Double limit of Riemann integrals

Question

Let $\{f_n(x)\}_{n\in\mathbb{N}}$ a family of infinitely differentiable functions, defined on the entire real line, vanishing outside $[a,b]$. Let $F(x)$ be any function defined on $\mathbb{R}$ with the property that $F$ and all its derivatives are $O(|x|^{-N})$ as $|x|\to\infty$, for every $N$.

By hypothesis, the following limit:

$$\lim_{n\to\infty}\int_{x=a}^b f_n(x)F(x)\mathrm{d}x$$

exists for any $F$ of the type specified above.

Is it true or false that $\lim_{|\xi|\to\infty}\left(\lim_{n\to\infty}\int_{x=a}^b f_n(x)F(x-\xi)\mathrm{d}x\right)=0$ ?

For me it is intuitively true, because $F(x-\xi)$ 'slip away' to infinity as $|\xi|\to\infty$, leaving inside $[a,b]$ a smaller and smaller contribution, but I can't formalize this argument.

Answer 1

We make use of the theorem below.

Theorem. Let $(T_n)_{n\in\mathbb{N}} \subset \mathcal{D}'(\mathbb{R})$ be a sequence of distributions on $\mathbb{R}$ such that the following conditions hold:

1. There exists a compact interval $K \subset \mathbb{R}$ such that each $T_n$ is supported on $K$.
2. $(T_n)_{n\in\mathbb{N}}$ converges to a linear functional $T$ on $C^{\infty}_c(\mathbb{R})$ under weak*-topology, that is, $$ \lim_{n\to\infty} \langle T_n, \varphi \rangle = \langle T, \varphi \rangle $$ holds for any $\varphi \in C^{\infty}_c(\mathbb{R})$.

Then $T$ is also a distribution on $\mathbb{R}$ that is supported on $K$.

Proof. We may regard each $T_n$ as a linear functional on $C^{\infty}(K)$ and we do so. Since $K$ is compact, $C^{\infty}(K)$ is a Fréchet space with respect to the topology induced by seminorms $\varphi \mapsto \sup_{x\in K}|\varphi^{(m)}(x)|$, $m\geq 0$. Then a version of uniform boundedness principle ([1], [2]) shows that the family $(T_n)_{n\in\mathbb{N}}$ is equicontinuous, hence the pointwise limit $T$ is also continuous. $\square$

Now let $(f_n)_{n\in\mathbb{N}}$ be the sequence of smooth functions as in OP's hypothesis, and let $T$ be the linear functional on $C^{\infty}_c(\mathbb{R})$ defined by the relation

$$ \langle T, \varphi \rangle = \lim_{n\to\infty} \int_{a}^{b} f_n(x) \varphi(x) \, \mathrm{d}x $$

for all $\varphi \in C^{\infty}_c(\mathbb{R})$. Then the theorem tells that $T$ is a distribution supported on the compact interval $[a, b]$. Since every compactly supported distribution had finite order, this implies that there exist an integer $k \geq 0$ and a finite constant $C \geq 0$ satisfying

$$ |\langle T, \varphi \rangle| \leq C \sup_{x \in K} |\varphi^{(k)}(x)| $$

for any $\varphi \in C^{\infty}(\mathbb{R})$. Consequently, for any $F$ satisfying OP's condition,

\begin{align*} \lim_{n\to\infty} \int_{a}^{b} f_n(x)F(x-\xi) \, \mathrm{d}x &= \langle T, F(\cdot - \xi) \rangle = \mathcal{O}\left( \sup_{x \in K} |F^{(k)}(x - \xi)| \right) \end{align*}

Now it is easy to check that this vanishes as $|\xi| \to \infty$.

Answer 2

You can start by proving that $$\lim_{n\to \infty} \int_{a}^{b} f_n(x)\mathrm d x$$ is finite. For that let $$h(x) = \begin{cases} \exp\left({-\frac{1}{\left(e^x - 1\right)^2}}\right) & \text{if $x > 0$}\\ 0 & \text{if $x \le 0$} \end{cases}$$

This $h$ function is infinitly differentiate function such that $$\lim_{x\to \infty} e^x \left(1-h(x)\right) = 0$$

Let $$F(x) = \begin{cases} 1-h(a-x) & \text{if $a < x$}\\ 1 & \text{if $a \le x \le b$}\\ 1-h(b-x) & \text{if $b < x$} \end{cases}$$

It is easy to prove that $F$ satisfy the conditions mentionned in the OP so you have that the limit $$\lim_{n\to \infty} \int_a^bf_n(x)F(x)\mathrm d x$$ exists.