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Thread: Lipschity function proof

  1. #1
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    Lipschity function proof

    Show that $\displaystyle f(x)=\sqrt{a^2+x^2}$ is a Lipschity function.

    My proof:

    Let u,v be in R.

    then $\displaystyle f(u)=\sqrt{a^2+u^2}$ and $\displaystyle f(v)=\sqrt{a^2+v^2}$

    Now $\displaystyle |\sqrt{a^2+u^2}-\sqrt{a^2+v^2}|$
    $\displaystyle \leq |\sqrt{a^2+u^2}-\sqrt{a^2+v^2}||\sqrt{a^2+u^2}+\sqrt{a^2+v^2}|$
    $\displaystyle = |a^2+u^2-a^2-v^2| = |u^2-v^2| = |u-v| |u+v|$

    So pick $\displaystyle C \geq |u+v|$, then we have $\displaystyle |f(u)-f(v)| \leq C|u-v|$

    Is that right?

    Thanks.
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  2. #2
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    Say that $\displaystyle x,y\not = 0$ (if they are there is nothing to proof so it is safe to say there are not).

    $\displaystyle |f(x)-f(y)| = |\sqrt{a^2+x^2} - \sqrt{a^2+y^2}| \cdot \frac{\sqrt{a^2+x^2}+\sqrt{a^2+y^2}}{\sqrt{a^2+x^2 }+\sqrt{a^2+y^2}} = \frac{|x^2-y^2|}{\sqrt{a^2+x^2}+\sqrt{a^2+y^2}}$$\displaystyle \leq \frac{|x-y||x+y|}{|x|+|y|} \leq \frac{|x-y||x+y|}{|x+y|} = |x-y|$.
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