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Thread: how can u prove this

  1. #1
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    how can u prove this

    $\displaystyle \int_{0}^{1}{|f|}=0\Rightarrow f=0 $
    given f is continuous on [0,1]
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by szpengchao View Post
    $\displaystyle \int_{0}^{1}{|f|}=0\Rightarrow f=0 $
    given f is continuous on [0,1]
    this is not true in general. more conditions are needed for this to be true. i believe flyingsquirrel answered a similar question for you earlier. baring that in mind, i don't think it would be hard for you to come up with a counter-example to this
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  3. #3
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    .

    that question is about integrable one.

    this is a continuous function f!!
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by szpengchao View Post
    that question is about integrable one.

    this is a continuous function f!!
    you said continuous on [0,1], that is not enough, as the function $\displaystyle f: \mathbb{R} \mapsto \mathbb{R}$ defined by

    $\displaystyle f(x) = \begin{cases} 0 &\mathrm{if}\,\,0\leq x \le1 \\ 1& \mathrm{otherwise}\end{cases}$

    is continuous on [0,1] but is not identically zero, yet $\displaystyle \int_0^1 |f(x)|~dx = 0$
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  5. #5
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    ..

    should I make it more clear ?

    $\displaystyle \int_{0}^{1}{|f|}=0\Rightarrow f(x)=0 \forall x\in [0,1] $
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by szpengchao View Post
    should I make it more clear ?

    $\displaystyle \int_{0}^{1}{|f|}=0\Rightarrow f(x)=0 \forall x\in [0,1] $
    ok, now that makes sense.

    try a proof by contradiction. assume the integral is zero but f is not. this would mean f is not continuous, a contradiction
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  7. #7
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    。。

    $\displaystyle

    \int_{0}^{1}{|f|}=0\Rightarrow \exists q\in[0,1], f(q)=0. $

    assume there exists p in [0,1] , $\displaystyle f(p)\neq0, $ Then we can find a $\displaystyle \epsilon=|f(p)|>0 , |p-q|<\delta $ there discontinuous.
    contradicts.
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  8. #8
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    If $\displaystyle f$ is continuous on $\displaystyle \left[ {0,1} \right]$ so is $\displaystyle \left| f \right|$.
    Suppose that $\displaystyle \left( {\exists c \in \left[ {0,1} \right]} \right)\left[ {\left| {f(c)} \right| > 0} \right]$ then $\displaystyle \left( {\exists \left[ {u,v} \right] \subseteq \left[ {0,1} \right]} \right)\left[ {\left( {\forall z \in \left[ {u,v} \right]} \right)\left[ {\left| {f(z)} \right| > 0} \right]}\right]$.
    What can we say about $\displaystyle \int\limits_u^v {\left| {f(z)} \right|dz} $.
    Do you see the contradiction in that?
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