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If $f_n(x)=nxe^{-nx^2}~\forall~n=1,2,\cdots$ and $x$ real, show that $$\lim_{n \rightarrow \infty} \int_0^1 f_n(x)dx \ne \int_0^1\lim_{n \rightarrow \infty}f_n(x) dx$$

Attempt:

By the $Mn$ Test, it comes out that $f_n$ is not uniformly convergent.

We also know that if $\{f_n\}$ is a uniformly convergent sequence of functions and if $f$ is the limiting function, then :

$$\lim_{n \rightarrow \infty} \int_0^1 f_n(x)dx = \int_0^1\lim_{n \rightarrow \infty}f_n(x) dx$$

However, this is just a sufficient condition and a non uniformly convergent sequence of functions may also exhibit the same properties.

In such a case, how do we prove the above relation in this and other problems? Is direct integration the only method to follow in such problems?

Thank you very much for your help in this regard.

MathMan
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1 Answers1

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Firstly, $$\int_0^1 f_n(x)dx=n\int_0^1 xe^{-nx^2}dx=n\left[-\frac{1}{2n}e^{-nx^2}\right]_0^1=n\left(\frac{1}{2n}-\frac{1}{2n}e^{-n}\right)=\frac{1}{2}-\frac{1}{2}e^{-n}$$ and thus $$\lim_{n\to\infty }\int_0^1 f_n(x)dx=\frac{1}{2}.$$

Secondly, $$\lim_{n\to\infty }nxe^{-nx^2}=0$$ and thus $$\int_0^1\lim_{n\to\infty }f_n(x)dx=0.$$

We conclude that $$\lim_{n\to\infty }\int_0^1 f_n(x)dx\neq \int_0^1 \lim_{n\to\infty }f_n(x)dx.$$

idm
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  • Thank you. So, no concept involved using the properties of uniformily convergent function sequences right – MathMan Jan 28 '15 at 21:18