How to prove that, $\lim\limits_{k \rightarrow \infty} \int_{\Bbb{R}} f_k(x) g(x) dm(x) = 0$ $f$ and $g$ are measurable functions on a measure space $(\mathbb R, \mathcal{B}, m)$ where $m$ is Lebesgue measure. $f(x)$ satisfies $$\int_\mathbb R f^2(x) dm(x) < \infty$$ Where,  $$f_k(x) = \sqrt{k} f(kx).$$
Prove that if $g(x)$ satisfies $$\int_\mathbb R g^2 dm < \infty$$ then $$\lim\limits_{k \rightarrow \infty} \int_\mathbb R f_k(x) g(x) dm(x) = 0.$$
I can prove that $ f_k(x) g(x) $ is integrable, but I didn't figure out how to get the final result.
 A: 
Lemma: Continuous function of compact support are dense in $L^2(dm)$. Hence,

For,  $\varepsilon>0$ we know that there exist a $f_\varepsilon,g_\varepsilon\in C_c(\Bbb R)$ a continuous function of compact support such that, 
$$ \|g-g_\varepsilon\|_2\le \varepsilon \|f\|_2^{-1} $$
and 
$$ \|f-f_\varepsilon\|_2\le \varepsilon \|g_\varepsilon\|_2^{-1} $$
Then,using Cauchy Schwartz inequalty we have, 
\begin{align} \left|\int_\mathbb R f_k(x) g(x) dm(x)\right|  &=  \left|\int_\mathbb R f_k(x) [g(x)-g_\varepsilon(x)] dm(x)+\int_\mathbb R f_k(x) g_\varepsilon(x) dm(x)\right|\\  
&\le  \|g-g_\varepsilon\|_2 \left( \int_\mathbb R  k f^2(kx)dm(x)\right)^{1/2} +\left|\int_\mathbb R \sqrt{k}f(kx) g_\varepsilon(x) dm(x)\right| 
\\& \overset{\color{red}{u=kx}}{=} \|g-g_\varepsilon\|_2 \|f\|_2 + \left|\int_\mathbb R\frac{1}{ \sqrt{k}}f(x) g_\varepsilon(\frac{x}{k}) dm(x)\right|\\ & \le
\varepsilon +  \frac{1}{ \sqrt{k}}\left|\int_\mathbb Rf(x) g_\varepsilon(\frac{x}{k}) dm(x)\right| \end{align}
Similarly we have, $$ \left|\int_\mathbb Rf(x) g_\varepsilon(\frac{x}{k}) dm(x)\right| \le \varepsilon+ \left|\int_\mathbb Rf_\varepsilon(x) g_\varepsilon(\frac{x}{k}) dm(x)\right| \\= \varepsilon +\left|\int_{\text{supp }f_\varepsilon}f_\varepsilon(x) g_\varepsilon(\frac{x}{k}) dm(x)\right|$$

Since, $f_\varepsilon$ $h_\varepsilon$ are continuous functions  with compact support, we get
 $$\color{blue}{\lim_{k\to \infty}\int_{\text{supp }f_\varepsilon}f_\varepsilon(x) g_\varepsilon(\frac{x}{k}) dm(x)=\int_{\text{supp }f_\varepsilon}f_\varepsilon(x) g_\varepsilon(0) dm(x)}$$
Whence, 
$$ \lim_{k\to \infty} \left|\int_\mathbb R f_k(x) g(x) dm(x)\right| \le\lim_{k\to \infty} \left[\varepsilon+\frac{\varepsilon}{ \sqrt{k}}+ \frac{1}{ \sqrt{k}}\left|\int_{\text{supp }f_\varepsilon}f_\varepsilon(x) g_\varepsilon(\frac{x}{k}) dm(x)\right| \right]=\varepsilon $$
That is for all $\varepsilon>0,$ we have, $$ \limsup_{k\to \infty} \left|\int_\mathbb R f_k(x) g(x) dm(x)\right| \le \varepsilon $$
letting  $\varepsilon\to 0$  yields the require result.
$$ \color{red}{\lim_{k\to \infty}  \int_\mathbb R f_k(x) g(x) dm(x)=0.} $$
