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Thread: A proof involving continuity?

  1. #1
    Member mfetch22's Avatar
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    A proof involving continuity?

    Okay, so heres the problem:

    Suppose $\displaystyle f$ is the function satisfying $\displaystyle |f(x)| \leq |x|$ for all $\displaystyle x$. Show that $\displaystyle f$ is continuous at $\displaystyle 0$.

    Okay, this is my work so far:



    We need to show $\displaystyle \lim_{x \to 0} f(x) = f(0)$, I started by figuring the value of $\displaystyle f(0)$. To find that value, I proceeded with:

    Since:

    $\displaystyle x = 0$

    and:

    $\displaystyle |f(x)| \leq |x|$

    then:

    $\displaystyle |f(0)| \leq |0| \;\; \Rightarrow \;\; |f(0)| \leq 0$

    but:

    $\displaystyle |a| \geq 0 \;\;\;\; for \;\; all \;\; a$

    So we must have:

    $\displaystyle |f(0)| \geq 0$

    But we also have the restriction that:

    $\displaystyle |f(0)| \leq 0$

    Therefore it must be true that:

    $\displaystyle |f(0)| = 0 \;\; \Rightarrow \;\; f(0) = 0$

    So, now i just need to show that:

    $\displaystyle 0 < |x-0| < \delta \;\; \Rightarrow \;\; |f(x) - f(0)| < \epsilon$

    but since $\displaystyle f(0) = 0$; we need only show:

    $\displaystyle 0 < |x| < \delta \;\; \Rightarrow \;\; |f(x)| < \epsilon$

    This is where i'm stuck. I betting the answer is right under my nose, and is really obivous, and I'm just not seeing it. I appreciate any help that anybody can give.
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  2. #2
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    Just let $\displaystyle \varepsilon = \delta$
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  3. #3
    Member mfetch22's Avatar
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    Quote Originally Posted by Plato View Post
    Just let $\displaystyle \varepsilon = \delta$
    I'm not questioning your answer, I was just wondering if you could show me what about:

    $\displaystyle 0 < |x-0| < \delta$

    and letting:

    $\displaystyle \delta = \epsilon$

    Implies that $\displaystyle |f(x) - f(0)| = |f(x)| < \epsilon$

    ??

    I'm guessing the answer is probably blatantly obvious or something, but I'm new to these proofs so I'm still learning. Thanks
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Prove something stronger:

    If $\displaystyle f(x)$ defined on $\displaystyle (a,b)$, and suppose there exist $\displaystyle L\in\mathbb{R}$ so to every two points $\displaystyle x_1$ and $\displaystyle x_2$ in $\displaystyle (a,b)$,

    $\displaystyle |f(x_1)-f(x_2)|\leq L|x_1-x_2|$

    then $\displaystyle f(x)$ continuous on $\displaystyle (a,b)$
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  5. #5
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    $\displaystyle \left| {f(x) - f(0)} \right| = \left| {f(x) - 0} \right| = \left| {f(x)} \right| \leqslant \left| x \right| < \delta = \varepsilon $
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  6. #6
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    Quote Originally Posted by mfetch22 View Post
    I'm not questioning your answer, I was just wondering if you could show me what about:

    $\displaystyle 0 < |x-0| < \delta$

    and letting:

    $\displaystyle \delta = \epsilon$

    Implies that $\displaystyle |f(x) - f(0)| = |f(x)| < \epsilon$

    ??

    I'm guessing the answer is probably blatantly obvious or something, but I'm new to these proofs so I'm still learning. Thanks
    $\displaystyle |f(x)| \le |x| < \delta$

    EDIT: Darn you Plato. Beaten again.
    Last edited by mr fantastic; Aug 10th 2010 at 06:50 PM.
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