Does $x_n(t) = nte^{-nt}$ converge in $(C[0,1],d_\infty)$? Let $X = C_b[0,1]$ be the set of bounded, continuous functions on $[0,1]$, and $d_\infty (x,y) = \max_{0 \leq t \leq 1} \lvert x(t) - y(t) \rvert$ be a metric on $X$. Does $x_n(t) = nte^{-nt}$ converge in $(X,d_\infty)$?
My instinct tells me no. What I have done is that $x_n$ converges pointwise to $0$, and that 
$$
d_\infty(x_n(t),0) = \dots = e^{-1} \in X
$$
with reservation for mistakes. What I actually want is that the function it is converging to is either unbounded or not continuous (so the sequence does not converge). I am very confused if this is leading me nowhere, so any hints and directions are appreciated.
 A: You are correct, $(x_n)$ does not converge in $(X,d_\infty)$. All that's left for you to show is that $\lim_{n\to \infty} d_\infty(x_n,0) \neq 0$. Fix $n\ge 1$. The function $x_n(t)$ has a unique critical point at $t = 1/n$ and $$x_n''(1/n) = n^2e^{-nt}(-2 + nt)\bigg|_{t = 1/n} = -n^2e^{-1} < 0$$ so by the second derivative test, $x_n(t)$ has a relative maximum at $t = 1/n$. Since $x_n(0) = 0$ and $x_n(1) = ne^{-n} < e^{-1}$, $x_n(t)$ has a global maximum at $t = 1/n$ with max value $e^{-1}$. Hence $d_\infty(x_n,0) = e^{-1}$, which does not converge to $0$.
A: While that metric space is closed, your sequence of functions is not a Cauchy sequence.
As you point out, the pointwise limit is the zero function. However, $f_n$ does not converge in the $d_\infty$ metric to $0$ since $d_\infty (x_n,0) = e^{-1}$ for all $n$.
A: Since it converges pointwise to $0$, the only function it could possibly converge to in the $d_\infty$ metric is $0$.  But, as you sat, the distance from $x_n$ to $0$ is $e^{-1}$, which does not go to $0$.  So that tells you it does not converge in the $d_\infty$ metric.
