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Thread: Lipschitz Continuity

  1. #1
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    Lipschitz Continuity

    Hey guys, I need to do the following.

    Let $\displaystyle c > 0$ and let $\displaystyle f(x) = e^x$. Show that $\displaystyle f(x)$ is Lipschitz continuous for $\displaystyle x \in [-c,c]$.

    My idea is to set $\displaystyle L = e^c$, but it is hard to prove that this is a valid Lipschitz constant.
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  2. #2
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    Quote Originally Posted by EinStone View Post
    Hey guys, I need to do the following.

    Let $\displaystyle c > 0$ and let $\displaystyle f(x) = e^x$. Show that $\displaystyle f(x)$ is Lipschitz continuous for $\displaystyle x \in [-c,c]$.

    My idea is to set $\displaystyle L = e^c$, but it is hard to prove that this is a valid Lipschitz constant.
    Mean value theorem: $\displaystyle f(y) - f(x) = (y-x)f'(z)$, for some z lying between x and y. So you can take the Lipschitz constant to be the maximum value of |f'(z)| in the given interval (which in this case is $\displaystyle e^c$, as you suspected).
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by EinStone View Post
    Hey guys, I need to do the following.

    Let $\displaystyle c > 0$ and let $\displaystyle f(x) = e^x$. Show that $\displaystyle f(x)$ is Lipschitz continuous for $\displaystyle x \in [-c,c]$.

    My idea is to set $\displaystyle L = e^c$, but it is hard to prove that this is a valid Lipschitz constant.
    What Opalg is just a consequence of a much broader theorem, namely that if $\displaystyle f:E\mapsto\mathbb{R}$ is differentiable and $\displaystyle f'$ is bounded then $\displaystyle f$ is Lipschitz. (this is, in fact, an iff statement)

    And thus, if $\displaystyle E$ is compact and if $\displaystyle f$ is continuously differentiable we must have that $\displaystyle f'$ necessarily bounded and thus $\displaystyle f$ is Lipschitz.
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