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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.

3 Answers3

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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 $0$.

kobe
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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 $n$.

Community
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Alexis Olson
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  • My initial was to show that the sequence is not Cauchy, because by completeness it does not converge either, but I did not find any reasonable way of doing so. Is there a (reasonable) way? –  Sep 07 '16 at 18:56
  • You can show that the sequence ${x_n(t)}$ is not Cauchy. For any $N\in\mathbb{N}$, we can take $m=2n$ for $n > N$. Then consider $t=1/n$, and we have $d(x_m,x_n)\ge |x_n(1/n)-x_m(1/n)|=|e^{-1}-2e^{-2}|$. – user539658 Jan 03 '24 at 09:19
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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.

Robert Israel
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  • So I understand that the distance between the sequence and its pointwise limit is $e^{-1}$ independent of $n$. What bothers me is why does this necessarily mean that the sequence cannot converge to some other function: Could you explain your first sentence? –  Sep 07 '16 at 18:50