If I define the integral for ∫ a b F ( t ) d t as ( ∫ a b F 1 ( t ) d t , ∫ a b F 2 ( t ) d t , … , ∫ a b F n ( t ) d t ). How do I then show that | ∫ a b F ( t ) d t | ≤ ∫ a b | F ( t ) | d t ?If we write out the inequality we have ( ∫ a b F 1 ( t ) d t ) 2 + ( ∫ a b F 2 ( t ) d t ) 2 + ⋯ + ( ∫ a b F n ( t ) d t ) 2 ≤ ∫ a b F 1 ( t ) 2 + F 2 ( t ) 2 + ⋯ + F n ( t ) 2 d t .attempt:If I use Jenssen's inequality I get for the left side: ( ∫ a b F 1 ( t ) d t ) 2 + ( ∫ a b F 2 ( t ) d t ) 2 + ⋯ + ( ∫ a b F n ( t ) d t ) 2 ≤ ∫ a b F 1 2 ( t ) d t + ∫ a b F 2 2 ( t ) d t + ⋯ + ∫ a b F n 2 ( t ) d t = ∑ i = 1 n ∫ a b F i 2 ( t ) d t = ∫ a b ∑ i = 1 n F i 2 ( t ) d t .The problem I have now is that the square root function is concave, so if I use Jenssen again, I get the opposite inequality as the one I need. Any idea on how to solve this?

Question

If I define the integral for       ∫    a    b        F    (  t  )  d  t as   (      ∫    a    b        F    1    (  t  )  d  t  ,      ∫    a    b        F    2    (  t  )  d  t  ,  …  ,      ∫    a    b        F    n    (  t  )  d  t  ). How do I then show that      |          ∫      a      b              F        (    t    )    d    t    |    ≤      ∫    a    b        |          F        (    t    )    |    d  t  ?If we write out the inequality we have                    (                  ∫          a          b                          F          1                (        t        )        d        t        )            2        +                  (                  ∫          a          b                          F          2                (        t        )        d        t        )            2        +    ⋯    +                  (                  ∫          a          b                          F          n                (        t        )        d        t        )            2        ≤      ∫    a    b              F      1        (    t          )      2        +          F      2        (    t          )      2        +    ⋯    +          F      n        (    t          )      2        d  t  .attempt:If I use Jenssen&#039;s inequality I get for the left side:                    (                  ∫          a          b                          F          1                (        t        )        d        t        )            2        +                  (                  ∫          a          b                          F          2                (        t        )        d        t        )            2        +    ⋯    +                  (                  ∫          a          b                          F          n                (        t        )        d        t        )            2        ≤            ∫      a      b              F      1      2        (    t    )    d    t    +          ∫      a      b              F      2      2        (    t    )    d    t    +    ⋯    +          ∫      a      b              F      n      2        (    t    )    d    t      =            ∑              i        =        1            n              ∫      a      b              F      i      2        (    t    )    d    t    =            ∫      a      b              ∑              i        =        1            n              F      i      2        (    t    )    d    t    .The problem I have now is that the square root function is concave, so if I use Jenssen again, I get the opposite inequality as the one I need. Any idea on how to solve this?

nuvolor8 · Accepted Answer

The linearity of the integral, that of the inner product and the Schwarz inequality imply, for any             v      →        ∈            R        n    ,,                    (1)                              ⟨                      ∫            a            b                    F          (          t          )          d          t          ,                                    v              →                                ⟩                =                  ∫          a          b                ⟨        F        (        t        )        ,                              v            →                          ⟩        d        t        ≤                  ∫          a          b                          |                ⟨        F        (        t        )        ,                              v            →                          ⟩                  |                d        t        ≤                  ∫          a          b                          |                F        (        t        )                  |                                      v            →                                    |                          |                d        t        ≤                  |                                      v            →                                    |                          ∫          a          b                          |                F        (        t        )                  |                d        t        .                                (2)                    Taking                                      v            →                          =                  ∫          a          b                F        (        t        )        d        t        ,                we get                          |                                      v            →                                                |                    2                ≤                  |                                      v            →                                    |                          ∫          a          b                          |                F        (        t        )                  |                d        t                                (3)                    which is                          |                      ∫            a            b                    F          (          t          )          d          t          |                ≤                  ∫          a          b                          |                F        (        t        )                  |                d        t        .

Kassandra Ross · Accepted Answer

This is slightly overkill, but this theorem is a consequence of the Minkowski integral inequality.

If I define the integral for <msubsup> ∫ a b </msubsup> <m

Answered question

Answer & Explanation

New Questions in Pre-Algebra