Norm of convolution of $n$ Gaussians

Question

If

$$f(x)=e^{-(\pi x)^2}$$ and

$$\psi_n(x)=(f* f*\dots*f)(x)$$ ($n$ times convolution).

Show that

$$\lVert \psi_n(x)\rVert = 1$$ (norm in $L^1(\mathbb{R})$).

I've tried using the Fourier Transform property but I'm stuck! Thanks

Answer 1

Taking the Fourier transform, $\hat{\psi}_n=\hat{f}\cdots\hat{f}$ with $n$ factors. We also know that the Fourier transform of a Gaussian is a Gaussian, in this case $\hat{f}(\xi)=e^{-(\pi\xi)^2}$. Therefore $$\hat{\psi}_n(0)=\hat{f}(0)\cdots\hat{f}(0)=1^n=1.$$ Finally note $\hat{\psi}_n(0)=\int \psi_n(x)\;dx = \|\psi_n\|_1$ as the integrand is nonnegative.