Let $f, g\in L^{2}(\mathbb R)$ and $\{x_{n}\} \subset (0, \infty)$ such that $x_{n}\to \infty$ as $n\to \infty.$
Assume that $\int_{\mathbb R} \left|\frac{1}{\sqrt{x_{n}}}f(y/x_{n})-g(y)\right|^2 dy \to 0$ as $n\to \infty.$
Then my Question is: Can we conclude that $f=0$?
My attempt: Take $z=\frac{y}{x_{n}}.$ Then by change of variable, we have $\int_{\mathbb R} \left|f(z)- \sqrt{x_{n}}g(x_{n}z)\right|^2 dz \to 0$ as $n\to \infty$
We first show that if $h$ is an element of $\mathbb L^2(\mathbb R)$, then $h_n:=h(\cdot /x_n) /\sqrt{x_n}\to 0$ weakly in $\mathbb L^2$. To see this, since the sequence $(h_n)$ is bounded in $\mathbb L^2$, it suffices to prove that for each continuous function $v$ with compact support, $$\tag{*} \int_{\mathbb R} h_n(y)v(y)\mathrm dy\to 0.$$ First assume that $h$ is continuous with compact support. Then $(*)$ holds by dominated convergence, as $\left|h_n(y) v(y)\right|\leqslant \sup_t \left|h(t)\right |\mathbf 1_{\operatorname{supp}v }(y)\cdot v(y)/\min_{n\geqslant 1}x_n $. We then conclude that (*) holds by a density argument.
Now, in the context of the problem,we apply the previous result to $h=f$ to deduce that $f(\cdot /x_n) /\sqrt{x_n}\to 0$ weakly in $\mathbb L^2$. Since it also converges to $g$ (strongly) in $\mathbb L^2$, we get that $g=0$ and the $\mathbb L^2$-norm of $f(\cdot /x_n) /\sqrt{x_n}$ is the same as that of $f$, hence $f=0$.