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Math Help - is this norm lipschitz equivalent to uniform norm?

  1. #1
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    is this norm lipschitz equivalent to uniform norm?

     \|f\|_{1}=\int_{0}^{1}{|f|} \ \ on \ \ C[0,1]

    uniform norm:  \|f\|_{\infty}=\sup \{|f(x)| : x\in [0,1] \}

    i think this is obvious:

     \|f\|_{1}\leq\|f\|_{\infty}

    but i cant find a positive real number a,  a\|f\|_{\infty}\leq\|f\|_{1}
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  2. #2
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    Quote Originally Posted by szpengchao View Post
     \|f\|_{1}=\int_{0}^{1}{|f|} \ \ on \ \ C[0,1]

    uniform norm:  \|f\|_{\infty}=\sup \{|f(x)| : x\in [0,1] \}

    i think this is obvious:

     \|f\|_{1}\leq\|f\|_{\infty}

    but i cant find a positive real number a,  a\|f\|_{\infty}\leq\|f\|_{1}
    This is because there exists no such a: for every \varepsilon>0 you can easily find f\in\ C([0,1]) such that f(0)=1 and \int_0^1 |f|<\varepsilon: sketch the graph of a piecewise linear function that quickly decreases to 0 and then is constant equal to 0. (Or choose f(x)=x^n for n large enough if you prefer to write formulas (then f(1)=1))
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  3. #3
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    ok

    ok. so they are not lipschitz equivalent then?
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  4. #4
    elm
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    but "a" can be arbitrarily small too...

    Quote Originally Posted by Laurent View Post
    This is because there exists no such a: for every \varepsilon>0 you can easily find f\in\ C([0,1]) such that f(0)=1 and \int_0^1 |f|<\varepsilon: sketch the graph of a piecewise linear function that quickly decreases to 0 and then is constant equal to 0. (Or choose f(x)=x^n for n large enough if you prefer to write formulas (then f(1)=1))
    i don't get it. "a" can be arbitrarily small...

    can't you do the following... let a -> 0+ and let A-> positive infinity. then it must be true that (a * uniform norm) <= norm1 <= (A * uniform norm) since 0 <= norm1 <= infinity, for all f.

    what's wrong with that?

    i understand there's a theory that all norms in Rn are Lipschitz equivalent to the Euclidian norm. then all should be Lipschitz equivalent to each other. seems applicable?
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