Lipschitz Continuity

**EinStone** · February 22nd 2010, 02:29 AM

Hey guys, I need to do the following.

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

My idea is to set $L = e^c$ , but it is hard to prove that this is a valid Lipschitz constant.

**Opalg** · February 22nd 2010, 04:41 AM

Originally Posted by EinStone

Hey guys, I need to do the following.

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

My idea is to set $L = e^c$ , but it is hard to prove that this is a valid Lipschitz constant.

Mean value theorem: $f(y) - f(x) = (y-x)f'(z)$ , 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 $e^c$ , as you suspected).

**Drexel28** · February 22nd 2010, 05:34 PM

Originally Posted by EinStone

Hey guys, I need to do the following.

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

My idea is to set $L = e^c$ , 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 $f:E\mapsto\mathbb{R}$ is differentiable and $f'$ is bounded then $f$ is Lipschitz. (this is, in fact, an iff statement)

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

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