How translation generates weak convergence? Let $\{x_n\} \subset \mathbb R^d$, define $f_n(x)= f(x-x_n)$. We note that $\|f_n\|_{L^p}= \|f\|_{L^p}$.
Assume that $x_n \to \infty$ as $n\to \infty.$

Question:
Let $f\in L^p, g \in L^{p'} (\frac{1}{p}+ \frac{1}{p'}=1).$ Can we expect  that $\int_{\mathbb R^{d}} f_{n}(x) g(x) dx \to 0$ as $n\to \infty$? If not, any counter example? (For simplicity, we may take $p=2.$)

 A: Without loss of generality, assume that $\Vert f\Vert_{L^p}=\Vert g\Vert_{L^{p'}}=1$.
Let's fix some arbitrarily small $\epsilon>0$.
First, there exists $r>0$ such that for $B_r=\{x\in\mathbb R^d:|x|\le r\}$
$$
\int_{B_r}|f(x)|^pdx>1-\epsilon^p,\quad\text{and}
\quad\int_{B_r}|g(x)|^{p'}dx>1-\epsilon^{p'}
$$
(it follows from the fact that $f(x,r,p)=|f(x)|^pI(|x|\le r)$ converges to $|f(x)|^p$ monotonically when $r\to+\infty$, so we can apply the monotone convergence theorem, and similar for $|g(x)|^{p'}$).
This implies that
$$
\int_{\mathbb R^d\setminus B_r}|f(x)|^pdx<\epsilon^p,\quad\text{and}
\quad\int_{\mathbb R^d\setminus B_r}|g(x)|^{p'}dx<\epsilon^{p'}
$$
Second, since $x_n\to\infty$, there exists $N\in\mathbb N$ such that for any $n>N$ the norm $|x_n|>2r$, so $B_r$ and shifted $B_r+x_n$ are disjoint.
Finally, using Holder's inequality,
$$
\begin{aligned}
\left|\int_{\mathbb R^d}f_n(x)g(x)dx\right|&=\left|\int_{B_r}f(x-x_n)g(x)dx+\int_{B_r+x_n}f(x-x_n)g(x)dx+\int\int_{\mathbb R^d\setminus(B_r\cup B_r+x_n)}f(x-x_n)g(x)dx\right|\\
&\le\left|\int_{B_r}f(x-x_n)g(x)dx\right|+\left|\int_{B_r+x_n}f(x-x_n)g(x)dx\right|+\left|\int\int_{\mathbb R^d\setminus(B_r\cup B_r+x_n)}f(x-x_n)g(x)dx\right|\\
&\le\left(\int_{B_r}|f(x-x_n)|^pdx\right)^{1/p}\left(\int_{B_r}|g(x)|^{p'}dx\right)^{1/p'}+
\left(\int_{B_r+x_n}|f(x-x_n)|^pdx\right)^{1/p}\left(\int_{B_r+x_n}|g(x)|^{p'}dx\right)^{1/p'}+
\left(\int_{\mathbb R^d\setminus(B_r\cup B_r+x_n)}|f(x-x_n)|^pdx\right)^{1/p}\left(\int_{\mathbb R^d\setminus(B_r\cup B_r+x_n)}|g(x)|^{p'}dx\right)^{1/p'}\\
&< \epsilon\cdot 1+1\cdot\epsilon+\epsilon^2
\end{aligned}
$$
which is arbitrary small. Hence
$$
\int_{\mathbb R^d}f_n(x)g(x)dx\to0\quad\text{when $n\to\infty$}
$$
