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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?

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  • $\begingroup$ 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, 2015 at 5:53
  • $\begingroup$ @GautamShenoy Thank you. How can I show that $f$ does not dominate $f_n$? $\endgroup$
    – kaos
    Jun 12, 2015 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, 2015 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, 2015 at 6:43

1 Answer 1

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[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 $x\in\mathbb{R}$

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