Suppose you have two sequences of complex numbers a i and b i indexed over the integer numbers such that they are convergent in l 2 norm and a has norm greater than b in the sense ∞ &gt; ∑ i | a i | 2 ≥ ∑ i | b i | 2 .Suppose moreover they are uncorrelated over any time delay, meaning ∑ i a i b i − n ¯ = 0 ∀ n ∈ Z .Is it true that the polinomial a ( z ) = ∑ i a i z − i is greater in absolute value than b ( z ) = ∑ i b i z − i for any unit norm complex number z?

Question

Suppose you have two sequences of complex numbers       a    i   and       b    i   indexed over the integer numbers such that they are convergent in       l    2   norm and a has norm greater than b in the sense  ∞  &amp;gt;      ∑    i        |        a    i              |        2    ≥      ∑    i        |        b    i              |        2    .Suppose moreover they are uncorrelated over any time delay, meaning      ∑    i        a    i              b              i        −        n              ¯    =  0    ∀  n  ∈      Z    .Is it true that the polinomial   a  (  z  )  =      ∑    i        a    i        z          −      i       is greater in absolute value than   b  (  z  )  =      ∑    i        b    i        z          −      i       for any unit norm complex number z?

Simone Ali · Accepted Answer

Step 1Since we&#039;re interested only in   z  ∈      S    1   , I&#039;ll call   z  =      e          i      x       and I&#039;ll work in       L                  C              2    [  0  ,  2  π  ] (since you also don&#039;t seem to be actually interested in polynomials). Call   Ξ  (  x  )  =      {                            0                          if           x          ≤          0          ∨          x          ≥          1                                      exp          ⁡                      1                                          x                2                            −              x                                                if           0          &amp;lt;          x          &amp;lt;          1                         and call   Φ  (  x  )  =            (              1                  2          π                            ∫        0        1            |      Ξ      (      t      )              |        2                  d      t      )              −      1              /            2        Ξ  (  x  ) . Then consider the Fourier series in       L    2    [  0  ,  2  π  ]                          a        (                  e                      i            x                          )        :=        5        Φ        (        x        )        =                  ∑                      k            ∈                          Z                                                a          k                          e                      i            k            x                                                    b        (                  e                      i            x                          )        :=        Φ        (        x        −        1        )        =                  ∑                      k            ∈                          Z                                                b          k                          e                      i            k            x                              You have                                            ∑                      k            ∈                          Z                                                a          k                                      b            ¯                                k            −            n                          =                  5                      2            π                                    ∫          0                      2            π                                                              e                              n                i                x                                      Φ            (            x            −            1            )                    ¯                Φ        (        x        )                d        x        =        0                                                    ∑                      k            ∈                          Z                                      |                  a          k                          |          2                =                  25                      2            π                                    ∫          0                      2            π                          |        Φ        (        x        )                  |          2                        d        x        =        25                                                    ∑                      k            ∈                          Z                                      |                  b          k                          |          2                =                  1                      2            π                                    ∫          0                      2            π                          |        Φ        (        x        −        1        )                  |          2                        d        x        =        1            However,   |  a  (      e          i      x        )  |  =  0  &amp;lt;  |  b  (      e          i      x        )  | for   1  &amp;lt;  x  &amp;lt;  2 .Notice that inequalities of norm and absolute value hold eventually for the partial sums, because these Fourier series converge uniformly on   [  0  ,  2  π  ] . However, the condition of &quot;uncorrelation over any time delay&quot; doesn&#039;t hold and, in fact, it can&#039;t hold for non-zero trigonometric polynomials.

Suppose you have two sequences of complex numbers <msub> a i </msub> and <msu

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