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.

<strike>EDIT: Also, if my math is correct, we can check this formula via the Fourier transform:

$\widehat{((f*g)')}(r) = ir\widehat{f*g}(r) = ir\widehat{f}(r)\widehat{g}(r) = \widehat{f'}(r)\widehat{g}(r) = \widehat{(f'*g)}(r)$</strike>

Apologies, this only applies in the case that $f$ is compactly supported. Back to the drawing board...