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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!

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2 Answers 2

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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.

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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 $0$.

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