# Prove that convolution smooths function

I want to prove that convolution of a function with itself smooths the function in some sense.

Given a function $$f:\mathbb{R}^n \to \mathbb{R}$$, the roughness $$R = \int \Delta f$$ where we integrate the Laplacian over all values of $$f$$.

Can we show that $$R(f * f) < R(f)$$ where $$*$$ is convolution. If not, is there another measure of smoothness we could use.

I'm not sure the "roughness" you define is a good measure of smoothness. For instance, if $$n=1$$, and $$f$$ is twice differentiable, then $$\int_{\mathbb R} \Delta f=\lim_{+\infty}f^\prime-\lim_{-\infty}f^\prime$$ when those limits exist. It only depends on the behavior of $$f^\prime$$ at the infinities.
Instead, you can look at something like a Sobolev norm $$\|f\|_S^2=\int |f|^2 + \int |\Delta f|^2$$ You see, if $$f$$ is not super smooth, then the Laplacian will have a lot of energy, and the norm will be high. So small Sobolev norm means smooth function.
Take $$f \in L^1(\mathbb R^n)\cap L^2(\mathbb R^n)$$. Then $$f$$ has a Fourier transform. Thanks to Parseval's theorem, the Sobolev norm can be expressed in the Fourier domain: $$\|f\|_S^2=\int_{\mathbb R^n}(1+\|\omega\|^2)|\hat{f}(\omega)|^2d\omega$$ Also, remember that the Fourier transform maps convolutions to products, so $$\|f*f\|_S^2=\int_{\mathbb R^n}(1+\|\omega\|^2)|\hat{f}(\omega)|^4d\omega$$ Now because $$f$$ is integrable, its Fourier transform is bounded by its $$L^1$$ norm $$\|f\|_1$$ (easy for you to check). So $$\|f*f\|_S^2\leq\|f\|_{1}^2\int_{\mathbb R^n}(1+\|\omega\|^2)|\hat{f}(\omega)|^2d\omega$$ In other words $$\|f*f\|_S\leq \|f\|_1\cdot\|f\|_S$$ Here $$\|f\|_1$$ acts as a normalization factor. That's necessary because $$\|f*f\|_S$$ is homogeneous of degree 2, while $$\|f\|_S$$ is of degree 1. So to make it easier to see the effect of the convolution, just focus on functions that are normalized such that $$\|f\|_1=1$$, then $$\boxed{\|f*f\|_S\leq \|f\|_S}$$ Convolving decreases the Sobolev norm, that is, the resulting function is smoother.