Let $T\colon X\to Y$ be a bounded linear operator. Then it is known that it is continuous as a map between $X$ and $Y$ endowed with their weak topologies.
Let me address a spacial case where $X=C(K)$, the space of continuous functions on a compact space $K$. Suppose that $T\colon C(K)\to Y$ is a bounded linear operator. Is it continuous as a map
$$T\colon (C(K), \text{topology of pointwise convergence})\to (Y, \text{weak})?$$
Take $K = [0,1], Y = L^1[0,1]$, and $T(f) = f$, and $\varphi \in Y^{\ast}$ given by $$ \varphi(f) = \int_0^1 f(x)dx $$ Now take $f_n$ to be a rectangle over $(0,1/n]$ with area 1. Then $f_n\to 0$ pointwise, but $$ \varphi\circ T(f_n) = 1 \quad\forall n\in \mathbb{N} $$ so $\{T(f_n)\}$ does not converge to $0$ in the weak topology of $Y$.
If $T:C(K)\to Y$ and $\varphi \in Y^{\ast}$, then $\varphi\circ T : C(K) \to \mathbb{C}$ is a bounded linear functional, so $\exists$ a complex measure $\mu$ such that $$ \varphi(T(f)) = \int_K fd\mu $$ Now what you are looking for is whether pointwise convergence implies the convergence of these integrals for all such complex measures $\mu$, which seems unlikely in most cases.