= Solution
The integral is conjugate-linear in $f$, linear in $g$, and conjugate-symmetric, by the corresponding properties of the integral and of <complex conjugation>. Also,
$$
\langle f,f\rangle=\int_{-\infty}^{\infty}|f(x)|^2e^{-x^2}\,dx\geq0.
$$
If a <polynomial> $f$ is not zero, there is a real point $x_0$ with $f(x_0)\ne0$: a nonzero <polynomial> cannot vanish at every real point. By <continuity>, $|f|$ has a positive lower bound on a sufficiently small closed interval about $x_0$. The continuous function $e^{-x^2}$ also has a positive lower bound there. The integral over this interval is strictly positive. Hence
$$
\boxed{\langle f,f\rangle=0\ \Longleftrightarrow\ f=0.}
$$
Together with the finiteness allowed in the question, these properties prove that the formula defines an <inner product> on $V=\mathbb C[x]$.
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