- #1
tom_rylex
- 9
- 0
This class is making my head hurt. I could use some help.
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.
Definition of distribution:
\langle f , \phi \rangle = \int_{R_n} f(x) \phi(x) dx
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?
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?
