Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The problem: Show that if $a>0$, then the sequence $f_n(x) = (n^2x^2e^{-nx})$ converges uniformly on the interval $[a,\infty)$ but not on $[0,\infty)$

My Solution: On the interval of $[0,\infty)$ we notice that $f_n(x)$ attains a maximum at $2/n$ by taking the derivative and setting it equal to zero. Since $f_n(2/n) = 4e^{-2}$, we know that our function cannot converge uniformly on $[0,\infty)$ since $||f_n||\not\rightarrow 0$. On the interval of $[a,\infty)$ $f_n$ does converge uniformly, because for large $n$, $2/n \rightarrow 0$, and therefore $a>2/n$ and thus $||f_n|| = n^2a^2e^{-na}$ which tends towards zero, and thus converges uniformly.

My question : It seems that the problem is at $x=0$ because removing this point allows for uniform convergence. What I dont get is why $x=0$ causes a problem to begin with. If we analyze $f_n$ pointwise at $x=0$, dont we have that $f_n(0) = 0$ for all $n\in \mathbb{N}$? Im conceptually lost as to where the problem arises by retaining the $x=0$ value.

Thanks for your help!

share|cite|improve this question
up vote 6 down vote accepted

The problem is not the value at $0$ precisely. For instance, the sequence does not converge uniformly on $(0,\infty)$ either, for the same reason you gave: there is always a point (namely $2/n$) at which $f_n$ takes the constant value $4/e^2$. And if you exclude some interval $[0,a)$, then $2/n$ ends up below $a$, so we don't have to care about it anymore.

I'm not sure what else to say.

share|cite|improve this answer

As the maximum of the norm is reached at $2/n$, the problem is not exactly at $0$ but at at neighborhood of $0$.

When we take $x\geqslant a$, these problems are avoid as we are not concerned of what happens in a neighborhood of $0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.