If $X$ is an infinite dimensional separable Banach space (like $C^0(\Omega)$, for a compact metric space $\Omega$ ), and $\{y _ j \} _ {j\ge 1}$ is a dense sequence in its unit ball, one considers the norm on $X^*$ defined by $$ ||| u |||:=\sum _ {j=1} ^\infty 2 ^ {-j} |\langle u, y _ j \rangle |\\ ,$$$$ ||| u |||:=\sum _ {j=1} ^\infty 2 ^ {-j} |\langle u, y _ j \rangle |\, ,$$ which is weaker than the dual norm $\| \cdot \|$, since $ |||u|||\le \| u\|$. On the space $ X ^ * $, the $|||\cdot|||$ norm topology and the $w^*$ topology differ, because the latter is not metrizable. However, they induce the same topology on any (dual norm) bounded subset.
rmk. If $Y$ denotes the dense linear subspace spanned by $\{y _ j \} _ {j\ge 1}$ we may also consider the weak topology $\sigma(X ^ *,Y)$ on $ X ^ * $: that is, the smallest TVS topology that makes continuous any evaluation map at points $y\in Y$, that is $ X ^ * \ni u\mapsto \langle u, y\rangle $. With this topology, $X ^ * $ is a locally convex, Hausdorff and first countable space, though metrizable, yet not normable as not locally bounded. So this is strictly weaker than the above $|||\cdot|||$ norm-topology; it is also strictly weaker than the weak $^*$ topology $\sigma(X ^ *,X)$, since their dual spaces are the evaluations at points of $Y$, respectively of $X$. Again, on $\|\cdot\|$-bounded subsets they induce the same topology: indeed, if $u_n\to u$ in $\sigma(X ^ *,Y)$ with $\|u_n\|\le R$, then $ ||| u - u _ n |||=\sum _ {j=1} ^\infty 2 ^ {-j} |\langle u - u_n , y _ j \rangle |$ also converges to $0$, since the terms of the series converge to $0$ while dominated by the series $ \sum _ {j=1} ^\infty 2 ^ {-j} (2R) $. Moreover, it is also true that $u_n\to u$ in $\sigma(X ^ *,X)$, because for any $x\in X$ and $y\in Y$ $|\langle u - u_n , x \rangle |\le |\langle u - u_n , y \rangle | + 2R\| x - y \|$, whence $\limsup _ { n \to \infty} |\langle u - u _ n , x \rangle | \le 2R\| x - y \| $, and the RHS can be made arbitrarily small.