$$ h(x) = (f * g)(x) $$

whereas if $\{a_n\}$ and $\{b_n\}$ are two sequences, the convolution can be written as

$$ {c[n]} = {a[n]}*{b[n]}. $$

What should not be done in any circumstance is to write the convolution of two functions as

$$ h(t) = f(t) * g(t). $$

Why not? Well, because the convolution operator $*$ does not operate on value but on functions; $f(t)$ and $g(t)$, however, are values. That can be easily seen by inserting $t=3$ in above equation. We get

$$ h(3) = f(e) * g(3) $$

which obviously does not make any sense. The equation

$$ h(3) = (f*g)(3) $$

is the only appropriate way of writing this. In stark contrast, take a different unitary operator, like addition. For addition one can easily argue that the value of the sum of two functions at a given point equals the sum of the values of the individual functions: Addition works on values, i.e.,

$$ h(t) = (f+g)(t) = f(t)+g(t). $$

Still, many excellent textbooks on signal processing (even the ones from Oppenheim and Schafer and from Vetterli) make use of this "abuse of notation" for convolution. The reason is that as soon a Dirac (or Kronecker) delta enters the game, the famous convolution property pops up somewhere. In incorrect notation, this property can be stated as follows:

$$ h(t)=f(t) * \delta(t-T) = f(t-T) $$

Finding a correct notation is not easy. I found a few helpful comments on stackexchange. I will add my idea at the bottom, please judge yourself which you like the most. I have to admit that actually none of these is really satisfactory...

- One commenter suggested to introduce the shift operator $T_T$, i.e., $T_T$ is the function mapping $t$ to $t-T$. Then, $\delta(t-T)=\delta(T_T(t))=(\delta\circ T_T)(t)$ and we get $h(t)=(f*(\delta\circ T_T))(t)$.
- Another commenter proposed to write $h(t) = (f*\delta(\cdot-T))(t)$.
- My idea is to define an ensemble of functions $\delta_T(t):=\delta(t−T)$ and then write the convolution as $h(t)=(f*\delta_T)(t)$.

While my comment might give the shortest notation, it is not as general as the one with the shift operator. I would have problems, e.g., to write the convolution of a function with its time-reversal, i.e., (in bad notation), $h(t)=f(t)*f(-t)$. If we introduce a time-reversal operator $R$ for which $R(t)=-t$, we could write the convolution easily as $h(t)=(f*f\circ R)(t)$.