 3,..., x 1 + x x r. (b) Let T (x, ω) denote the trgonometrc polynomal Show that the functon T (x, ω) = x 1 + x 2 cos ω + x 3 cos 2ω + + x n cos(n 1)ω. f(x) = 2π 0 log T (x, ω) dω s convex on {x R n T (x, ω) > 0, 0 ω 2π}. Soluton. The functon g(x, ω) = log(x 1 + x 2 cos ω + x 3 cos 2ω + + +x n cos(n 1)ω) s convex n x for fxed ω. Therefore s convex n x. f(x) = 2π 0 g(x, ω)dω 3.20 Composton wth an affne functon. Show that the followng functons f : R n R are convex. (a) f(x) = Ax b, where A R m n, b R m, and s a norm on R m. Soluton. f s the composton of a norm, whch s convex, and an affne functon. (b) f(x) = (det(a 0 + x 1A x na n)) 1/m, on {x A 0 + x 1A x na n 0}, where A S m. Soluton. f s the composton of the convex functon h(x) = (det X) 1/m and an affne transformaton. To see that h s convex on S m ++, we restrct h to a lne and prove that g(t) = det(z + tv ) 1/m s convex: g(t) = (det(z + tv )) 1/m = (det Z) 1/m (det(i + tz 1/2 V Z 1/2 )) 1/m m = (det Z) 1/m ( (1 + tλ )) 1/m where λ 1,..., λ m denote the egenvalues of Z 1/2 V Z 1/2. We have expressed g as the product of a negatve constant and the geometrc mean of 1 + tλ, = 1,..., m. Therefore g s convex. (See also exercse 3.18.) (c) f(x) = tr (A 0 + x 1A x na n) 1, on {x A 0 +x 1A 1 + +x na n 0}, where A S m. (Use the fact that tr(x 1 ) s convex on S m ++; see exercse 3.18.) Soluton. f s the composton of tr X 1 and an affne transformaton x A 0 + x 1A x na n Pontwse maxmum and supremum. Show that the followng functons f : R n R are convex.

5 Exercses (a) f(x) = max,...,k A () x b (), where A () R m n, b () R m and s a norm on R m. Soluton. f s the pontwse maxmum of k functons A () x b (). Each of those functons s convex because t s the composton of an affne transformaton and a norm. (b) f(x) = r x on R n, where x denotes the vector wth x = x (.e., x s the absolute value of x, componentwse), and x s the th largest component of x. In other words, x 1, x 2,..., x n are the absolute values of the components of x, sorted n nonncreasng order. Soluton. Wrte f as f(x) = r x = max 1 1 < 2 < < x x r r n whch s the pontwse maxmum of n!/(r!(n r)!) convex functons Composton rules. Show that the followng functons are convex. (a) f(x) = log( log( m eat x+b )) on dom f = {x m eat x+b < 1}. You can use the fact that log( n ey ) s convex. Soluton. g(x) = log( m eat x+b ) s convex (composton of the log-sum-exp functon and an affne mappng), so g s concave. The functon h(y) = log y s convex and decreasng. Therefore f(x) = h( g(x)) s convex. (b) f(x, u, v) = uv x T x on dom f = {(x, u, v) uv > x T x, u, v > 0}. Use the fact that x T x/u s convex n (x, u) for u > 0, and that x 1x 2 s convex on R Soluton. We can express f as f(x, u, v) = u(v x T x/u). The functon h(x 1, x 2) = x 1x 2 s convex on R 2 ++, and decreasng n each argument. The functons g 1(u, v, x) = u and g 2(u, v, x) = v x T x/u are concave. Therefore f(u, v, x) = h(g(u, v, x)) s convex. (c) f(x, u, v) = log(uv x T x) on dom f = {(x, u, v) uv > x T x, u, v > 0}. Soluton. We can express f as f(x, u, v) = log u log(v x T x/u). The frst term s convex. The functon v x T x/u s concave because v s lnear and x T x/u s convex on {(x, u) u > 0}. Therefore the second term n f s convex: t s the composton of a convex decreasng functon log t and a concave functon. (d) f(x, t) = (t p x p p) 1/p where p > 1 and dom f = {(x, t) t x p}. You can use the fact that x p p/u p 1 s convex n (x, u) for u > 0 (see exercse 3.23), and that x 1/p y 1 1/p s convex on R 2 + (see exercse 3.16). Soluton. We can express f as ( )) 1/p ( ) 1/p f(x, t) = (t p 1 t x p p = t 1 1/p t x p p. t p 1 t p 1 Ths s the composton of h(y 1, y 2) = y 1/p 1 y 1 1/p 2 (convex and decreasng n each argument) and two concave functons g 1(x, t) = t 1 1/p, g 2(x, t) = t x p p t p 1.

6 (e) f(x, t) = log(t p x p p) where p > 1 and dom f = {(x, t) t > x p}. You can use the fact that x p p/u p 1 s convex n (x, u) for u > 0 (see exercse 3.23). Soluton. Express f as f(x, t) = log t p 1 log(t x p p/t p 1 ) = (p 1) log t log(t x p p/t p 1 ). The frst term s convex. The second term s the composton of a decreasng convex functon and a concave functon, and s also convex Perspectve of a functon. (a) Show that for p > 1, f(x, t) = x1 p + + x n p t p 1 = x p p t p 1 s convex on {(x, t) t > 0}. Soluton. Ths s the perspectve functon of x p p = x 1 p + + x n p. (b) Show that f(x) = Ax + b 2 2 c T x + d s convex on {x c T x + d > 0}, where A R m n, b R m, c R n and d R. Soluton. Ths functon s the composton of the functon g(y, t) = y T y/t wth an affne transformaton (y, t) = (Ax + b, c T x + d). Therefore convexty of f follows from the fact that g s convex on {(y, t) t > 0}. For convexty of g one can note that t s the perspectve of x T x, or drectly verfy that the Hessan 2 I/t y/t 2 g(y, t) = y T /t y T y/t 3 s postve semdefnte, snce T v I/t y/t 2 v w y T /t y T y/t 3 w for all v and w. = tv yw 2 2/t Some functons on the probablty smplex. Let x be a real-valued random varable whch takes values n {a 1,..., a n} where a 1 < a 2 < < a n, wth prob(x = a ) = p, = 1,..., n. For each of the followng functons of p (on the probablty smplex {p R n + 1 T p = 1}), determne f the functon s convex, concave, quasconvex, or quasconcave. (a) E x. Soluton. E x = p 1a p na n s lnear, hence convex, concave, quasconvex, and quasconcave (b) prob(x α). Soluton. Let j = mn{ a α}. Then prob(x α) = n p, Ths s a lnear =j functon of p, hence convex, concave, quasconvex, and quasconcave. (c) prob(α x β). Soluton. Let j = mn{ a α} and k = max{ a β}. Then prob(α x β) = k p. Ths s a lnear functon of p, hence convex, concave, quasconvex, =j and quasconcave.

7 Exercses we have so p >q (p q ) = ( p q (p q ) d mp(p, q) = (1/2) (p q ) (1/2) (p q ) p >q p q = (1/2) p q = (1/2) p q 1. Ths makes t very clear that d mp s convex. The best way to nterpret ths result s as an nterpretaton of the l 1-norm for probablty dstrbutons. It states that the l 1-dstance between two probablty dstrbutons s twce the maxmum dfference n probablty, over all events, of the dstrbutons More functons of egenvalues. Let λ 1(X) λ 2(X) λ n(x) denote the egenvalues of a matrx X S n. We have already seen several functons of the egenvalues that are convex or concave functons of X. The maxmum egenvalue λ 1(X) s convex (example 3.10). The mnmum egenvalue λ n(x) s concave. The sum of the egenvalues (or trace), tr X = λ 1(X) + + λ n(x), s lnear. The sum of the nverses of the egenvalues (or trace of the nverse), tr(x 1 ) = n 1/λ(X), s convex on Sn ++ (exercse 3.18). The geometrc mean of the egenvalues, (det X) 1/n = ( n λ(x))1/n, and the logarthm of the product of the egenvalues, log det X = n log λ(x), are concave on X S n ++ (exercse 3.18 and page 74). In ths problem we explore some more functons of egenvalues, by explotng varatonal characterzatons. (a) Sum of k largest egenvalues. Show that k λ(x) s convex on Sn. Hnt. HJ85, page 191 Use the varatonal characterzaton k λ (X) = sup{tr(v T XV ) V R n k, V T V = I}. Soluton. The varatonal characterzaton shows that f s the pontwse supremum of a famly of lnear functons tr(v T XV ). (b) Geometrc mean of k smallest egenvalues. Show that ( n =n k+1 λ(x))1/k s concave on S n ++. Hnt. MO79, page 513 For X 0, we have ( n ) 1/k λ (X) = 1 k nf{tr(v T XV ) V R n k, det V T V = 1}. =n k+1 Soluton. f s the pontwse nfmum of a famly of lnear functons tr(v T XV ). (c) Log of product of k smallest egenvalues. Show that n log λ(x) s concave =n k+1 on S n ++. Hnt. MO79, page 513 For X 0, { n k } λ (X) = nf (V T XV ) V R n k, V T V = I. =n k+1 ),

8 Soluton. f s the pontwse nfmum of a famly of concave functons log (V T XV ) = log(v T XV ) Dagonal elements of Cholesky factor. Each X S n ++ has a unque Cholesky factorzaton X = LL T, where L s lower trangular, wth L > 0. Show that L s a concave functon of X (wth doman S n ++). Hnt. L can be expressed as L = (w z T Y 1 z) 1/2, where Y z z T s the leadng submatrx of X. Soluton. The functon f(z, Y ) = z T Y 1 z wth dom f = {(z, Y ) Y 0} s convex jontly n z and Y. To see ths note that Y z (z, Y, t) ep f Y 0, z T 0, t so ep f s a convex set. Therefore, w z T Y 1 z s a concave functon of X. Snce the squareroot s an ncreasng concave functon, t follows from the composton rules that l kk = (w z T Y 1 z) 1/2 s a concave functon of X. Operatons that preserve convexty 3.28 Expressng a convex functon as the pontwse supremum of a famly of affne functons. In ths problem we extend the result proved on page 83 to the case where dom f R n. Let f : R n R be a convex functon. Defne f : R n R as the pontwse supremum of all affne functons that are global underestmators of f: f(x) = sup{g(x) g affne, g(z) f(z) for all z}. (a) Show that f(x) = f(x) for x nt dom f. (b) Show that f = f f f s closed (.e., ep f s a closed set; see A.3.3). Soluton. (a) The pont (x, f(x)) s n the boundary of ep f. (If t were n nt ep f, then for small, postve ɛ we would have (x, f(x) ɛ) ep f, whch s mpossble.) From the results of 2.5.2, we know there s a supportng hyperplane to ep f at (x, f(x)),.e., a R n, b R such that w a T z + bt a T x + bf(x) for all (z, t) ep f. Snce t can be arbtrarly large f (z, t) ep f, we conclude that b 0. Suppose b = 0. Then a T z a T x for all z dom f whch contradcts x nt dom f. Therefore b > 0. Dvdng the above nequalty by b yelds t f(x) + (a/b) T (x z) for all (z, t) ep f. Therefore the affne functon g(z) = f(x) + (a/b) T (x z) s an affne global underestmator of f, and hence by defnton of f, f(x) f(x) g(x). However g(x) = f(x), so we must have f(x) = f(x).

9 (a) Let A = conv B. Snce B A, we obvously have S B(y) S A(y). Suppose we have strct nequalty for some y,.e., y T u < y T v for all u B and some v A. Ths leads to a contradcton, because by defnton v s the convex combnaton of a set of ponts u B,.e., v = θu, wth θ 0, θ = 1. Snce y T u < y T v for all, ths would mply y T v = θ y T u < θ y T v = y T v. We conclude that we must have equalty S B(y) = S A(y). (b) Follows from (c) Follows from S A+B(y) = sup{y T (u + v) u A, v B} = sup{y T u u A} + sup{y T v u B} = S A(y) + S B(y). S A B(y) = sup{y T u u A B} = max{sup{y T u u A}, sup{y T v u B} = max{s A(y), S B(y)}. (d) Obvously, f A B, then S A(y) S B(y) for all y. We need to show that f A B, then S A(y) > S B(y) for some y. Suppose A B. Consder a pont x A, x B. Snce B s closed and convex, x can be strctly separated from B by a hyperplane,.e., there s a y 0 such that y T x > y T x for all x B. It follows that S B(y) < y T x S A(y). Conjugate functons 3.36 Derve the conjugates of the followng functons. (a) Max functon. f(x) = max,...,n x on R n. Soluton. We wll show that { f 0 f y 0, 1 T y = 1 (y) = otherwse. We frst verfy the doman of f. Frst suppose y has a negatve component, say y k < 0. If we choose a vector x wth x k = t, x = 0 for k, and let t go to nfnty, we see that x T y max x = ty k, so y s not n dom f. Next, assume y 0 but 1 T y > 1. We choose x = t1 and let t go to nfnty, to show that x T y max x = t1 T y t

10 Exercses s unbounded above. Smlarly, when y 0 and 1 T y < 1, we choose x = t1 and let t go to nfnty. The remanng case for y s y 0 and 1 T y = 1. In ths case we have x T y max x for all x, and therefore x T y max x 0 for all x, wth equalty for x = 0. Therefore f (y) = 0. (b) Sum of largest elements. f(x) = r x on R n. Soluton. The conjugate s { f 0 0 y 1, 1 T y = r (y) = otherwse, We frst verfy the doman of f. Suppose y has a negatve component, say y k < 0. If we choose a vector x wth x k = t, x = 0 for k, and let t go to nfnty, we see that x T y f(x) = ty k, so y s not n dom f. Next, suppose y has a component greater than 1, say y k > 1. If we choose a vector x wth x k = t, x = 0 for k, and let t go to nfnty, we see that x T y f(x) = ty k t, so y s not n dom f. Fnally, assume that 1 T x r. We choose x = t1 and fnd that x T y f(x) = t1 T y tr s unbounded above, as t or t. If y satsfes all the condtons we have x T y f(x) for all x, wth equalty for x = 0. Therefore f (y) = 0. (c) Pecewse-lnear functon on R. f(x) = max,...,m(a x + b ) on R. You can assume that the a are sorted n ncreasng order,.e., a 1 a m, and that none of the functons a x + b s redundant,.e., for each k there s at least one x wth f(x) = a k x + b k. Soluton. Under the assumpton, the graph of f s a pecewse-lnear, wth breakponts (b b +1)/(a +1 a ), = 1,..., m 1. We can wrte f as ( f (y) = sup x ) xy max (ax + b),...,m We see that dom f = a 1, a m, snce for y outsde that range, the expresson nsde the supremum s unbounded above. For a y a +1, the supremum n the defnton of f s reached at the breakpont between the segments and + 1,.e., at the pont (b +1 b )/(a +1 a ), so we obtan f y a (y) = b (b +1 b ) a +1 a where s defned by a y a +1. Hence the graph of f s also a pecewse-lnear curve connectng the ponts (a, b ) for = 1,..., m. Geometrcally, the epgraph of f s the epgraphcal hull of the ponts (a, b ).

11 (d) Power functon. f(x) = x p on R ++, where p > 1. Repeat for p < 0. Soluton. We ll use standard notaton: we defne q by the equaton 1/p + 1/q = 1,.e., q = p/(p 1). We start wth the case p > 1. Then x p s strctly convex on R +. For y < 0 the functon yx x p acheves ts maxmum for x > 0 at x = 0, so f (y) = 0. For y > 0 the functon acheves ts maxmum at x = (y/p) 1/(p 1), where t has value y(y/p) 1/(p 1) (y/p) p/(p 1) = (p 1)(y/p) q. Therefore we have f (y) = { 0 y 0 (p 1)(y/p) q y > 0. For p < 0 smlar arguments show that dom f = R ++ and f (y) = p q ( y/p)q. (e) Geometrc mean. f(x) = ( x ) 1/n on R n ++. Soluton. The conjugate functon s { ( f (y) = 0 f y 0, ( y)) 1/n 1/n otherwse. We frst verfy the doman of f. Assume y has a postve component, say y k > 0. Then we can choose x k = t and x = 1, k, to show that x T y f(x) = ty k + y t 1/n s unbounded above as a functon of t > 0. Hence the condton y 0 s ndeed requred. Next assume that y 0, but ( ( y))1/n < 1/n. We choose x = t/y, and obtan ( 1/n x T y f(x) = tn t ( )) 1 y as t. Ths demonstrates that the second condton for the doman of f s also needed. Now assume that y 0 and ( ( y)) 1/n 1/n, and x 0. The arthmetcgeometrc mean nequalty states that k x T y n ( ) 1/n ( ) 1/n ( y x ) 1 x, n.e., x T y f(x) wth equalty for x = 1/y. Hence, f (y) = 0. (f) Negatve generalzed logarthm for second-order cone. f(x, t) = log(t 2 x T x) on {(x, t) R n R x 2 < t}. Soluton. f (y, u) = 2 + log 4 log(u 2 y T y), dom f = {(y, u) y 2 < u}. We frst verfy the doman. Suppose y 2 u. Choose x = sy, t = s( x 2 + 1) > s y 2 su, wth s 0. Then y T x + tu > sy T y su 2 = s(u 2 y T y) 0,

12 Exercses so y x + tu goes to nfnty, at a lnear rate, whle the functon log(t 2 x T x) goes to as log s. Therefore y T x + tu + log(t 2 x T x) s unbounded above. Next, assume that y 2 < u. Settng the dervatve of y T x + ut + log(t 2 x T x) wth respect to x and t equal to zero, and solvng for t and x we see that the maxmzer s 2y x = u 2 y T y, t = 2u u 2 y T y. Ths gves f (y, u) = ut + y T x + log(t 2 x T x) = 2 + log 4 log(y 2 u t u) Show that the conjugate of f(x) = tr(x 1 ) wth dom f = S n ++ s gven by f (Y ) = 2 tr( Y ) 1/2, dom f = S n +. Hnt. The gradent of f s f(x) = X 2. Soluton. We frst verfy the doman of f. Suppose Y has egenvalue decomposton Y = QΛQ T = λ q q T wth λ 1 > 0. Let X = Q dag(t, 1,..., 1)Q T = tq 1q1 T + n =2 qqt. We have tr XY tr X 1 = tλ 1 + λ 1/t (n 1), whch grows unboundedly as t. Therefore Y dom f. Next, assume Y 0. If Y 0, we can fnd the maxmum of =2 tr XY tr X 1 by settng the gradent equal to zero. We obtan Y = X 2,.e., X = ( Y ) 1/2, and f (Y ) = 2 tr( Y ) 1/2. Fnally we verfy that ths expresson remans vald when Y 0, but Y s sngular. Ths follows from the fact that conjugate functons are always closed,.e., have closed epgraphs Young s nequalty. Let f : R R be an ncreasng functon, wth f(0) = 0, and let g be ts nverse. Defne F and G as F (x) = x 0 f(a) da, G(y) = y 0 g(a) da. Show that F and G are conjugates. Gve a smple graphcal nterpretaton of Young s nequalty, xy F (x) + G(y). Soluton. The nequalty xy F (x) + G(y) has a smple geometrc meanng, llustrated below.