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I would like you guys to verify my proof for the following exercise:

Let $$f_n:[-1,1] \to \mathbb R; \quad f_n(x) := nx(1-x^2)^n$$ Is $f_n$ pointwise (uniformly) continuous? And in case it is, specify $\lim_{n\to \infty} f_n.$

My solution:

$f_n$ converges pointwise towards the $0$-function, i.e. $f_n \to 0$ pointwise. Proof:

$f_n(1) = f_n(-1) = 0$, so let $x\in ]-1,1[$. For fixed $x$ with $|x| < 1$ we have that $$|f_n(x) - 0| = n|x||1-x^2|^n \leq n |1-x^2|^n \to 0$$ since $nq^n \to 0$ when $|q| < 1$. Therefore $f_n \to 0$ pointwise.

$f_n$ does not uniformly converge. Pick $x_n = 1/n$. We then have $$ \lim_{n\to \infty}f_n(x_n) = \lim_{n\to \infty} n \frac{1}{n} \left(1-\frac{1}{n}\right)^n\left(1+\frac{1}{n}\right)^n = e^{-1}e = 1 \neq 0$$ as desired.

EDIT: Changed $[0,1]$ to $[-1,1]$.

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  • $\begingroup$ Looks fine to me. $\endgroup$
    – DonAntonio
    Commented Apr 22, 2017 at 14:06
  • $\begingroup$ Don't write $f_n(-1)$ . $-1$ is not in the domain of $f_n$ . You should be only concerned about the points in $[0,1$ . Everything other than this is okay . $\endgroup$ Commented Apr 22, 2017 at 14:08
  • $\begingroup$ Oh, it was actually a typo, I meant the interval $[-1,1]$. $\endgroup$
    – Staki42
    Commented Apr 22, 2017 at 14:21
  • $\begingroup$ @lappen68 Here is an other approach $\endgroup$ Commented Apr 22, 2017 at 14:44

2 Answers 2

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We have pointwise convergence towards zero, but we cannot have uniform convergence, since that would imply $\int_{0}^{1}f_n(x)\,dx \to 0$, but $$ \int_{0}^{1} f_n(x)\,dx = \frac{n}{2n+2}\to \frac{1}{2}\color{red}{\neq }0.$$ As an alternative, it is enough to notice that $f_n\left(\frac{1}{\sqrt{n}}\right)\sim\frac{\sqrt{n}}{e}$.

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$f_n $ is an odd function. we restrict the study to $[0,1] $.

For enough great $n $,

$$|f_n|'(x)=$$ $$n (1-x^2)^{n-1}(1-x^2-2nx^2) $$

the maximum is attained at $$x_n=\frac {1}{\sqrt {2n+1}} $$

$$|f (x_n)|=\frac{n}{\sqrt{2n+1}}(\frac {2n}{2n+1})^n$$

which goes to $+\infty$.

the convergence is not uniform.

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