0
$\begingroup$

The function $f_n$: $\mathbb{R}\rightarrow \mathbb{R}$ is defined by $f_n(x):=\chi_{[0,\infty)}(x)\frac{1}{n}\exp(-\frac{x}{n})$, where $n\in \mathbb{N}$.

$f(x):=\lim\limits_{n \rightarrow \infty}{f_n(x)}$ is lebesgue integrable.

And $\int_\mathbb{R} f(x) d\lambda(x) \neq \lim\limits_{n \rightarrow \infty}{\int_\mathbb{R} f_n(x) d\lambda(x)}$.

Does this violate the dominated convergence theorem?

$\endgroup$
4
  • $\begingroup$ For Dominated Convergence theorem, you need a dominating function $g$. i.e. $|f_n| \leq g$ and $\int g < \infty$. $f$ does not qualify since it does not dominate $f_n$. $\endgroup$
    – Gautam Shenoy
    Jun 12 '15 at 5:53
  • $\begingroup$ @GautamShenoy Thank you. How can I show that $f$ does not dominate $f_n$? $\endgroup$
    – kaos
    Jun 12 '15 at 6:00
  • $\begingroup$ Ah ok, I think I got it. We need a $|f_n| \leq g$, but $f$ does not qualify, because it is $0$ for all $x\in\mathbb{R}$. $\endgroup$
    – kaos
    Jun 12 '15 at 6:22
  • $\begingroup$ @kaos: Exactly. In fact what you have is a good example of the need of "dominating function" hypothesis in DCT. $\endgroup$
    – Gautam Shenoy
    Jun 12 '15 at 6:43
0
$\begingroup$

[All credit goes to Gautam Shenoy]

$\int_\mathbb{R} f(x) d\lambda(x) \neq \lim\limits_{n \rightarrow \infty}{\int_\mathbb{R} f_n(x) d\lambda(x)}$ does not violate the DCT, because you need a dominating function $g$ with $|f_n|\leq g$.

$f$ does not qualify for this, as it equals $0$ for all $x\in\mathbb{R}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.