If $\mu$ is a positive, finite measure on $\Bbb R$, define $Q=\{f\in L^2(\Bbb R,\mu): \int_{\Bbb R} |x| |f|^2\text d\mu<\infty\}$ and $q(f,g)=\int_{\Bbb R} x\bar f g\text d \mu$. Suppose there exists $M\ge 0$ s.t. $q(f,f)\ge -M||f||_{L^2}$ and let $$||f||_+^2=q(f,f)+(M+1)||f||_{L^2}^2$$ Is it true that $Q$ is complete under $||\cdot||_+$?
Since $||f||_+\ge ||f||_{L^2}$ and $L^2$ is complete, any Cauchy $f_n\in Q$ converges in the $L^2$ norm to some $f\in L^2$. It remains to show that $f\in Q$ and $q(f_n-f,f_n-f)\to 0$. In order to show that $\int |x||f|^2<\infty$ I thought about using Fatou's lemma but I have not managed to prove that $\liminf \int |x||f_n|^2<\infty$. The hint in the book says to use the dominated convergence theorem but I have no idea which dominating function to pick. Maybe the fact that $\mu(\Bbb R)<\infty$ and Egorov's theorem can help, but I have not been able to connect the dots, is there something I'm missing?
My idea is to split the problem into two parts. We separate the positive and the negative real halfaxis so that we can control the stupid sign hidden in $q$.
Introduce the short notation $g_-= 1_{(-\infty, 0]}$ and $g_+= 1_{[0, \infty)}$ and $\nu(dx)= \vert x \vert \mu(dx)$, $\nu_-(dx)= \vert x \vert \cdot 1_{(-\infty, 0]}(x) \mu(dx)$, $\nu_+(dx)= \vert x \vert \cdot 1_{[0,\infty)}(x) \mu(dx)$ and $\mu_+(dx)= 1_{[0, \infty)}(x) \mu(dx)$.
We have $$ \Vert g \Vert_+^2 = \Vert g_+ \Vert_+^2 + \Vert g_- \Vert_+^2. $$ This reduces the problem to show convergence on the positive and on the negative reals.
On the negative reals: We have $$ \Vert g_- \Vert_{L^2(\mu)}^2 \leq \Vert g_- \Vert_+^2 = - \Vert g_- \Vert_{L^2(\nu)}^2 + (M+1) \Vert g_-\Vert_{L^2(\mu)} \leq (M+1) \Vert g_- \Vert_{L^2(\mu)} $$ Hence, if $(f_n)$ is a Cauchy sequence in $\Vert \cdot \Vert_+$, then $((f_n)_-)$ is a Cauchy in $L^2(\mu)$ and (as $L^2(\mu)$ is complete) we get that it converges to some $f_-$. Thus, $$\Vert f_-- (f_n)_-\Vert_+^2 \leq \Vert f_- - (f_n)_- \Vert_{L^2(\mu)}^2 \rightarrow 0.$$ Hence, $$ \Vert f_- - (f_n)_- \Vert_+^2 \rightarrow 0. $$
On the positive reals: We have $$ \Vert g_+ \Vert_+^2 = \Vert g_+ \Vert_{L^2(\nu)}^2 + (M+1) \Vert g_+ \Vert_{L^2(\mu)}^2. $$ Hence, if $(f_n)$ is a Cauchy sequence in $\Vert \cdot \Vert_+$, then $((f_n)_+)$ is a Cauchy sequence in both $L^2(\mu)$ and $L^2(\nu)$. As both of them are complete, there exists limiting functions $f_+$ and $\tilde{f}_+$.
Recall that $L^2$ convergence implies pointwise convergence almost everywhere along a subsequence. Hence, we can pick a common subsequence which we call - by abuse of notation - again $((f_n)_+)$. This means $((f_n)_+)$ converges pointwise a.e. for both $\mu$ and $\nu$. However, as $(f_n)_+$ are supported on $[0,\infty)$, we get that $((f_n)_+)$ converges pointwise a.e. also in $\mu_+$ and $\nu_+$. But they have same zero sets ($\mu_+ \ll \nu_+ \ll \mu_+$) and thus, we get $f_+ = \tilde{f}_+$ both $\nu_+$ and $\mu_+$ a.e. Again as $f_+, \tilde{f}_+$ are supported in $[0,\infty)$ this means $f_+ = \tilde{f}_+$ $\mu$-a.e. and $\nu$-a.e.
Hence, we get that $$ \Vert f_+ - (f_n)_+ \Vert_+^2 = \Vert f_+ - (f_n)_+\Vert_{L^2(\nu)}^2 + (M+1)\Vert f_+ - (f_n)_+\Vert_{L^2(\mu)}^2 = \Vert \tilde{f}_+ - (f_n)_+\Vert_{L^2(\nu)}^2 + (M+1)\Vert f_+ - (f_n)_+\Vert_{L^2(\mu)}^2 \rightarrow 0. $$
On the whole real axis:
Now we define $f:= f_- + f_+$, then we get $$ \Vert f - f_n \Vert_+^2 = \Vert f_- - (f_n)_-\Vert_+^2 + \Vert f_+ - (f_n)_+ \Vert_+^2 \rightarrow 0. $$