The logarithmic mean of two convex functionals

Proposition 3.1

Let A , B ∈ ℬ ( H ) . Then the following assertions hold:

1. Assume that A and B are self-adjoint. Then, Q A ≤ ( ≥ ) Q B if and only if A ≤ ( ≥ ) B .

2. Q A + Q B = Q A + B and α Q A = Q α A for any real number α .

3. Q A is continuous. Furthermore, Q A is convex if and only if A is positive.

4. Assume that A ∈ ℬ + ⁎ (H). Then the conjugate of Q A is given by

   ∀ x ⁎ ∈ E ⁎ Q A ⁎ ( x ⁎ ) = 1 2 〈 A − 1 x ⁎ , x ⁎ 〉 ,

   or, in short,

   (3.1) Q A ⁎ = Q A − 1 .

5. Q A is G-differentiable at any x ∈ H . If further A is self-adjoint, then ∇ Q A ( x ) = A x . So, ∂ f ( x ) = { A x } whenever A is (self-adjoint) positive.

6. Let A , B ∈ ℬ + ⁎ (H). Then Q A □ Q B = Q A / / B , where A / / B ≕ ( A − 1 + B − 1 ) − 1 is called the parallel sum of A and B.

Proof

The proofs of (i), (ii), (iii) and (v) are straightforward. For the proof of (iv), see [17] for instance. For more details about (vi) we can consult [18,19].□

Remark 3.2

By (3.1) we immediately deduce that Q I ⁎ = Q I ≕ 1 2 ∥ ⋅ ∥ 2 . That is, 1 2 ∥ ⋅ ∥ 2 is self-conjugate. By using the Fenchel inequality (2.1) it is not hard to check that 1 2 ∥ ⋅ ∥ 2 is the unique self-conjugate functional defined on a Hilbert space.

Theorem 3.3

Let Φ : Γ 0 ( H ) × Γ 0 ( H ) → ℝ ˜ H and Ψ : Γ 0 ( H ) × Γ 0 ( H ) → ℝ ˜ H be two binary maps such that Φ ( f , g ) ≤ Ψ ( f , g ) for any f , g ∈ Γ 0 ( H ) . Assume that, for any A , B ∈ ℬ + ⁎ ( H ) we have

where θ ( A , B ) , γ ( A , B ) ∈ ℬ ( H ) are self-adjoint. Then

Proof

Since A , B ∈ ℬ + ⁎ ( H ) , by Proposition 3.1(iii) we have Q A , Q B ∈ Γ 0 ( H ) . It follows that Φ ( Q A , Q B ) ≤ Ψ ( Q A , Q B ) and so Q θ ( A , B ) ≤ Q γ ( A , B ) , with θ ( A , B ) and γ ( A , B ) ∈ ℬ ( H ) are self-adjoint. By Proposition 3.1(i), we conclude that θ ( A , B ) ≤ γ ( A , B ) and the proof is complete.□

Example 3.4

Let A , B ∈ ℬ + ⁎ ( H ) . Assume that A ≤ B , then by Proposition 3.1(i), we have Q A ≤ Q B . By the point-wise decrease monotonicity of the Fenchel duality we infer that Q B ⁎ ≤ Q A ⁎ and by (3.1) we deduce that Q B − 1 ≤ Q A − 1 . Again by Proposition 3.1(i) we conclude that B − 1 ≤ A − 1 . This means that the map X ↦ X − 1 , for X ∈ ℬ + ⁎ ( H ) , is operator monotone (increasing).

Example 3.5

For fixed t ∈ [0, 1] , we set

Remark 4.1

1. The λ -weighted functional geometric mean can be written as follows:

   f # λ g = ∫ 0 1 f ! t g d ν λ ( t ) , λ ∈ ( 0 , 1 ) ,

   where ν λ ( t ) defines a family of probability measures on the interval ( 0 , 1 ) defined by

   (4.2) d ν λ ( t ) ≕ sin ( π λ ) π t λ − 1 ( 1 − t ) λ d t , λ ∈ ( 0 , 1 ) .

2. Although the previous functional means can be defined, by the same expressions, even f , g ∉ Γ 0 ( E ) , we restrict ourselves throughout this paper to assume that f , g ∈ Γ 0 ( E ) . In this case, f ∇ λ g , f ! λ g and f # λ g belong to Γ 0 ( E ) provided that dom f ∩ dom g ≠ ∅ .

We extend the previous functional means on the whole interval [0,1] by setting:

We pay attention that these latter relations cannot be deduced from (4.1), by virtue of the convention 0 . ( + ∞ ) = + ∞ . The previous functional means satisfy the following relationships:

for any λ ∈ [ 0 , 1 ] . The two first relationships of (4.4) are immediate and for the third one we can consult [7]. In particular, if λ = 1 / 2 , the three previous functional means are symmetric in f and g. Note that f ∇ λ f = f ! λ f = f # λ f = f . Furthermore, the following inequalities hold, see [7]

and f ∇ λ g , f ! λ g , f # λ g ∈ Γ 0 ( E ) provided that dom f ∩ dom g ≠ ∅ . Denoting by m λ one of any mean among ∇ λ , ! λ , # λ and utilizing (2.2), we can easily see that, for any α > 0 ,

Otherwise, for any A , B ∈ ℬ + ⁎ ( H ) , we have the following relationships:

where

stands for the λ -weighted operator arithmetic mean, the λ -weighted operator harmonic mean and the λ -weighted operator geometric mean of A and B, respectively. For λ = 1 / 2 , they are also simply denoted by A ∇ B , A ! B and A # B , respectively. The relationships (4.7) justify that the previous functional means are, respectively, extensions of their related operator means. Furthermore, according to Theorems 3.3, (4.5) and (4.7) immediately imply that the following operator inequalities

hold for any A , B ∈ ℬ + ⁎ ( H ) and λ ∈ [0, 1] . It is worth mentioning that (4.9), which are well known in the operator mean theory, are here again obtained in a simultaneous manner and under a convex point of view that does not need to refer to the techniques of functional calculus.

Proposition 5.1

Let f , g ∈ Γ 0 ( E ) and λ ∈ [0, 1] . Then the two binary maps ( f , g ) ↦ f ! λ g and ( f , g ) ↦ f # λ g are both separately point-wisely increasing.

Proof

Let f 1 , f 2 ∈ Γ 0 ( E ) be such that f 1 ≤ f 2 . By the point-wise decrease monotonicity of the map f ↦ f ⁎ , we deduce ( 1 − λ ) f 2 ⁎ + λ g ⁎ ≤ ( 1 − λ ) f 1 ⁎ + λ g ⁎ and again ( 1 − λ ) f 1 ⁎ + λ g ⁎ ⁎ ≤ ( 1 − λ ) f 2 ⁎ + λ g ⁎ ⁎ , i.e., f 1 ! λ g ≤ f 2 ! λ g , which means that ( f , g ) ↦ f ! λ g is point-wisely increasing with respect to the first argument f. By virtue of (4.4) we then deduce that ( f , g ) ↦ f ! λ g is point-wisely increasing with respect to the second argument g, too. This, with the relation of f # λ g given in (4.1) and the linearity of the integral, implies that ( f , g ) ↦ f # λ g is separately point-wisely increasing. The proof is complete.□

Theorem 5.2

Let ( f , g ) ∈ W and λ ∈ [0, 1] . Then the two binary maps ( f , g ) ↦ f ! λ g and ( f , g ) ↦ f # λ g are both point-wisely concave.

Proof

By the same reasons as in the proof of the previous proposition, we need to prove that ( f , g ) ↦ f ! λ g is point-wisely concave with respect to the first argument f. By definition of f ! λ g , with the help of (2.2) and (2.4), we can write

Since ( f , g ) ∈ W , it is easy to verify that condition (2.6) is satisfied here, i.e., int dom ( 1 − λ ) . f ⁎ ∩ dom λ . g ⁎ ≠ ∅ . It follows that f . ( 1 − λ ) □ g . λ ∈ Γ 0 ( E ) and so (5.1) becomes

Now, let ( f 1 , g 1 ) , ( f 2 , g 2 ) ∈ W and t ∈ (0, 1) . By (5.2) and the definition of the inf-convolution, we have for any x ∈ E

Hence, the desired result.□

Remark 5.3

Let ( f , g ) ∈ W . Then, for any λ ∈ (0, 1) we have

Indeed, it is easy to see that dom ( f ∇ λ g ) = dom f ∩ dom g for any λ ∈ (0, 1) . Otherwise, by using (5.2) it is not hard to check that

This, when combined with (4.5), yields (5.3).

Proposition 5.4

Let f , g ∈ Γ 0 ( E ) be fixed. Then the map t ↦ f ! t g is point-wisely convex on [0, 1] .

Proof

By definition we have f ! t g = ( 1 − t ) f ⁎ + t g ⁎ ⁎ . Since the map ϕ ↦ ϕ ⁎ is point-wisely convex and the map t ↦ ( 1 − t ) f ⁎ + t g ⁎ is point-wisely affine, the desired result follows immediately.□

Theorem 5.5

With the above, the following assertions are met:

1. The map s ↦ G s ( f , g ; λ ) is point-wisely convex on [0, 1] .

2. For any s ∈ [0, 1] , we have

   (5.5) f ! λ g ≤ G s ( f , g ; λ ) ≤ ( f ! λ g ) ∇ s ( f # λ g ) ≤ f # λ g ( ≤ f ∇ λ g ) ,

   which refines the left inequality in (4.5).

3. We have

   (5.6) inf s ∈ [ 0 , 1 ] G s ( f , g ; λ ) = f ! λ g and sup s ∈ [ 0 , 1 ] G s ( f , g ; λ ) = f # λ g ,

   where the infimum and supremum are taken for the point-wise order.

4. The map s ↦ G s ( f , g ; λ ) is point-wisely monotone increasing.

Proof

1. Since the map t ↦ f ! t g is point-wisely convex on [0, 1] and the real-function s ↦ s t + ( 1 − s ) λ ∈ [ 0 , 1 ] is affine, we deduce the desired result.

2. By the point-wise convexity of t ↦ f ! t g , we get

   f ! s t + ( 1 − s ) λ g ≤ s f ! t g + ( 1 − s ) f ! λ g .

   If we multiply this latter inequality by d ν λ ( t ) and we integrate over t ∈ [0, 1] we get the middle inequality of (5.5). The right inequality of (5.5) is obvious by virtue of the inequality f ! λ g ≤ f # λ g . Now, let us show the left inequality of (5.5). For the sake of simplicity for the reader, we fix f , g ∈ Γ 0 ( E ) and we set Φ ( s ) = f ! s g . By Proposition 5.4, Φ is point-wisely convex on [0, 1] . With this, (5.4) takes the following form:

   G s ( f , g ; λ ) = ∫ 0 1 Φ ( s t + ( 1 − s ) λ ) d ν λ ( t ) .

   We can apply the integral Jensen inequality to this latter equality [22], and we then obtain

   (5.7) G s ( f , g ; λ ) ≥ Φ ∫ 0 1 ( s t + ( 1 − s ) λ ) d ν λ ( t ) .

   We have

   ∫ 0 1 ( s t + ( 1 − s ) λ ) d ν λ ( t ) = s ∫ 0 1 t d ν λ ( t ) + ( 1 − s ) λ ∫ 0 1 d ν λ ( t ) = s ∫ 0 1 t d ν λ ( t ) + ( 1 − s ) λ .

   In another part, let us denote by Γ and B the standard special gamma and beta functions, respectively. By (4.2), we get

   ∫ 0 1 t d ν λ ( t ) = sin ( π λ ) π ∫ 0 1 t λ ( 1 − t ) − λ d t = sin ( π λ ) π B ( 1 + λ , 1 − λ ) = sin ( π λ ) π Γ ( 1 + λ ) Γ ( 1 − λ ) Γ ( 2 ) = sin ( π λ ) π λ Γ ( λ ) Γ ( 1 − λ ) = λ .

   Substituting these in (5.7) we obtain

   G s ( f , g ; λ ) ≥ Φ ( s λ + ( 1 − s ) λ ) = Φ ( λ ) ≕ f ! λ g ,

   and the left inequality of (5.5) is obtained.

3. Since G 0 ( f , g ; λ ) = f ! λ g and G 1 ( f , g ; λ ) = f # λ g , (5.6) is immediate from (5.5).

4. If 0 ≤ s 1 < s 2 ≤ 1 , the point-wise convexity of s ↦ G s ( f , g ; λ ) implies that

This, with G 0 ( f , g ; λ ) = f ! λ g and (5.6), yields the desired result. The proof is complete.□

Definition 6.1

Let f , g ∈ Γ 0 ( E ) . The expression

is called the logarithmic mean of f and g.

Proposition 6.2

Let f , g ∈ Γ 0 ( E ) . Then the following assertions hold:

1. L ( f , f ) = f and L ( f , g ) = L ( g , f ) .

2. For any c , d ∈ ℝ , L ( f + c , g + d ) = L ( f , g ) + c ∇ d , where c ∇ d ≕ c + d 2 denotes the extension of the arithmetic mean for any two real numbers.

3. Let α , β > 0 , then

where α # β ≕ α β is the scalar geometric mean of α and β . In particular, one has