Interesting, but I'm not seeing what this has to do with Fourier or Laplace. Mathematically, the statement is: if x is L^{\infty} and h is L^1, then y=x*h is L^

Interesting, but I'm not seeing what this has to do with Fourier or Laplace. Mathematically, the statement is: if x is L^{\infty} and h is L^1, then y=x*h is L^{\infty}. The proof of this is very routine... just write down the definition of the convolution and stare at it. Now it's true that the Fourier transform is not well-behaved on L^{\infty}. Any book on harmonic analysis will discuss this point in great lengths. But the context in which these questions are discussed (singular integral operators on spaces like BMO, which contains L^{\infty}) doesn't tend to include the Laplace transform in a useful manner. Someday I'll have to properly learn harmonic analysis to sort out these questions for myself.