Verify convolution theorem, f = g = e^{-x²}

Convention. F{f}(s) = ∫_{−∞}^{∞} f(x) e^{−isx} dx and (f * g)(x) = ∫ f(u) g(x − u) du. The theorem states F{f * g} = F(s) · G(s). We verify it for f = g = e^{−x²}.

Right side: F(s) · G(s)

Using the standard result ∫ e^{−x²} e^{−isx} dx = √π e^{−s²/4}:

F(s) = G(s) = √π e^{−s²/4}

F(s) G(s) = π e^{−s²/2}

Left side: F{f * g}

Step 1: convolution.

(f * g)(x) = ∫ e^{−u²} e^{−(x−u)²} du

Combine the exponents: u² + (x − u)² = 2u² − 2xu + x² = 2(u − x/2)² + x²/2. So

(f * g)(x) = e^{−x²/2} ∫ e^{−2(u − x/2)²} du
           = e^{−x²/2} · √(π/2)

(using ∫ e^{−2t²} dt = √(π/2)).

Step 2: transform of the convolution. The transform of e^{−x²/2} is √(2π) e^{−s²/2}. So

F{f * g} = √(π/2) · √(2π) e^{−s²/2} = π e^{−s²/2}

Comparison

Left side = π e^{−s²/2} = right side. Hence the convolution theorem is verified for f = g = e^{−x²}. ∎

(Numerical check at x = 0.8: ∫ e^{−u²−(0.8−u)²} du = 0.91009 = √(π/2) e^{−0.32} ✓.)