I believe we get the stronger statement that the convolution of an infinitely differentiable integrable function $f$ with any integrable function $g$ will result in $f*g$ being infinitely differentiable. In fact, $(f*g)' = f'*g$ (This can be found in The Fourier Transform and Its Applications, Bracewell) Thus the convolution of two infinitely differentiable, integrable functions will necessarily be (infinitely) differentiable and so also continuous.