Frequency + Fourier Transform Notation Question
They are both angular frequencies and indicate whether you are dealing with a continuous or discrete signal/system. It's perhaps instructive to relate these to the s-transform and z-transform.

For a continuous signal or TF we might write, e.g., $$h(t) = e^{-at}\rightarrow H(s)= \frac{1}{a+s}$$ and the steady-state frequency response would be obtained by substituting \$s=j\Omega\$, thus $$H(j\Omega)=\frac{1}{a+j\Omega}$$

Note that the s-domain and z-domain are related by \$z\rightarrow e^{sT}\$ (\$e^{-sT}\$ is the Laplace time-delay operator, delaying a signal by \$T\$ sec, and performs the same action as \$z^{-1}\$, which delays a discrete signal by one sample). Hence, for a purely discrete signal or TF, say, $$h(n)=a^n\rightarrow H(z)=\frac{z}{z-a},  \lvert a\rvert &lt;1$$ we may normalise by setting \$T=1\$ and the frequency response is obtained by the substitution \$z=e^{j\omega}\$, thus $$H(e^{j\omega})=\frac{e^{j\omega}}{(e^{j\omega}-a)}= \frac{\cos(\omega) +j\sin(\omega)}{[\cos(\omega)-a] +j\sin(\omega)}$$

In filter design it's common to work in normalised form, even if the \$h(n)\$ are samples of a real analogue signal. This makes the analysis less cluttered, reduces calculation errors, and the results can be scaled back to normal time at the end of the analysis