= Solution
Put
$$
H(u)=\frac{4\sin^2(u/2)}{u^2},
\qquad H(0)=1.
$$
This function is bounded near zero and is $O(u^{-2})$ at infinity, so $H\in L^1(\mathbb R)$ and the defining integral for $g$ converges absolutely.
For the <triangular function>
$$
h(x)=(1-|x|)\mathbf1_{[-1,1]}(x),
$$
direct integration gives
$$
\begin{aligned}
\widehat h(u)
&=\int_{-1}^1(1-|x|)e^{iux}\,dx\\
&=2\int_0^1(1-x)\cos(ux)\,dx\\
&=\frac{2(1-\cos u)}{u^2}
=\frac{4\sin^2(u/2)}{u^2}=H(u).
\end{aligned}
$$
The <Fourier inversion theorem>, or equivalently the <Fourier transform of a triangular function>, therefore yields
$$
\boxed{g(x)=(1-|x|)\mathbf1_{[-1,1]}(x).}
$$
In particular $g(x)=0$ whenever $|x|>1$. Finally,
$$
\boxed{\|g\|_2^2
=2\int_0^1(1-x)^2\,dx
=\frac23.}
$$
This also agrees with the <Plancherel theorem> applied to $h$ and $H$.
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