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Can we apply the Dominated convergence theorem of Lebesgue to the sequence of functions $ f_n(x) =\frac{n}{x^2 +n^2}, \quad x\in \mathbb{R}$ ?

It is $ f_n(x) =\frac{n}{x^2 +n^2} \leq \frac{n}{2nx} =\frac{1}{2x}$ but the function $1/x$ is not Lebesgue integrable. How can I find a function $ g \in L^+( \mathbb{R})$ such that $|f_n| \leq g, n\in \mathbb{N}$ almost everywhere in $\mathbb{R}$ so I can use Lebesgue's theorem?

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    $\begingroup$ Is the conclusion of the Dominated Convergence Theorem true in this case? What is $\int_{-\infty}^\infty f_n(x)\ dx$? $\endgroup$ Commented Feb 2, 2012 at 19:50

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No, dominated convergence cannot be applied here. Indeed, we have

$$\int_{-\infty}^\infty \frac{n\, dx}{x^2 + n^2} = \int_{-\infty}^\infty \frac{ d(x/n)}{(x/n)^2 + 1} = \int_{-\infty}^\infty \frac{dx}{x^2 + 1} = \pi$$

eventhough pointwise the sequence converges to $0$.

Note also that $f_y(x) = \frac{y}{x^2+y^2}$ is the Poisson kernel for the upper half-plane. As such, it "converges to the dirac delta distribution" as $y \to 0$.

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  • $\begingroup$ Thank's for the answer Sam! It was easy. I did't look it carefully. In other cases is there a way to find the function $g$? $\endgroup$
    – passenger
    Commented Feb 2, 2012 at 20:07
  • $\begingroup$ I mean to find the minimum function $g$. $\endgroup$
    – passenger
    Commented Feb 2, 2012 at 21:42
  • $\begingroup$ @passenger: There is no general way of doing this. You can of course define $g(x) = \sup_{n} |f_n(x)|$ (which is the minimal function dominating all $f_n$), but this won't help at all, since you will still be left with proving that $g \in L^1$. Many times, however, there will be an obvious choice for $g$ (or indeed several possibilities). Other times you will need to do some work. $\endgroup$
    – Sam
    Commented Feb 2, 2012 at 22:30
  • $\begingroup$ I knew the way define $g(x)=\sup_n |f_n(x)|$. Anyway thank's again! $\endgroup$
    – passenger
    Commented Feb 2, 2012 at 22:40

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