(a) Show that $ f_n: [0, \infty) \to \mathbb R, f_n (x) = x^2e^{-nx} $ converges uniformly on $[0, \infty)$.

(b) Calculate $\lim_{n \to \infty} \int_0^b x^2 e^{-nx}dx$ for each $b \gt 0$.

(c) Show that $n^2x^2e^{-nx}$ do not converge uniformly at $[0, \infty)$.

My attempt

(a) $f_n'(x)=-x(nx - 2)e^{-nx}$ and that it follows from this that $f_n$ is increasing on $[0,\frac{2}{n}]$ and decreasing on $[\frac{2}{n},\infty)$. Therefore,

$$\max f_n=f_n(\frac{2}{n})=\frac{4}{n^2e^2}.$$

Since $\lim_{n\to\infty}\max f_n=0$, given $\epsilon \gt 0$. There is some $n \in \mathbb N$ such that $n \ge \mathbb N \to\max f_n \lt \epsilon$.

But this is the same thing that $n\ge \mathbb N \to\max|f_n−0|\lt \epsilon$, then it follows that, if $n \ge \mathbb N$ and if $x \in [0,\infty)$, $∣f_n(x)−0∣ \lt $max$|f_n−0|\lt \epsilon$ so $∣f_n(x)−0∣\lt \epsilon$.

Therefore convergence is uniform.

(b) $$\lim_{n \to \infty} \int_0^b x^2 e^{-nx}dx =\lim_{n \to \infty} \frac{2}{n^3}-\frac{(b^2n^2+2bn+2)e^{-bn}}{n^3}$$ but $$\lim_{n \to \infty} \frac{2}{n^3}=0$$ and $$\lim_{n \to \infty}\frac{(b^2n^2+2bn+2)e^{-bn}}{n^3}= 0$$ then $$\lim_{n \to \infty} \int_0^b x^2 e^{-nx}dx = 0$$

(c) My plan was to use the same argument as in item (a) and arrive in a contradiction.But if I use the same argument I get the same answer because $f_n'(x)=−n^2x(nx−2)e^{−nx}$

So my argument is wrong (a)? in (c) ? or both?

Thanks in advance for any help.


1 Answer 1


Part a) is correct. As regards c) note that the pointwise limit of $n^2f_n(x)$ is always zero for any $x\in [0, \infty)$ but $$n^2f_n(2/n)=n^2(2/n)^2e^{-2}= 4e^{-2}\not\to0$$ which implies that $n^2f_n$ does not converge uniformly in $[0, \infty)$.

As regards b) we may also compute the limit without the explicit evaluation of the integral: for $n\geq 2/b$, $$0\leq \int_0^b x^2 e^{-nx}dx\leq b \max_{[0,b]} f(x)=bf_n(2/n)=b(2/n)^2e^{-2}\to 0.$$


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