# Continuity of convolution on L¹

Question

Let $f,g\in L^1(\mathbb{R}^n)$ be nonnegative and assume $\int f\,g \; dx <\infty$. Let $\Phi_\varepsilon$ be the heat kernel at time $\varepsilon$. Denote with $f_\varepsilon = f \ast \Phi_\varepsilon$ and $g_\varepsilon = g \ast \Phi_\varepsilon$ the convolved functions.

Is the functional $F(f,g) := \int f \, g \; dx$ continuous with respect to the convolution, that is $$F(f_\varepsilon,g_\varepsilon) \to F(f,g) \qquad \text{as}\qquad \varepsilon \to 0\qquad ?$$

Background

The question can be rephrased in terms of shifts $$F(f_\varepsilon,g_\varepsilon) - F(f,g) = \int \int f(x) \left( g(x-y) - g(x) \right) \Phi_{2\varepsilon}(y) \; dy \; dx .$$ If in addition $f\in L^\infty(\mathbb{R}^n)$ the statement follows, because the right hand side vanishes as $\varepsilon \to 0$ as $$\int \left| \int \left( g(x-y) - g(x) \right) \Phi_{2\varepsilon}(y) \; dy \right | dx \to 0 \qquad\text{as} \qquad \varepsilon \to 0 .$$ For only $f\in L^1(\mathbb{R}^n)$ the situation is less clear. Let $$h(y) = \int f(x) g(x-y) \; dx .$$ Then $h\in L^1(\mathbb{R}^n)$, since $\int h(y) \, dy = \Vert f \Vert_{L^1} \Vert g \Vert_{L^1}$. In addition, we have $h(0) = F(f,g)<\infty$. Hence the question is answered, if we can show that $0$ is a Lebesgue point of $h$.

No. Take $C=\bigcup_{n\in\mathbb N} [1/(2n+1),1/(2n))$ and let $D=[0,1]\setminus C$. Now take take $f(x)=x^{-1/2}1_C$ and $g(x)=x^{-1/2}1_D$. Now $f_\epsilon$ and $g_\epsilon$ are both approximately $\epsilon^{-1/2}$ on $[0,\epsilon]$, so that there is no convergence of the integrals.