Let $\{f_n\}_{n\in \mathbb N}$ be a sequence of uniformly convergent functions on the interval [0,1]. Find the limit of the following integral: $$\lim_{n\to \infty} \int_0^1nf_n(t)e^{-nt}\,dt$$

Here is my thought process so far. In isolation l'hospital's rule would indicate that $\lim_{n\to \infty} \frac{n}{e^{nt}}$ would tend to zero, and so I expect that this exponential term will be more dominating.

Normally Lebesgue's Dominated convergence theorem (or at the very least his monotone convergence theorem) would allow us to say that this uniformly convergent sequence of functions would converge to f...however I am guessing the surrounding terms sufficiently mess this up to the point where we no longer have convergence. My guess is the point of this question is to prove that this is the case.

We know that $\lim_{n\to \infty}\{f_n\}_{n\in \mathbb N}$ is measurable, but If I can show that $g_n= \frac{n}{e^{nt}}$ is not measurable for $t\in[0,1]$ (because it tends to zero), then can I say this limit does not exist? I'm pretty confused, as is abundantly clear hahahaha.

  • 1
    $\begingroup$ Note that $\int_0^1 ne^{-nt} f(t) \, dt = \int_0^\infty e^{-u} f(u/n) \mathbf{1}_{[0,n]} \, du \to f(0)$ by the DCT for continuous $f$. What other conditions on $f_n$ might you have? $\endgroup$
    – RRL
    Commented Oct 28, 2020 at 5:00
  • 1
    $\begingroup$ If the $f_n$ are continuous, then so is $f$ and $f_n(u/n) \to f(0)$ -- when there is uniform convergence. $\endgroup$
    – RRL
    Commented Oct 28, 2020 at 5:08
  • $\begingroup$ That is a very clever thing to notice! However, I don't think it lines up with what I am supposed to be learning from this question. I'll update my attempt soon, but I'm working more with the Lebesgue definition of the integral and some basic convergence theorems... $\endgroup$
    – Chair
    Commented Oct 28, 2020 at 5:13
  • $\begingroup$ This can be the Lebesgue integral. Just with $f$ instead of $f_n$ the limit may not exist if $ \lim_{z \to 0+} f(z)$ does not exist. So I'm wondering if you are missing some assumptions. Why only the very strong assumption of uniform convergence? $\endgroup$
    – RRL
    Commented Oct 28, 2020 at 5:17
  • $\begingroup$ they are continuous! I'm sure your answer is correct, but I wonder if there is a way that follows more from the definitions and does not rely on a transform...just to help me understand it in terms of the fundamentals. I also think, part of it is to show the limitations behind EVEN uniform convergence...maybe(?) $\endgroup$
    – Chair
    Commented Oct 28, 2020 at 5:47

1 Answer 1


Let $(f_n)$ be a sequence of continuous functions that converges uniformly to $f$ on $[0,1]$. By the triangle inequality, we have

$$\left|\int_0^1 n e^{-nt} f_n(t) \, dt - f(0) \right|\leqslant \left|\int_0^1 n e^{-nt} (f_n(t) - f(t)) \, dt \right|+ \left|\int_0^1 n e^{-nt} f(t) \, dt - f(0) \right|$$

For the first term on the RHS, we have

$$\left|\int_0^1 n e^{-nt} (f_n(t) - f(t)) \, dt \right| \leqslant \int_0^1 n e^{-nt}|f_n(t) - f(t)|\, dt$$

As $f_n \to f$ uniformly, for any $\epsilon > 0$ there exists $N_1 \in \mathbb{N}$ such that if $n > N_1$, then $|f_n(t) - f(t)| < \frac{\epsilon}{2}$ for all $t \in [0,1]$, and

$$\left|\int_0^1 n e^{-nt} (f_n(t) - f(t)) \, dt \right| \leqslant \frac{\epsilon}{2} \int_0^1ne^{-nt} \, dt = \frac{\epsilon}{2} (1 - e^{-n}) < \frac{\epsilon}{2} $$

Thus, for all $n > N_1$,

$$\left|\int_0^1 n e^{-nt} f_n(t) \, dt - f(0) \right|\leqslant \frac{\epsilon}{2}+ \left|\int_0^1 n e^{-nt} f(t) \, dt - f(0) \right|$$

Since $f$ is the uniform limit of a sequence of continuous functions it is continuous, and the proof given here shows there exists $N_2$ such that if $n > N_2$

$$\left|\int_0^1 n e^{-nt} f(t) \, dt - f(0) \right| < \frac{\epsilon}{2}$$

Altogether this proves that

$$\lim_{n \to \infty}\int_0^1 n e^{-nt} f_n(t) \, dt = f(0)$$


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