Integrability and Lipschitz continuity

[AxiomOfChoice]

(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 $f$ is Lipschitz (on all of $\mathbb R$), then

$$\int_{-\infty}^\infty f^2(x) e^{-\frac{x^2}{2t}}dx < \infty \quad ?$$

I've tried a number of different approaches - boundedness of the derivative and integration by parts, bounded/dominated/monotone convergence theorem, approximation of $f^2$ by polynomials - but I haven't gotten any of them to work. Apparently, just because $f$ is Lipschitz doesn't mean $f^2$ is Lipschitz. (Just look at $f(x) = x$.)

Can anyone think of a Lipschitz function for which the integral above is infinite?

 

[CompuChip]

They only thing I can think of that you haven't tried yet is going back to definitions.
For example, when you write out the integral as the limit of a Riemann sum, you will probably get summands like
$$|f^2(x + \Delta x) e^{-(x + \Delta x)^2 / 2t} - f^2(x) e^{-x^2/2t}| \le \cdots \le |f(x + \Delta x)|^2 e^{-x^2/2t} \le (K {\Delta x})^2 e^{-x^2/2t}$$
and eventually, maybe, rewrite it to
$$\cdots \le C \int_{-\infty}^\infty e^{-x^2/2t} \, dx$$
or something else for which you know it is finite.