Is this sequence bounded ? (An open problem between my schoolmates !)

Solution 1:

The sequence $\{A_n\}$ need not to be bounded. To see this, one could for example as $f(t,T)$ choose something that approximates a derivative of a delta distribution as $T\to+\infty$. I wish to give credits to my colleague Tomas Persson who came up with that idea.

I will give such an approximating example. My example is non-smooth, but that is just to make the calculations more transparent.

Let $$ g(t,T)= \begin{cases} \frac{T}{2} & |t|\leq\frac{1}{T}\\ 0 & |t|>\frac{1}{T}. \end{cases} $$ This is an approximation of the delta distribution as $T\to+\infty$. We then let $f$ be the following difference quotient: $$ f(t,T)=\frac{g(t-1/T,T)-g(t-2/T,T)}{1/T} $$ It is then a simple matter to calculate the integral $$ \int_0^1 e^{-nt}f(t,T)\,dt=\frac{T^2}{2n}\Bigl(1+e^{-3n/T}-e^{-2n/T}-e^{-n/T}\Bigr) $$ Hence, $$ A_n=\lim_{T\to+\infty}\int_0^1 e^{-nt}f(t,T)\,dt = n, $$ which of course is unbounded.

Update Let me, for completeness, add a smooth function $f$ that also gives $A_n=n$: $$ f(t,T)=(T^2-T^3t)e^{-Tt}. $$ The argument is the same, it approximates a derivative of the delta distribution.

Solution 2:

$0 \lt \mathrm e^{-nt} \le 1 \quad \forall x \in [0,1] \\ \implies A_n \lt \lim_{T \to \infty} \int_0^1 f(t,T) \,\mathrm d t$

so it's bounded above if the limit exists.

A similar result gives a lower bound if the limit is negative.