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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 $p=2.$)

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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 $$ \int_{\mathbb R^d}f_n(x)g(x)dx\to0\quad\text{when $n\to\infty$} $$

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    @SG: Thanks. Please would you tell me how you are getting $|\int_{B_r} |f(x-x_n)|^p dx| < \epsilon$? In other words, how you got the last inequality? – 2017-02-27
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    $\int_{B_r}|f(x-x_n)|^pdx=\int_{B_r+x_n}|f(x)|^pdx\le\int_{\mathbb R^d\setminus B_r}|f(x)|^pdx<\epsilon^p$. The second transition is possible because $B_r+x_n\subset\mathbb R^d\setminus B_r$ since $B_r$ and $B_r+x_n$ are disjoint. – 2017-02-27