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Integral norm proving

  1. Nov 16, 2012 #1
    1. The problem statement, all variables and given/known data
    show that ||f||1 = ∫|f| (integral from 0 to 1) does define a norm on the subspace C[0,1] of continuous functions

    and also the same for ||f||= ∫t|f(t)|dt is a norm on C[0,1]


    2. Relevant equations

    (there are 3 conditions , i just dont know how to prove that ||v||>0,||v||=0 implies v=0)
     
  2. jcsd
  3. Nov 16, 2012 #2

    LCKurtz

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    If a continuous function is nonzero (or positive) at a point, what about its value nearby that point, and why?
     
  4. Nov 16, 2012 #3
    yes if it is not zero ,then |f(x)-f(c)|<esillope then choose esillope to be f(c)/2 then it will get |f(x)|>sth..... but how can u ensure that f(c) >0 for esillope has to be >0
     
  5. Nov 16, 2012 #4

    LCKurtz

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    You have this property of integrals to work with: If ##f,g\in C[a,b]## and ##f(x)> g(x)## on [a,b] then ##\int_a^b f(x)\, dx > \int_a^b g(x)\, dx##
     
  6. Nov 17, 2012 #5
    yes,i worked out the first one now,but how about the secound one for t f(t) i dont know how to construct the inequality that u have mentioned above..
     
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