$\int_{\mathbb R^d} f_n (x) g(x) \to 0$ if $g\in L^{\infty}$?

Question

Let $f\in C(\mathbb R^d)$ and support of $f$ is $ \subset (-1,1)$, that is, $f \in C_c(-1,1).$ Define $f_n(x) =n^{1/p} f(nx).$ ($1\leq p \leq \infty$)

We note that $f_n \in C_{c} (-1/n, 1/n)$ and $f_n(0) \to \infty.$

My Question: Can we say $\int_{\mathbb R^d} f_n (x) g(x) \to 0$ (as $n\to \infty$) if $g\in C_{c} (\mathbb R^d)$? Can we say $\int_{\mathbb R^d} f_n (x) g(x) \to 0$ if $g\in L^{\infty}$?

Answer 1

Using the change of variables $y = nx$,

$$ \left|\int_{\mathbb R^d} f_n(x) g(x) \; dx\right| = n^{1/p-d} \left|\int_{\mathbb R^d} f(y) g(y/n)\; dy\right| \le n^{1/p-d} \|f\|_1 \|g\|_\infty$$

Unless $p=1$ and $d=1$, this does go to $0$. If $p=1$ and $d=1$, then no: e.g. take $f$ and $g$ to be nonnegative, and $1$ in a neighbourhood of $0$.