Given a function $f \in L^1(\mathbb{R})$, suppose I want to evaluate the following integral

$$\lim_{n\to\infty} \int_{\mathbb{R}}e^{-nx^2}f(x)dx$$

It looks to me like one could use dominated convergence here. If we let $f_n(x) = e^{-nx^2}f(x)$, then clearly $|f_n(x)| \leq |f(x)|$ for $n \geq 0$, and we know that $|f(x)|$ is an integrable function. The issue I am having is showing that $f_n(x)$ itself is integrable - it seems clear to me that it would be, but not sure how to show it (maybe it just follows from the above inequality since $|f(x)|$ is integrable?)

Also, suppose we can show dominated convergence theorem applies, the limit is a little bit interesting as it depends on $x$. Clearly, if $x\neq 0$, then we have

$$\lim_{n\to\infty} e^{-nx^2}f(x) = 0,$$ but if $x = 0$, then what happens?


  • $\begingroup$ $\{0\}$ is of measure $0$, so it doesn't matter what happens there. $\endgroup$ – saulspatz Dec 4 '18 at 19:13

$f_n$ is integrable because $|f_n| \le |f|$, as you mentioned.

We have $$\lim_{n \to \infty} f_n(x) = \begin{cases} 0 & x \ne 0 \\ f(0) & x = 0. \end{cases}$$

The integral of this function is simply zero. So if you show that you can use the dominated convergence theorem, then $\lim_{n \to \infty} \int f_n = 0$.

  • $\begingroup$ What else needs to be shown to use dominated convergence? I thought it was that the sequence $f_n$ had to be bounded above by an integrable function and be integrable themselves and then conditions were satisfied. $\endgroup$ – Sorey Dec 4 '18 at 19:17
  • $\begingroup$ @Sorey Sure, that's enough. $\endgroup$ – angryavian Dec 4 '18 at 20:25

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