Imagine the Lebesgue space $L^2_\mu(\mathbb{R},\mathbb{C})$, a space of functions $f:\mathbb{R}\to\mathbb{C}$, where the measure function is $\mu=\frac{e^{-x^2}}{\sqrt{\pi}}$. If a function $f$ belong to this space then the following condition is satisfied:
$$\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty} |f(x)|^2e^{-x^2}dx<\infty$$
I want to demonstrate that $L^2_{\mu}$ is a vector space. So, it must satisfy the following conditions:
(i) Existence of origin $(0)$, i.e.
$$0\in L^2_{\mu}$$
(ii) Closed for sum of elements, i.e.
$$f,g\in L^2_{\mu}\Rightarrow f+g \in L^2_{\mu}$$
(iii) Closed for multiplication by scalars, i.e.
$$a\in \mathbb{K}, f\in L^2_{\mu} \Rightarrow af\in L^2_{\mu}$$
The (i) and (iii) conditions are easy to demonstrate. My problem is on (ii) condition where I did the following procedure:
$$\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty} |f(x)+g(x)|^2e^{-x^2}dx=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty} \left(f(x)+g(x)\right).\left(f(x)+g(x)\right)^*e^{-x^2}dx= $$
$$\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty} |f(x)|^2+|g(x)|^2+2\text{Re}\{f(x)g^*(x)\}e^{-x^2}dx$$
I know that $\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty} |f(x)|^2e^{-x^2}dx<\infty$ and $\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty} |g(x)|^2e^{-x^2}dx<\infty$ because $f,g\in L^2_{\mu}$ but I don't know what to with the term $\text{Re}\{f(x)g^*(x)\}$ on this integration.
What would you do?
Hint: $$|f(x) + g(x)|^2 \le 2|f(x)|^2 + 2|g(x)|^2$$
Proof of the hint: $0 \le (a - b)^2 = a^2 + b^2 -2ab$. Rearranging we have $$2ab\le a^2 + b^2. \tag 1$$
Now, as you noticed in the comment below $$|f(x) + g(x)|^2 \le |f(x)|^2 + |g(x)|^2 + 2|f(x)||g(x)|.$$ Apply $(1)$ with $a = |f(x)|$, $b = |g(x)|$ to the last term on the RHS to get the desired result.