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

(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.

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 $$n\geq 2/b$$, $$0\leq \int_0^b x^2 e^{-nx}dx\leq b \max_{[0,b]} f(x)=bf_n(2/n)=b(2/n)^2e^{-2}\to 0.$$