= Solution
The direct squared <L2 norm> is the length of the interval, namely $2$. The <Fourier transform> is
$$
\widehat f(k)=\int_{-1}^1e^{-ikx}dx=\frac{2\sin k}{k},\qquad \widehat f(0)=2.
$$
For the transformed norm, let $I=\int_0^\infty\sin^2k/k^2\,dk$. <Integration by parts>, with vanishing endpoint term $-\sin^2k/k$, gives
$$
I=2\int_0^\infty\frac{\sin k\cos k}{k}\,dk=\frac\pi2.
$$
The supplied oscillatory integral is understood as an <improper integral>. Therefore
$$
\boxed{\frac1{2\pi}\int_{-\infty}^{\infty}\left|\frac{2\sin k}{k}\right|^2dk=\frac4\pi I=2,}
$$
which agrees with the direct norm and verifies the <Parseval identity> with the specified normalization.
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