Let L : ( − 1 2 , 1 2 ) → M ( n × n ) be a continuous matrix-valued function such that each L(t) is symmetric and positive definite.Define M k = ∫ − 1 2 1 2 t k L ( t ) d tI want to prove or disprove that M 2 ≥ M 1 M 0 − 1 M 1 In other words, for all x ∈ R n we have ⟨ M 2 x , x ⟩ ≥ ⟨ M 1 M 0 − 1 M 1 x , x ⟩This is obviously true for n=1 since by Cauchy-Schwarz inequality we have ( ∫ − 1 2 1 2 t L ( t ) d t ) 2 ≤ ( ∫ − 1 2 1 2 t 2 L ( t ) d t ) ( ∫ − 1 2 1 2 L ( t ) d t ) That is M 1 2 ≤ M 2 M 0 .

Question

Let   L  :  (  −      1    2    ,      1    2    )  →  M  (  n  ×  n  ) be a continuous matrix-valued function such that each L(t) is symmetric and positive definite.Define      M    k    =      ∫          −              1        2                            1        2                  t    k    L  (  t  )  d  tI want to prove or disprove that      M    2    ≥      M    1        M    0          −      1            M    1  In other words, for all   x  ∈            R        n   we have  ⟨      M    2    x  ,  x  ⟩  ≥  ⟨      M    1        M    0          −      1            M    1    x  ,  x  ⟩This is obviously true for n=1 since by Cauchy-Schwarz inequality we have            (            ∫          −              1        2                            1        2              t  L  (  t  )  d  t                    )              2    ≤            (            ∫          −              1        2                            1        2                  t    2    L  (  t  )  d  t            )                  (            ∫          −              1        2                            1        2              L  (  t  )  d  t            )      That is       M    1    2    ≤      M    2        M    0    .

Leo Robinson · Accepted Answer

First note that       M    0   and       M    2   are positive definite. By the Schur complement characterization of block positive semidefinite matrices your matrix inequality is equivalent to  Z  =      (                                        M            0                                                M            1                                                            M            1                                                M            2                                )    ≥  0    .So it suffices to show   Z  ≥  0 and we have                    Z                            =                  (                                                                      ∫                                      −                    1                                          /                                        2                                                        1                                          /                                        2                                                  L                (                t                )                                  d                                t                                                              ∫                                      −                    1                                          /                                        2                                                        1                                          /                                        2                                                  t                L                (                t                )                                  d                                t                                                                                      ∫                                      −                    1                                          /                                        2                                                        1                                          /                                        2                                                  t                L                (                t                )                                  d                                t                                                              ∫                                      −                    1                                          /                                        2                                                        1                                          /                                        2                                                                    t                  2                                L                (                t                )                                  d                                t                                              )                                                  =                  ∫                      −            1                          /                        2                                1                          /                        2                                    (                                                    L                (                t                )                                            t                L                (                t                )                                                                    t                L                (                t                )                                                              t                  2                                L                (                t                )                                              )                          d                t                                          =                  ∫                      −            1                          /                        2                                1                          /                        2                                    (                                                    I                                                                    t                I                                              )                L        (        t        )                  (                                                    I                                            t                I                                              )                          d                t                                          =                  ∫                      −            1                          /                        2                                1                          /                        2                          N        (        t        )        L        (        t        )        N        (        t                  )          T                          d                t            where on the final line we defined   N  (  t  )  =      (                            I                                      t          I                      )  . Note that as L(t)&amp;gt;0 for all t we have   N  (  t  )  L  (  t  )  N  (  t      )    T    ≥  0 for all t and thus   Z  ≥  0 and we are done.

Let L:(−1/2,1/2) -> M(n xx n) be a continuous matrix-valued function such that each L(t) is symmetric and positive definite. Define M_k=int_(-1/2)^(1/2) t^k L(t)dt I want to prove or disprove that M_2 >= M_1M_0^(−1) M_1

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