- #1
AxiomOfChoice
- 533
- 1
(I've been lighting this board up recently; sorry about that. I've been thinking about a lot of things, and my professors all generally have better things to do or are out of town.)
Is there an easy way to show that if [itex]f[/itex] is Lipschitz (on all of [itex]\mathbb R[/itex]), then
[tex]
\int_{-\infty}^\infty f^2(x) e^{-\frac{x^2}{2t}}dx < \infty \quad ?
[/tex]
I've tried a number of different approaches - boundedness of the derivative and integration by parts, bounded/dominated/monotone convergence theorem, approximation of [itex]f^2[/itex] by polynomials - but I haven't gotten any of them to work. Apparently, just because [itex]f[/itex] is Lipschitz doesn't mean [itex]f^2[/itex] is Lipschitz. (Just look at [itex]f(x) = x[/itex].)
Can anyone think of a Lipschitz function for which the integral above is infinite?
Is there an easy way to show that if [itex]f[/itex] is Lipschitz (on all of [itex]\mathbb R[/itex]), then
[tex]
\int_{-\infty}^\infty f^2(x) e^{-\frac{x^2}{2t}}dx < \infty \quad ?
[/tex]
I've tried a number of different approaches - boundedness of the derivative and integration by parts, bounded/dominated/monotone convergence theorem, approximation of [itex]f^2[/itex] by polynomials - but I haven't gotten any of them to work. Apparently, just because [itex]f[/itex] is Lipschitz doesn't mean [itex]f^2[/itex] is Lipschitz. (Just look at [itex]f(x) = x[/itex].)
Can anyone think of a Lipschitz function for which the integral above is infinite?