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Math Help - inequality with two metrics

  1. #1
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    inequality with two metrics

    I want to prove that sup \ x \in C[0,1] |f_n(x)-f(x) | \geq ({\displaystyle\int^1_0 |f_n(x)-f(x)|^2 \ dx})^{1/2}

    ({\displaystyle\int^1_0 |f_n(x)-f(x)|^2 \ dx})^{1/2}
    \leq (by c-s inequality)
    ({\displaystyle\int^1_0 |f_n(x)|^2 \ dx})^{1/2} + ({\displaystyle\int^1_0 |f(x)|^2 \ dx})^{1/2}
    \leq
    sup |f_n(x)| + sup|f(x)|

    However I am unsure where to go now
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  2. #2
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    Re: inequality with two metrics

    Quote Originally Posted by FGT12 View Post
    I want to prove that sup \ x \in C[0,1] |f_n(x)-f(x) | \geq ({\displaystyle\int^1_0 |f_n(x)-f(x)|^2 \ dx})^{1/2}

    ({\displaystyle\int^1_0 |f_n(x)-f(x)|^2 \ dx})^{1/2}
    \leq (by c-s inequality)
    ({\displaystyle\int^1_0 |f_n(x)|^2 \ dx})^{1/2} + ({\displaystyle\int^1_0 |f(x)|^2 \ dx})^{1/2}
    \leq
    sup |f_n(x)| + sup|f(x)|

    However I am unsure where to go now
    If M = \sup_{x\in[0,1]}|f_n(x)-f(x)| then |f_n(x)-f(x)|\leqslant M for all x in [0,1]. It follows that \int_0^1|f_n(x)-f(x)|^2dx\leqslant\int_0^1M^2dx = M^2. Take the square root of both sides to get the inequality that you want (no need for Cauchy–Schwarz).
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