Appendix A. Technical Lemmas
In this section we gather all technical lemmas needed to prove our main theorems.
Lemma A1. For any , let be the matrix such that: (i) if ; (ii) if either , or , with ; (iii) otherwise. We have that is a positive-semidefinite matrix.
Proof. To show the claim, we resort to the Sylvester’s criterion, stating that a symmetric matrix is positive-semidefinite if and only if the determinant of each principal minor of (i.e., each upper upper left h-by-h corner of ) is non-negative. Let be a matrix such that: (i) if ; (ii) if , and, either , or ; (iii) otherwise. We have that each principal minor of matrix is of type for some . Thus, it is sufficient showing that the determinant of matrix , denoted as , is non-negative for any .
We first show by induction on integers that for any fixed . If we trivially get . Now, we assume that holds for some , and we show that . We get , where the first equality comes from the Laplace expansion for computing the determinant, and the second equality comes from the inductive hypothesis.
By using the fact that holds for any and any integer , we have that for any , where the last inequality holds since quantity is always non-negative for any if . Thus each principal minor of has a non-negative determinant, and the claim follows. □
Lemma A2. Let be the function such that . For any such that and , it holds that .
Proof. At a first glance, in order to use standard arguments from calculus, we allow the 6-tuples to take values in the set of non-negative real numbers.
We first show that, in such an extended scenario, attains its minimum for 6-tuples such that and . Consider to this aim the 6-tuple , where . By definition of , we get
Hence, we do not lose in generality by restricting to 6-tuples of non-negative real values such that and . In this case becomes . Consider the two partial derivatives and . Since they are linear and decreasing in b and e, respectively, it follows that is minimized at one of the following four cases: , , and .
In the first case, becomes . Since , is minimized at . By substituting, becomes which is always non-negative for any .
In the second case, becomes . Since , is minimized at . By substituting, becomes which is always non-negative for any .
In the third case, becomes . Since , is minimized at . By substituting, becomes which is always non-negative for any .
In the fourth case, becomes . Since , is minimized at . By substituting, becomes which is always non-negative for any .
Hence, in order to complete the proof, we are left to settle the following cases: , , , , , and .
In the case , becomes which is always non-negative for any . In the case , becomes which is always non-negative for any . In the cases and , becomes which is always non-negative for any . In the cases and , becomes which is always non-negative for any . Finally, in the case , becomes which is always non-negative for any . Hence, we are only left to consider the case for which becomes . Since , it holds that which is always non-negative since for any such that . □
Lemma A3. Let be the function such that . For any such that , it holds that .
Proof. Since , is minimized at . By substituting, we get which is non-negative for any . □
Lemma A4. Let be the function such that . For any such that , it holds that .
Proof. Since , is minimized at . By substituting, we get which is non-negative for any . □
Lemma A5. Let be the function such that . For any , it holds that .
Proof. Since , is minimized at . By substituting, we get which is non-negative for any . □
Lemma A6. Let be the function such that . For any , it holds that .
Proof. Since , is minimized at . By substituting, we get which is non-negative for any . □
Lemma A7. Let be the function such that . For any such that and , it holds that .
Proof. At a first glance, in order to use standard arguments from calculus, we allow the 6-tuples to take values in the set of non-negative real numbers. Since and , is minimized at one of the following four cases: , , and .
In the first case, we get . Since , is minimized at which yields . Since , is minimized at which yields . The claim then follows for any by applying Lemma A3. For the leftover tuples of the form , we get which is always non-negative for any .
In the second case, we get . Since , is minimized at , which yields . The claim then follows for any by applying Lemma A3. For the leftover tuples of the form , we get which is always non-negative for any .
In the third case, we get . Since , is minimized at either or . For , we get . Since , is minimized at . This yields and the claim then follows by applying Lemma A5. For , we get . Since , is minimized at which yields and the claim then follows for any by applying Lemma A4. For the leftover tuples of the form , we get which is always non-negative for any . Hence, we are still left to prove what happens for the tuples of the form . In this case, we get which is always non-negative for any .
In the fourth case, we get . Since , is minimized at either or . For , we get . Since , is minimized at either or . The first case yields and the claim then follows by applying Lemma A6, while the second one yields and the claim then follows by applying Lemma A5. For , we get . Since , is minimized at which yields and the claim then follows by applying Lemma A5. □
Lemma A8. Let be the function such that . For any , it holds that .
Proof. Since , is minimized at either or .
In the first case, becomes which is always non-negative for any .
In the second case, becomes which is always non-negative for any . For the leftover case , becomes which is always non-negative for any . For the other case , becomes which is always non-negative for any . □
Lemma A9. Let be the function such that . For any such that and , it holds that .
Proof. Note first, that for 6-tuples of the form , it holds that , since and , for 6-tuples of the form , it holds that , since , and for 6-tuples of the form , it holds that , since . Hence, in the sequel of the proof, we avoid to consider the cases , and .
At a first glance, in order to use standard arguments from calculus, we allow the 6-tuples to take values in the set of non-negative real numbers. Since it holds that and , is minimized at one of the following four cases: , , and .
In the first case, we get . The claim follows by applying Lemma A8, since .
In the second case, we get . Since , is minimized at , which yields . Since , is minimized for . In this case, becomes . The claim follows by applying Lemma A8, since .
In the third case, we get . Since , is minimized for , which yields . Since , is minimized for . In this case, becomes . The claim follows by applying Lemma A8, since .
In the fourth case, we get . Since , is minimized at either or . For , becomes . Since , is minimized at either or . In these two cases, becomes, respectively, and which are always non-negative because of Lemma A8 and the fact that . For , becomes . Since , is minimized at either or . In these two cases, becomes, respectively, and which are always non-negative because of Lemma A8 and the fact that . □
Lemma A10. Let be the function such that . For any , it holds that .
Proof. Since , is minimized at either or .
In the first case, becomes which is always non-negative for any .
In the second case, becomes which is always non-negative for any . For the leftover case , becomes , which is non-negative for any . For the other case of , becomes which is non-negative for any . □
Lemma A11. Let be the function such that . For any , it holds that .
Proof. Since , is minimized at either or .
In the first case, becomes which is always non-negative for any .
In the second case, becomes which is always non-negative for any . For the leftover case of , becomes which is non-negative for any . □
Lemma A12. Let be the function such that . For any , it holds that .
Proof. Since , is minimized at either or .
In the first case, becomes which is always non-negative for any .
In the second case, becomes which is always non-negative for any . For the leftover case , becomes , which is non-negative for any . For the other case of , becomes which is non-negative for any . □
Lemma A13. Let be the function such that . For any such that and , it holds that .
Proof. At a first glance, in order to use standard arguments from calculus, we allow the 6-tuples to take values in the set of non-negative real numbers. Since it holds that and , is minimized at one of the following four cases: , , and .
In the first case, we get . Since , is minimized at , which yields . Since , is minimized for . In this case, becomes . The claim follows by applying Lemma A10.
In the second case, we get . Since , is minimized at , which yields . Since , is minimized at either and . In both cases becomes and the claim follows by applying Lemma A10.
In the third case, we get . Since , is minimized at , which yields . Since , is minimized at either or . For , becomes and the claim follows by applying Lemma A11. For , becomes and the claim follows for any 6-tuple such that by applying Lemma A12. Hence, we are left to consider the 6-tuples of the form . In this case becomes which is always non-negative since .
In the fourth case, we get . Since , is minimized at either or . For , becomes . Since , is minimized at either or . In these two cases, becomes, respectively, and which are always non-negative because of Lemma A10 and the fact the . For , becomes . Since , is minimized at either or . In both cases, becomes and the claim follows for any 6-tuple such that by applying Lemma A12. Hence, we are left to consider the 6-tuples of the form . In this case, becomes which is always non-negative since . □