Is it possible to have lim sup n → ∞ ‖ g h n ‖ 1 ‖ h n ‖ 1 = ∞?Consider a metric space ( Ξ , F , ν ) and g and { h n } n non negative, measurable functions in this space. Consider then the norm ‖ g ‖ 1 = ∫ Ξ | g ( ω ) | d ν ( ω )and that1. ‖ g ‖ 1 &lt; ∞2. ‖ h n ‖ 1 &lt; ∞ ∀ n3. h n ≤ 1, so we have that ‖ g h n ‖ 1 ≤ 1 ∀ n .4. lim n → ∞ ‖ h n ‖ = 0My question is, assuming that g doesn't depend on n, is it possible to have lim sup n → ∞ ‖ g h n ‖ 1 ‖ h n ‖ 1 = ∞ ?The result is true if g and h n were uncorrelated, or at least negativelly correlated (that is, ‖ g h n ‖ 1 ≤ ‖ g ‖ 1 ‖ h n ‖ 1 ). But I was wondering if the fact g doesn't depend on n is enough.

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Is it possible to have       lim sup          n      →      ∞                  ‖      g              h        n                    ‖        1                    ‖              h        n                    ‖        1              =  ∞?Consider a metric space   (  Ξ  ,      F    ,  ν  ) and   g and   {      h    n        }    n   non negative, measurable functions in this space. Consider then the norm  ‖  g      ‖    1    =      ∫          Ξ            |    g  (  ω  )      |    d  ν  (  ω  )and that1.   ‖  g      ‖    1    &amp;lt;  ∞2.   ‖      h    n        ‖    1    &amp;lt;  ∞    ∀  n3.       h    n    ≤  1, so we have that   ‖  g      h    n        ‖    1    ≤  1    ∀  n  .4.       lim          n      →      ∞        ‖      h    n    ‖  =  0My question is, assuming that   g doesn&#039;t depend on   n, is it possible to have      lim sup          n      →      ∞                  ‖      g              h        n                    ‖        1                    ‖              h        n                    ‖        1              =  ∞  ?The result is true if   g and       h    n   were uncorrelated, or at least negativelly correlated (that is,   ‖  g      h    n        ‖    1    ≤  ‖  g      ‖    1    ‖      h    n        ‖    1  ). But I was wondering if the fact   g doesn&#039;t depend on   n is enough.

glorietka4b · Accepted Answer

We will consider the space   [  0  ,  1  ]. The function   g  (  x  )  =      x          −      1              /            2       is integrable on   [  0  ,  1  ], so   ‖  g      ‖    1    &amp;lt;  ∞. Let       h    n   be the indicator function for the interval   [  0  ,      n          −      1        ]. Then       h    n    ≤  1 and   ‖      h    n        ‖    1    =  1      /    n  →  0. On the other hand,  ‖  g      h    n        ‖    1    =      ∫    0                  n                  −          1                          x          −      1              /            2          d  x  ≥      n          −      1              /            2        ,since       x          −      1              /            2        &amp;gt;      n          1              /            2       for   x  ∈  [  0  ,      n          −      1        ]. Therefore             ‖      g              h        n            ‖              ‖              h        n            ‖        ≥      n          1              /            2      , which diverges to   +  ∞

Is it possible to have <munder> <mo movablelimits="true" form="prefix">lim sup

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