Solution

Because the centered normal distribution density

g_{t}

is nonnegative and has integral one, Fubini's theorem and translation invariance give

∥ f * g_{t} ∥_{1} \leq \int_{R} \int_{R} ∣ f (x - y) ∣ g_{t} (y) d y d x = \int_{R} g_{t} (y) ∥ f ∥_{1} d y = ∥ f ∥_{1} .

Thus

f * g_{t} \in L^{1}

Use the angular-frequency convention

f (ξ) = \int_{R} f (x) e^{- i x ξ} d x .

The convolution theorem and the Gaussian transform give

f * g_{t} (ξ) = f (ξ) e^{- t ξ^{2} /2} .

Since

∣ f (ξ) ∣ \leq ∥ f ∥_{1}

, this product is integrable. Moreover, absolute integrability justifies exchanging the integrals:

\frac{1}{2 π} \int_{R} f (ξ) e^{- t ξ^{2} /2} e^{i x ξ} d ξ = \int_{R} f (z) [\frac{1}{2 π} \int_{R} e^{- t ξ^{2} /2} e^{i (x - z) ξ} d ξ] d z = \int_{R} f (z) g_{t} (x - z) d z = (f * g_{t}) (x) .

This proves the Fourier inversion theorem for the convolution directly.

Now additionally suppose

f \in L^{1}

. For every

x

lim_{t ↓ 0} (f * g_{t}) (x) = \frac{1}{2 π} \int_{R} f (ξ) e^{i x ξ} d ξ .

Fourier inversion identifies the right-hand side with

f (x)

x

f * g_{t} (x) ⟶ f (x) for almost every x .

Solved by gpt-5.6-sol high.

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