Prove that $x^n e^{-x} $ is uniformly continuous on $[0, \infty)$

Question

Thus far I have shown that $|f(x)-f(y)| = e^{-x} |x^n - y^n e^{x-y}|$

Answer 1

It is enough to prove that $f(x) = x^ne^{-x}$ is continuously differentiable and its derivative is bounded. Computing the first derivative you get $$f'(x) = x^{n-1}(n-x)e^{-x}$$ which is continuous on $[0,+\infty)$. To find its maximum you can compute \begin{align} f''(x) & = (n-1)x^{n-2}(n-x)e^{-x}-x^{n-1}e^{-x}-x^{n-1}(n-x)e^{-x}\\ & = x^{n-2}e^{-x}((n-1)(n-x)-x-x(n-x)) \\ & = x^{n-2}e^{-x}(x^2-2nx+n^2-n). \end{align} This is positive when $0 < x < n-\sqrt{n}$ or $x > n+\sqrt{n}$. Since $f'$ tends to $0$ when $x \to +\infty$, it has a maximum at $x = n-\sqrt{n}$, where $f'(n-\sqrt{n}) = \sqrt{n}(n-\sqrt{n})^{n-1}e^{-n+\sqrt{n}} > 0$. Hence $f$ is uniformly continuous.

Answer 2

hint

$f_n$ is unifomly continuous at $[0,+\infty)$ since $f_n$ is continuous at $[0,+\infty)$ and $\lim_{x\to+\infty}=0\in \Bbb R$. For a given $\epsilon>0,$ Use the fact that $f_n$ is uniformly continuous at some compact $[0,a]$ wher $a$ is such that

$$x\ge a \text{ and } y\ge a \implies $$ $$|f(x)-f(y)|<|f(x)|+|f(y)|<\epsilon$$

to prove that $f_n$ is uniformly continuous at $[a,+\infty)$.

Now, we use the continuity of $f_n$ at the point $x=a$.

there exist some $\eta$ such that

$$x,y\in ]a-\eta,a+\eta[\implies $$ $$|f(x)-f(y)|\le |f(x)-f(a)|+|f(y)-f(a)|<\epsilon$$

You can take it.

Answer 3

The function $f(x) = x^n e^{-x}$ is uniformly continuous on all compact intervals, as a continuous function, and tends to zero as $x \to \infty$ for all $n \in \mathbb{N}$. It is therefore uniformly continuous by this.