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Ess Sup Norm as limit ->oo of L^p norm

  1. Mar 16, 2009 #1


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    I am trying to show that the ess sup norm is the limit of the L^p

    norms as p-->oo . i.e., ess sup =lim_p->oo ( {Int f^p)^1/p

    Please tell me if this is correct:

    1) Def. ess sup f(t)=inf{M:m(t:f(t)>M)=0 }

    Then, f(t)>M only in the set S , with m(S)=0 , and f(t

    So Lim_p->oo ||f||_p =Lim_p->oo (Int_[0,1] ||f||^p)^1/p

    <= Lim_p->oo(Int[0,1]-S |M|^p +Int_S (M')^p )^1/p .

    Since m(S)=0 , integral on the right goes to 0

    (is this O.K if f is oo in S?) , so we get:

    Lim_p->oo(Int_([0,1]-S) (M)^p)^1/p) . Then, since m([0,1]-S)=1

    (Int_([0,1]-S)|M|^p )^1/p =M^p , so we get

    Lim_p->oo (M^p)^1/p .

    Then the sequence :{a_n}=(M^n)^1/n =M , is constant, with limit M.

    Does this work?

    P.S: Sorry, I am still learning Latex.
  2. jcsd
  3. Mar 27, 2009 #2
    This is not quite correct, but you rather have half of what is needed. If [tex]||f||_\infty =M[/tex], then [tex]\int_{ [0,1] - S } |f|^p \leq M^p[/tex], but equality need not hold. Therefore you have [tex]\lim_{ p \to \infty } ||f||_p } \leq M[/tex]. You just need to prove the reverse inequality.
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