Is $x^2 e^{-x}$ uniformly continuous in the domain $[0,\infty)$.

Question

I understood the fact that $e^{-x}$ is continuous in the given domain and $x^2$ is continuous in a fixed interval. Also product of uniformly continuous functions are continuous only if it is bounded in the domain. But these informations are not giving me idea to prove the uniform continuity of $x^2 e^{-x}$. Can someone help me with ideas?

Answer 1

Let $f(x)=x^2 \exp(-x)$. Fix any $\epsilon>0$. Since $f(x)\rightarrow 0$ as $x\rightarrow\infty$, there exist sufficiently large $M>0$ such that $|f(x)|<\epsilon/2$ for any $x\geq M$. Therefore, for any $x,y\geq M$, we have $|f(x)-f(y)|<\epsilon$ (Statement 1).

Since $f(x)$ is continuous on a closed interval $[0,M]$, it is uniformly continuous on $[0,M]$. (Statement 2).

Use both statements 1 and 2 to show that $f(x)$ is uniformly continuous on $[0,\infty)$.

Answer 2

Take $(x+\delta)^2e^ {-(\delta+x)}-(x)^2e^{-x}$, differentialte w.r.t x, and see if the dfference has a maximum, and if that maximum can be made smaller than any epsilon.

After the first derivative, you can cancel out exponential of x, and you will find a quadratic, which means you can solve for it. Checking the second derivative might seem slightly more tricky, but it is not. You will see that the second derivative is positive.

Perhaps there is an easier way for this problem. I wanted to propose a general method I use for such problems.