= Solution
Use the transform convention $F(\lambda)=\int_{\mathbb R}f(t)e^{-i\lambda t}\,dt$. The given <Fourier inversion theorem> gives
$$
f(t)=\frac1{2\pi}\int_{-\pi}^{\pi}F(\lambda)e^{it\lambda}\,d\lambda.
$$
Here $F$ is continuous, vanishes at both endpoints and belongs to $L^2[-\pi,\pi]$. It therefore defines a continuous periodic function. Its <Fourier coefficient> at index $-n$ is $f(n)$. By part (i),
$$
P_N(\lambda)=\sum_{|n|\leq N}f(n)e^{-in\lambda}\longrightarrow F(\lambda)
\quad\text{in }L^2.
$$
Pair this convergence with $e^{it\lambda}$. Since that function has normalized $L^2$ norm one, the <Cauchy-Schwarz inequality> yields, uniformly in $t$,
$$
\left|f(t)-\frac1{2\pi}\int_{-\pi}^{\pi}P_N(\lambda)e^{it\lambda}\,d\lambda\right|
\leq\|F-P_N\|_2\longrightarrow0.
$$
The elementary integral is the <sinc function>,
$$
\frac1{2\pi}\int_{-\pi}^{\pi}e^{i(t-n)\lambda}\,d\lambda
=D(t-n).
$$
Thus the <sampling expansion by periodic Fourier projection> is
$$
\boxed{f(t)=\sum_{n\in\mathbb Z}f(n)D(t-n).}
$$
There is no pointwise interchange with an unproved <Fourier series>: the calculation first uses finite sums and then an $L^2$ limit.
The convergence can also be made absolute. <Parseval's identity> gives $\sum_n|f(n)|^2=\|F\|_2^2$, and <Bessel's inequality> applied to $e^{it\lambda}$ gives $\sum_n|D(t-n)|^2\leq1$. Hence
$$
\sum_{|n|>N}|f(n)D(t-n)|
\leq\left(\sum_{|n|>N}|f(n)|^2\right)^{1/2}\longrightarrow0
$$
uniformly in $t$. At an integer argument, $D$ is one at zero and zero at the other integers, so the expansion interpolates the samples exactly.
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