LEIBNIZ SEMINORMS IN PROBABILITY SPACES

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1 LEIBNIZ SEMINORMS IN PROBABILITY SPACES ÁDÁM BESENYEI AND ZOLTÁN LÉKA Abstract. In this aer we study the (strong) Leibniz roerty of centered moments of bounded random variables. We shall answer a question raised by M. Rieel on the non-commutative standard deviation. 1. Introduction We say that a seminorm L on a unital normed algebra (A, ) is strongly Leibniz if (i) L(1 A ) = 0, (ii) the Leibniz roerty L(ab) a L(b) + b L(a) holds for every a, b A and, furthermore, (iii) for every invertible a, L(a 1 ) a 1 2 L(a) follows. Primary sources of strongly Leibniz seminorms are normed rst-order dierential calculi, see [8]. It is said that the coule (Ω, δ) is a normed rst-order dierential calculus over A if Ω is a normed bimodule over A and δ is a derivation from A to Ω. Now let us assume that Ω is acting boundedly over A; that is, the inequalities aω ω Ω a and ωa ω Ω a hold for every ω Ω and for every a A. From the derivation rule δ(ab) = δ(a)b + aδ(b), the Leibniz roerty of the seminorm L(a) = δ(a) Ω simly follows. Furthermore, we clearly have that δ(a 1 ) = a 1 δ(a)a 1, whenever a is invertible, hence (iii) follows as well. For instance, if we choose a (real or comlex) Banach sace X and B(X) denotes the normed algebra of its bounded linear oerators, ractically, we can easily get a rst-order dierential calculus. Actually, with the choice of Ω = B(X), which acts naturally over B(X) via the left and right multilications, the commutator δ(a) = [D, A] = DA AD for some xed D B(X) denes the required calculus. Consider a unital C -algebra A and denote B a C -subalgebra of A with a common unit. Rieel ointed out in [7, Theorem] that the factor norm inf b B a b obeys the strong Leibniz roerty, since it equals to a commutator norm. To get connection with the standard deviation, notice that K. Audenaert rovided shar estimate for dierent tyes of non-commutative (or quantum) deviations determined by matrices [1]. Not long ago Rieel extended these results to C - algebras with a comletely dierent aroach [8]. His theorem reads as follows: for any a A, max ω( a ω S(A) ω(a) 2 ) 1/2 = min a λ1 A, λ C 2000 Mathematics Subject Classication. Primary 46L53, 60E15 ; Secondary 26A51, 60B99. Key words and hrases. standard deviation, Leibniz seminorm, central moments, C -algebra. This study was artially suorted by the Hungarian NSRF (OTKA) grant no. K

2 2 ÁDÁM BESENYEI AND ZOLTÁN LÉKA where S(A) denotes the state sace of A; i.e. the set of ositive linear functionals of A with norm 1. For a short roof of this theorem, exloiting the BirkhoJames orthogonality in oerator algebras, the reader might see [2]. The factor norm on the left-hand side above indicates that 'the largest standard deviation' is a strongly Leibniz seminorm. Surrisingly, the standard deviation itself is a strongly Leibniz seminorm. Precisely, whenever σ2 ω (a) = ω( a ω(a) 2 ) 1/2, the seminorm σ2 ω on A is strongly Leibniz if ω is tracial [8, Proosition 3.4]. Moreover, if one denes the non-commutative standard deviation by the formula σ ω 2 (a) = ω( a ω(a) 2 ) 1/2 ω( a ω(a ) 2 ) 1/2, then σ ω 2 is strongly Leibniz for any ω S(A), see [8, Theorem 3.5] (without assuming that ω is tracial). Quite recently, the equality max ω( a ω S(A) ω(a) k ) 1/k = 2B 1/k k min a λ1 A λ C was roved in [4] for the kth central moments of normal elements, where k is even and B k denotes the largest kth centered moment of the Bernoulli distribution. From this result it follows that 'the largest kth moments' in commutative C -algebras are strongly Leibniz as well. The aim of the aer is to study whether general or higher-ordered centered moments ossess the (strong) Leibniz roerty in ordinary robability saces, or not. In the next section we shall give a rough estimate of the centered moments of roducts of bounded random variables which gives back Rieel's statement on the standard deviation. After that we shall resent some scattered Leibniz-tye result for dierent moments on dierent (discrete, general) robability saces. We leave oen the question whether all centered moments in general robability saces dene a strongly Leibniz seminorm. Lastly, in Section 3, we shall answer armatively Rieel's question on the standard deviation in non-commutative robability saces. 2. Leibniz seminorms in function saces In this section we shall study the Leibniz roerty and similar estimates in ordinary robability saces. Let (Ω, F, µ) be a robability sace. For any f : Ω C L (Ω, µ) and 1 <, let us dene ( ) σ (f; µ) = f 1/ f dµ dµ and Ω σ (f; µ) = ess su Ω f Ω f dµ. If no confusion can arise, we simly use the notation σ (f). Relying on [8], we know that the standard deviation is a strongly Leibniz seminorm; that is, the inequalities for f, g L (Ω, µ), and σ 2 (fg) g σ 2 (f) + f σ 2 (g) σ 2 (1/f) 1/f 2 σ 2 (f) whenever 1/f L (Ω, µ) hold. For the non-commutative analogues of the result, see [8]. We begin with an observation which shows that one can reduce the roblem of the strongly Leibniz roerty to that of the discrete uniform distributions. Proosition 2.1. Fix 1 <. The following statements are equivalent: (i) For any robability sace (Ω, F, µ), σ is a strongly Leibniz seminorm on L (Ω, µ).

3 LEIBNIZ SEMINORMS IN PROBABILITY SPACES 3 (ii) For every n Z +, σ is a strongly Leibniz seminorm on l n the uniform distribution. endowed with Proof. Obviously, (i) imlies (ii). To see the reverse imlication, choose airwise disjoint sets S k F (1 k n). As usual χ Sk denotes the characteristic function of the set S k. Let us consider the measurable simle functions f n = n k=1 a kχ Sk and g n = n k=1 b kχ Sk on Ω. Let us assume that n k=1 S k = Ω, so that the constants µ(s k ) dene a robability measure µ n on the set Z n = {1,..., n}. Then for any ε > 0 we can readily nd a robability measure ν n = ( 1,..., n ) such that i Q (1 i n) and the inequalities σ (f n ; µ n ) σ (f n ; ν n ) ε σ (g n ; µ n ) σ (g n ; ν n ) ε σ (f n g n ; µ n ) σ (f n g n ; ν n ) ε hold. Now let us choose the integers m and r i such that i = r i /m for every 1 i n. Then the ma Φ: (c 1,..., c n ) (c 1,..., c }{{} 1,..., c n,..., c n ) }{{} r 1 r n denes an isometric algebra homomorhism from l n into l m. Let λ m denote the uniform distribution on the set Z m. We clearly have, for instance, σ (f n ; ν n ) = σ (Φ(f n ); λ m ), hence σ (f n g n ; ν n ) f n σ (g n ; ν n ) + g n σ (f n ; ν n ) follows as well. Since ε can be arbitrary small, we obtain that σ is a Leibniz seminorm on l n (µ n ). Now if we choose sequences {f n } n=1 and {g n } n=1 of measurable simle functions such that f n f and g n g in L norm, furthermore, f n = f and g n = g hold for every n, we infer that σ has the Leibniz roerty. A very similar reasoning on the invertible elements gives that σ is actually strongly Leibniz on L (Ω, µ). Desite of the above equivalence, in arbitrary measure saces we do not know whether σ is strongly Leibniz or not. But later we will rove this roerty for σ in the real Banach sace L (Ω, µ; R) (see Theorem 2.6 below). Actually, the second art of the section deals with only real-valued functions. In the general situation, we have only a rough Leibniz-tye estimate as we shall see below. In any L (Ω, µ) (1 ) sace, the rojection P is given by the ma f Ef = f dµ. Then we are able to rove a slight generalization of Rieel's statement [8, Proosition 3.4] in robability saces. Proosition 2.2. For any 1 and f, g L (Ω, µ), we have that 2 I P + 1 fg E(fg) g f Ef + f g Eg. Proof. First, note that I P 1 (excet for the trivial case I = P ). Hence, without loss of generality, we can assume that fg E(fg) max( f g Eg, g f Ef ), otherwise the roof is done. Obviously, f(g Eg) E(f(g Eg)) I P f(g Eg). Ω

4 4 ÁDÁM BESENYEI AND ZOLTÁN LÉKA From the reversed triangle inequality we obtain that fg E(fg) EfEg feg f(g Eg) E(f(g Eg)), which imlies that fg E(fg) I P f g Eg + g f Ef. Changing the variables f, g and summing u the inequalities, we get the statement of the roosition. Remark 2.3. One can nd a non-trivial uer estimate of the constant I P. For instance, if Ω = {1,..., n} and µ is the uniform distribution on Ω, from the denition of the matrix -norms one can easily see that I P 1 = I P = 2 2 n and I P 2 = 1. As another examle, let us consider the Banach saces L [0, 1] endowed with the Lebesgue measure. Then a simle calculation shows that I P 1 = I P = 2. Moreover, I P is clearly an orthogonal rojection in L 2 [0, 1]; that is, I P 2 = 1. Now a straightforward alication of the Riesz Thorin interolation theorem gives that (see [6]) I P The rojection I P is actually the minimal rojection to the hyerlane X = {f L [0, 1] : Ef = 0}; i.e. it has the minimal norm among the rojections of range X. C. Franchetti showed in his aer [3] that I P = max 0 x 1 (x 1 + (1 x) 1 ) 1/ (x q 1 + (1 x) q 1 ) 1/q, where 1/ + 1/q = 1. Remark 2.4. One can aly a derivation aroach mentioned in the Introduction to obtain Leibniz-tye estimates of the moments of invertible functions. To do this, let us renorm the sace L (Ω, µ), 2 <, so that x, := Ex + x Ex. Let X denote the renormed sace. Dene the multilication oerator M f : x fx and the derivation δ(m f ) = [P, M f ] = P M f M f P. A straightforward calculation yields that M f x, f x + I P f x (1 + I P ) f x, ; that is, M f (1+ I P ) f. Moreover, δ(m f )Ex = Ex(Ef f) (I P )X, thus δ(m f ) P X = σ (f). On the other hand, δ(m f )(x Ex) = E(fx) EfEx = E((f Ef)(x Ex)). From Hölder's inequality we get that δ(m f )(x Ex), f Ef q x Ex (1/q + 1/ = 1) hence δ(m f ) (I P )X σ (f) follows. Since the oerator δ(m f ) interchanges the subsaces P X and (I P )X, we have δ(m f ) = σ (f). An alication of the derivation rules tells us that σ (1/f) (1 + I P ) 2 1/f 2 σ (f) holds whenever 1/f L (Ω, µ). For any 1, we can get a dierent estimate from the equality Hence we conclude that (I P )M 1/f (I P )f = (1/f E(1/f))Ef. Ef σ (1/f) I P 1/f σ (f).

5 LEIBNIZ SEMINORMS IN PROBABILITY SPACES 5 Much of the rest of the section is devoted to a study of the otimality of the above roosition. We begin with the following observation. Proosition 2.5. Let (Ω, F, µ) be a robability sace. For any real-valued f and x L (Ω, µ), the inequality holds. fex E(fx) x f Ef Proof. Without loss of generality, we can assume that Ef = 0 holds and f = 1. Note that the function f fex E(fx) is convex on the weak- comact, convex set L 0 (Ω) := {f L (Ω, µ): f 1 and Ef = 0} (L 1 (Ω, µ)). Hence, from the KreinMilman theorem, it is enough to rove the statement if f is an extreme oint of L 0 (Ω). We claim that the extreme oints of L 0 (Ω) are the functions with essential range { 1, 1, c} for some 1 < c < 1, (µ({f = c}) = 0 might be ossible) and (2.1) Ef = µ({f = 1}) µ({f = 1}) + cµ({f = c}) = 0. Let us choose a measurable subset A of Ω such that fχ A 1 ε < 1. If µ is non-atomic (A is not a singleton), we can nd a function g L 0 (Ω) satisfying g ε and g = 0 a.e. on Ω \ A. Since f = 1 2 (f + g) + 1 (f g), 2 f is an extreme oint if and only if µ(a) = 0. When µ is atomic, the set A might be a singleton, hence our claim follows. Now let f be an extreme oint of L 0 (Ω). Obviously, f Ef = 1. Furthermore, we have fex E(fx) = max( Ex E(fx), Ex + E(fx), cex E(fx) ) = max( E(x(1 f)), E(x(1 + f)), E(x(c f)) ) x max( 1 f 1, 1 + f 1, c f 1 ). It remains to show that max( 1 f 1, 1 + f 1, c f 1 ) = 1. Clearly, from (2.1) Similarly, 1 f 1 = 2µ({f = 1}) + 1 c µ({f = c}) = 1 µ({f = 1}) + µ({f = 1}) cµ({f = c}) = f 1 = 2µ({f = 1}) c µ({f = c}) and lastly we infer that The roof is comlete. = 1 + µ({f = 1}) µ({f = 1}) + cµ({f = c}) = 1, c f 1 = c 1 µ({f = 1}) + c + 1 µ({f = 1}) = µ({f = 1}) + µ({f = 1}) + c 2 µ({f = c}) 1. For the real Banach sace L (Ω, µ; R), we can simly rove that the seminorm σ is strongly Leibniz as we have seen before for the standard deviation. Theorem 2.6. Let (Ω, F, µ) be a robability sace. L (Ω, µ; R), σ (f) = f Ef is a strongly Leibniz seminorm. For the real Banach sace

6 6 ÁDÁM BESENYEI AND ZOLTÁN LÉKA Proof. From Proosition 2.5, it follows that fg E(fg) = f(g Eg) + (feg E(fg)) f g Eg + g f Ef, and 1 f E 1 f = 1 f which is what we intended to have. ( ( E f 1 ) fe 1 ) f f 1 f 1 f f Ef, Regarding the case of the uniform distributions seen above in Proosition 2.1, we are able to rove the analogue of Proosition 2.5 in very articular cases. Let λ n stand for the uniform distribution on Z n. Proosition 2.7. Fix 1 n 4. For 1 <, and any real-valued f, x l n (λ n ), we have fex E(fx) x f Ef. Proof. First note that the case Ω = Z 1 is trivial. On the other hand, in case of Ω = Z 2, one can have arbitrary distribution. Indeed, let µ(1) = 1 and µ(2) = 2 = 1 1. Then by simle calculation we obtain and f Ef = (f 1 f 2 ) ( 2, 1 ) fex E(fx) = (f 1 f 2 ) ( 2 x 2, 1 x 1 ) so the desired inequality follows immediately. To rove the remaining cases Ω = Z 3 and Ω = Z 4, let us rescale the inequality and assume that x = 1. Notice that the function x fex E(fx) is convex on the closed unit ball {x L (Ω, µ) : x 1}, therefore it suces to check the inequality only for its extreme oints. First, we turn to the case Ω = Z 3. Clearly, for x = (1, 1, 1) even equality holds, so after ossible rearrangement and multilication by constants we may assume that x = (1, 1, 1). Then and f Ef = 1 3 (2f 1 f 2 f 3, f 1 + 2f 2 f 3, f 1 f 2 + 2f 3 ) fex E(fx) = 1 3 (f 3 f 1, f 3 f 2, 2f 3 f 1 f 2 ). By using the notation a 1 = 2f 1 f 2 f 3 and a 2 = 2f 2 f 1 f 3, the inequality reduces to the form 2a 1 + a a 1 + 2a 2 3 a 1 + a 2, which is obviously true from the convexity of the function t t. Next, let Ω = Z 4. By symmetry arguments we can assume that x = (1, 1, 1, 1) or x = (1, 1, 1, 1). Set x = (1, 1, 1, 1). A simle calculation imlies that f Ef = 1 4 (a 1, a 2, a 3, a 4 ), where a j = 3f j i j f i.

7 LEIBNIZ SEMINORMS IN PROBABILITY SPACES 7 Moreover, +f 1 f 2 f 3 + f 4 a 2 a 3 fex E(fx) = 1 f 1 + f 2 f 3 + f 4 4 f 1 f 2 + f 3 + f 4 = 1 a 1 a 3 8 a 1 a 2. f 1 f 2 f 3 + 3f 4 2a 4 Therefore, it is enough to check that a 2 + a a 1 + a a 1 + a 2 2 a 1 + a 2 + a 3, which follows again by the convexity of the function t t. Lastly, consider the remaining case x = (1, 1, 1, 1). Then f 1 f 2 + f 3 + f 4 a 1 a 2 + a 3 + a 4 fex E(fx) = 1 f 1 f 2 + f 3 + f 4 4 f 1 f 2 + f 3 + f 4 = 1 a 1 a 2 + a 3 + a 4 16 a 1 a 2 + a 3 + a 4. f 1 f 2 + f 3 + f 4 a 1 a 2 + a 3 + a 4 Since 4 a 1 a 2 + a 3 + a 4 4 a 1 + a 2 + a 3 + a 4, by a convexity argument as seen before, we get the statement of the roosition. Examle 2.8. The statement of Proosition 2.7 does not hold in general. Let n 5 and = 1, for instance. Let x = (1,..., 1, 1) and f = (1, 0,..., 0, 1) in l n (λ n ). Obviously, Ef = 0, Ex = 1 2 n, E(fx) = 2 n, x = 1, furthermore, and Thus f Ef 1 = 2 n, ( fex E(fx) 1 = 1 4 n, 2 n,..., 2 ) 1 n, 1 = 4n 8 n 2 (n 5). fex E(fx) 1 = ( 2 4 ) f Ef 1 > f Ef 1. n Examle 2.9. In the case of non-uniform distributions, the inequality of Proosition 2.7 is not true even on Ω = {1, 2, 3}. To see this, dene the measure µ(1) = 1 8, µ(2) = 3 4, µ(3) = 1 8 and consider f = (1, 0, 1) and x = (1, 1, 1). Then Ef = 0, Ex = 3 4, E(fx) = 1 4, and f Ef 1 = 1 4, while ( 1) fex E(fx) 1 = 1 1 2, 1 4, = 3 8. As we have seen before in the roof of Theorem 2.6, we can infer the next statement on discrete measure saces. Corollary For 1 n 4 and 1 <, the seminorm σ is strongly Leibniz on the real l n endowed with uniform distribution. Surrisingly, we cannot rove or disrove the last statement on measure saces which contain more than 4 atoms. Comuter simulations suggest us that Corollary 2.10 might be true for any n which would imly that σ is a strongly Leibniz seminorm for every 1 < (see Proosition 2.1). Now we have only a very few articular results on general measure saces. Denote λ n the uniform distribution on the set Z n, as usual.

8 8 ÁDÁM BESENYEI AND ZOLTÁN LÉKA Proosition Let 1 < and f, g l n (λ n ) be such that the coordinates of f, g and fg have the same order. Then holds. fg E(fg) g f Ef + f g Eg Proof. We use the fact that the l norm with uniform distribution and 1 is a Schur-convex function [5, Ch. 3 Examle I.1]. Therefore, it suces to rove that the vector fg E(fg) is majorized by f (g Eg) + g (f Ef). To see this, we may assume without loss of generality that f 1 f 2 f n, thus we also have g 1 g 2 g n and f 1 g 1 f 2 g 2 f n g n. Then we have to verify that (f j g j E(fg)) ( g (f j Ef) + f (g j Eg)), for all 1 k n 1, and equality holds when k = n. The latter equality is obvious because both sides are zero if k = n. In the remainder of the roof, a simle calculation gives that n n n (f j Ef) = (n k) f j k f j = (f j f i ) and analogously n (f j g j E(fg)) = Therefore, it follows that = n i=k+1 n i=k+1 j=k+1 (f j g j f i g i ) i=k+1 (f j (g j g i ) + g i (f j f i )). ( f (g j Eg) + g (f j Ef) (f j g j E(fg))) = 1 n n i=k+1 (g j g i )( f f j ) + (f j f i )( g g i ) 0. Analogously to the roof of Proosition 2.1, we readily obtain the following corollaries. Corollary Let (Ω, F, µ) be a robability sace and 1 <. For any non-negative f L (Ω, µ), f 2 Ef 2 2 f f Ef. Corollary Let 1 < and µ be a robability measure on the interval [0, 1]. For any non-negative, bounded and monotone increasing (or decreasing) functions f and g, we have fg Efg g f Ef + f g Eg.

9 LEIBNIZ SEMINORMS IN PROBABILITY SPACES 9 3. Standard deviation in C -algebras In this section we shall comlete Rieel's argument on the standard deviation in non-commutative robability saces. Let A be a unital C -algebra and denote ω any faithful state of it. Denote L 2 (A, ω) the GNS Hilbert sace obtained by comleting A for the inner roduct a, b = ω(b a), as usual. Obviously, every a A has a natural reresentation; i.e. the left-regular reresentation L a, in the oerator algebra of L 2 (A, ω). Consider now the rojection (or Dirac oerator) E : a ω(a)1 A on L 2 (A, ω). Direct calculations for the norm of the commutator δ(l a ) = [E, L a ] = EL a L a E give that δ(l a ) = ω( a ω(a) 2 ) 1/2 ω( a ω(a ) 2 ) 1/2. Thus it immediately follows that Rieel's non-commutative standard deviation is a strongly Leibniz -seminorm, see [8, Theorem 3.7]. Moreover, an alication of the 'indeendent coies trick' in C -algebras gives that σ ω 2 (a) := ω( a ω(a) 2 ) is strongly Leibniz as well if one assumes that ω is tracial [8, Proosition 3.6]. Actually, the 'strong' art of the statement requires only the tracial assumtion. Comuter simulations for matrices indicate that σ2 ω might be strongly Leibniz for any state ω but the question remained oen in [8]. Now we shall rovide the armative answer by means of an elementary argument. Pick a faithful state ω of A. Let a 2 = ω( a 2 ) 1/2 denote the norm on L 2 (A, ω). We begin with Lemma 3.1. For any a and x A, ω(x)a ω(xa) 2 x a ω(a) 2. Proof. There is no loss of generality in assuming that ω(a) = 0. Denote E the orthogonal rojection from L 2 (A, ω) onto its subsace C1 A. Then ω(x)a ω(xa) 2 = ω(x)(i E)a Eω(xa) 2 = ω(x)a 2 + ω(xa). Notice that the CauchySchwarz inequality readily gives that Hence ω(xa) = ω((x ω(x))a) a 2 x ω(x ) 2. ω(x)a ω(xa) 2 = ω(x)a 2 + ω(xa) ω(x ) a 2 + a 2 x ω(x ) 2 = x 2 a 2 x a 2 = x a 2, and the roof is nished. Now the main theorem of the section reads as follows. Theorem 3.2. For any invertible a A, the inequality a 1 ω(a 1 ) 2 a 1 2 a ω(a) 2 holds. Proof. We clearly have that xa 2 x a 2 for any x A. In fact, ω( xa 2 ) = ω(a x 2 a) ω(a x 2 a) = x 2 ω( a 2 ).

10 10 ÁDÁM BESENYEI AND ZOLTÁN LÉKA Combining the revious inequality with Lemma 3.1, it follows that and the roof is comlete. a 1 ω(a 1 ) 2 = a 1 (ω(a 1 a) ω(a 1 )a)) 2 = a 1 (ω(a 1 a) ω(a 1 )a)) 2 a 1 ω(a 1 a) ω(a 1 )a) 2 a 1 2 a ω(a) 2, With [8, Proosition 3.4] at hand, we immediately obtain the following Theorem 3.3. Let A be a unital C -algebra. For any faithful state ω of A, σ ω 2 (a) is a strongly Leibniz seminorm. Alternatively, for any faithful tracial state ω, we can dene a derivation on a Banach algebra to infer the above corollary. In fact, let us consider the Banach sace A L 2 (A, ω) endowed with the norm (x, y) = max( x, y 2 ). The linear oerators E : (x, y) (0, ω(x)1 A ) and T a : (x, y) (xa, ya) on A L 2 (A, ω) dene a strongly Leibniz seminorm L on A via the norm of the derivation L(a) = δ(t a ) = [E, T a ]. From Lemma 3.1, we have that δ(t a )(x, y) = (0, ω(x)a ω(xa)) a ω(a) 2 x a ω(a) 2 (x, y). With the choice of (1 A, 0), we get δ(t a ) = a ω(a) 2. Since ω is tracial, xa 2 a x 2. Hence it clearly follows that T a = a. Notice that T ab = T a T b. Now a direct alication of the derivation rules gives that δ(t a ) = L(a) = σ2 ω (a) is a strongly Leibniz seminorm. References [1] K.M.R. Audenaert, Variance bounds, with an alication to norm bounds for commutators, Linear Algebra Al., 432 (2010), [2] T. Bhattacharyya and P. Grover, Characterization of BirkhoJames orthogonality, J. Math. Anal. Al., 407 (2013), [3] C. Franchetti, The norm of the minimal rojection onto hyerlanes in L [0, 1] and the radial constant, Boll. Un. Mat. Ital. B, (7) 4, (1990), [4] Z. Léka, A note on central moments in C -algebras, htt://arxiv.org/abs/ , to aear in J. Math. Inequal. [5] A. W. Marshall, I. Olkin, B. C. Arnold, Inequalities: Theory of Majorization and Its Alications, second edition, Sringer, [6] S. Rolewicz, On minimal rojections of the sace L [0, 1] on 1-codimensional subsace, Bull. Polish Acad. Sci. Math., 34 (1986), [7] M.A. Rieel, Leibniz seminorms and best aroximation from C -subalgebras, Sci. China Math., 54 (2011), [8] M.A. Rieel, Standard deviation is a strongly Leibniz seminorm, New York J. Math., 20 (2014), Deartment of Alied Analysis, Eötvös Loránd University, H-1117 Budaest, Pázmány P. sétány 1/C, Hungary address: badam@cs.elte.hu Alfréd Rényi Institute of Mathematics, 1053 Budaest, Reáltanoda u address: leka.zoltan@renyi.mta.hu