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

Let $f$ be a smooth function (say $\mathcal{C}^{\infty}$) in its two real variables ($t$ and $T$). I consider the following sequence defined by $$A_n:=\lim_{T \to \infty} \int_{0}^{1} e^{-n t} f(t,T)dt \quad \quad (n\geq 1)$$ where we suppose that this limit exists and is finite for all integer $n$. I would like to conclude that the sequence $(A_n)$ is bounded. Is that true ?

This is not an obvious problem : assuming certain conditions on $f$ then the exercise becomes classic. However, very little can be said about $(A_n)$ in the general case. In fact I deeply think that the answer is no and that we can find a $f$ such that $A_n$ tends to infinity.

If one takes $f(t,T)=2\sin(tT)/\pi t$, then $(A_n)$ is the constant sequence equal to $1$. This shows that $f$ does not necessarily tend to $0$.

Thanks for your help !

• I'm sorry, it isn't trivial to me that the sequence $(|A_n|)$ is decreasing. Can you develop your claim ? (Thanks !) Jan 4, 2015 at 15:43
• I think @TobiasHurth is assuming $f(t,T)\geq 0$ for all $t,T$. Another scenario when $\{A_n\}_{n=1}^{\infty}$ is obviously bounded is when $\int_0^1 |f(t,T)|dt \leq C$ for all $T$, for some finite constant $C$. Jan 5, 2015 at 23:06
• To see that $|A_1|$ can be less than $|A_n|$ for some $n$, define $f(t,T)=c-t$ for a constant $c$ such that $\int_0^1 e^{-t}(c-t)=0$. Then $A_1=0$, but there are $n>1$ such that $|A_n| \neq 0$. Jan 5, 2015 at 23:24
• @Stabilo is this integral a Lebesgue or improper Riemann integral?
– user159356
Jan 5, 2015 at 23:36
• I think that the statement is false in general but one has to find a counter example : If we build a $g(t,T)$ ($\mathcal{C}^{\infty}$) such that $g(0,T)=g(1,T)=0$ and $g(t,T)$ tends to Dirac in the sense of distributions (when $T$ grows up), then by taking $f(t,T)$ an antiderivitive of $g$ (in $t$) it seems that one conclude that the sequence $(A_n)$ tends to infinity. Jan 8, 2015 at 13:14

2 Answers

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.

• Isn't it the same idea as proposed by @Stabilo in his comment to the OP? Jan 13, 2015 at 19:01
• Almost, it seems to be the opposite. If I read that comment correctly $f'$ is delta there. Here $f$ is the derivative of delta. Jan 13, 2015 at 19:09
• Thanks for your answer ! And yes I made a mistake in my previous comment : I meant the contrary... But I was not able to find such an example. Otherwise this result is really counter-intuitive ! Jan 14, 2015 at 8:06

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

so it's bounded above if the limit exists.

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

• It seems that you assumed the positivity of $f$ which does not answer the question in generality. Can you develop your answer ? Jan 12, 2015 at 21:05