I have a question: if { f n } ⊂ L 2 ( X , d μ ) with | | f n | | L 2 ≤ 1, then f n ( x ) / n → 0 for a.e. x. Here we are considering a measure space ( X , M , μ ). Intuitively, I would think we prove this by | | f n n | | L 2 = ( ∫ X | f n ( x ) n | 2 d μ ( x ) ) 1 2 = 1 | n | ( ∫ X | f n ( x ) | 2 d μ ( x ) ) 1 2 ≤ 1 | n | ⇒ lim n → ∞ f n n = 0However, this can't be true because the same proof could be used to show that the result holds in the case where we consider the L 1 norm, which is false. What am I missing here?

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I have a question: if             {            f    n              }        ⊂      L    2    (  X  ,  d  μ  ) with       |        |        f    n        |              |                      L        2              ≤  1, then       f    n    (  x  )      /    n  →  0 for a.e. x. Here we are considering a measure space   (  X  ,      M    ,  μ  ). Intuitively, I would think we prove this by            |                  |                  f      n        n              |                          |                            L        2              =            (            ∫    X              |                          f        n            (      x      )        n                      |              2    d  μ  (  x  )                    )                            1        2              =      1                  |            n              |                        (            ∫    X        |        f    n    (  x  )            |        2    d  μ  (  x  )                    )                            1        2              ≤      1                  |            n              |              ⇒      lim          n      →      ∞                  f      n        n    =  0However, this can&#039;t be true because the same proof could be used to show that the result holds in the case where we consider the       L    1   norm, which is false. What am I missing here?

aletantas1x · Accepted Answer

For a fixed   k  &amp;gt;  0, note that  μ  (  |      f    n    |      /    n  &amp;gt;  1      /    k  )  ≤            ‖              f        n                    /            n              ‖        2        2                    (      1              /            k              )        2              ≤            k      2              n      2      by Markov&#039;s inequality. This way,  μ  (  |      f    n    |      /    n  &amp;gt;  1      /    k   for infinitely many   n  )  =  0thanks to Borel-Cantelli. (Note that this uses that   ∑  1      /        n    2    &amp;lt;  ∞, which is the key difference of using       L    2   instead of       L    1  .) At last, we get that  μ      (    x    :    ∃    k    ,    |          f      n        |          /        n    &amp;gt;    1          /        k     for infinitely many     n    )    =  0by countable additivity. You can easily check that the complement consists of points   x such that for all   k, there exists some       n    0   such that   |      f    n    (  x  )      /    n  |  &amp;lt;  1      /    k for   n  ≥      n    0  , hence       f    n    (  x  )      /    n  →  0.

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