Convergence of Slowly Growing Functions in Distribution Theory

[tom_rylex]

This class is making my head hurt. I could use some help.

Homework Statement

Show that if f(x) is a function of slow growth on the real line,
$$\lim}_{\substack \varepsilon \rightarrow0^+} \langle f(x)e^{- \varepsilon |x|}, \phi (x) \rangle = \langle f, \phi \rangle$$
where $\phi (x)$ is a test function.

Homework Equations

Definition of distribution:
$$\langle f , \phi \rangle = \int_{R_n} f(x) \phi(x) dx$$

The Attempt at a Solution

If I look at the function $f_\varepsilon (x) = f(x)e^{-\varepsilon | x |}$, I can say that the function is locally integrable, (actually $L_1 (-\infty, \infty)$). Can't I just invoke the Lebesgue Dominated Convergence theorem here? That is, since $f_\varepsilon$ is locally integrable, and $f_\varepsilon \rightarrow f$ pointwise, that $f_\varepsilon \rightarrow f$ in a distributional sense.

Or am I missing something here. Are there other considerations to take into account?

 

[Dick]

I think dominated convergence theorem works, doesn't it? You know f*phi is integrable. e^(-epsilon*|x|)*f(x)*phi(x) converges pointwise to f*phi and is dominated by |f*phi|. It all sounds kosher to me.