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 Incoming links: Fourier convolution of two periodic L2 functions

Use the inner product convention

⟨ f, h ⟩ = \int_{- π}^{π} f \overset{ˉ}{h}

. For

h = 1_{[- π /2, π /2]}

, integration gives

⟨ e_{0}, h ⟩ = π /2, ⟨ e_{k}, h ⟩ = 2/ π \frac{s i n ( kπ /2 )}{k} (k \neq = 0) .

The nonzero even coefficients vanish, while

sin ((2 n + 1) π /2) = (- 1)^{n}

. The preceding polarized Parseval identity therefore yields

\int_{- π /2}^{π /2} f (t) d t = \frac{π}{2} ⟨ f, e_{0} ⟩ + \frac{2}{π} n \in Z \sum (- 1)^{n} \frac{⟨ f , e _{2 n + 1} ⟩}{2 n + 1} .

Absolute convergence follows from square summability of the Fourier coefficients and of

((2 n + 1)^{- 1})_{n \in Z}

For the convolution identity, fix any real

t

h_{t} (s) = \overline{g (t - s)}

h_{t} \in L^{2} (- π, π)

with

∥ h_{t} ∥_{2} = ∥ g ∥_{2}

. A change of variable and integration over one full period give

⟨ e_{n}, h_{t} ⟩ = \frac{1}{2 π} \int_{- π}^{π} e^{in s} g (t - s) d s = e^{in t} ⟨ g, e_{n} ⟩ .

Applying polarized Parseval identity to

f, h_{t}

gives, for every fixed

t

\int_{- π}^{π} f (s) g (t - s) d s = n \in Z \sum ⟨ f, e_{n} ⟩ ⟨ g, e_{n} ⟩ e^{in t} .

The integral exists by the Cauchy-Schwarz inequality. The same inequality gives

\sum_{n} ∣ ⟨ f, e_{n} ⟩ ⟨ g, e_{n} ⟩ ∣ \leq ∥ f ∥_{2} ∥ g ∥_{2}

, so the displayed series converges absolutely and uniformly in

t

. In particular the equality holds at every

t