I am familiar with the weak topology and a linear functional on a Banach space being weakly continuous, but in PDE I often see a statement along the lines of "$f$ is a weakly continuous solution to the PDE" or "$f$ is a weakly continuous map from the interval $(0,T)$ to $X$" where $X$ is some appropriate function space (usually Hilbert or Banach). What is the precise definition of weakly continuous here?
One definition I thought of is that this means $f$ is continuous with respect to the weak topology on $(0,T)$ and the weak topology on $X$ (i.e. weak-weak continuity). I saw a few older questions asking something similar and the answers are all in terms of convergence, i.e. $f$ is weakly continuous if whenever $x_n \rightarrow x$ weakly then $f(x_n) \rightarrow f(x)$ weakly. But as others pointed out, this is stronger than $f$ being weakly continuous and in fact states that $f$ is sequentially continuous.
Best Answer
First of all: There is no weak topology on $(0,T)$, because it is not even a vector space.
Therefore weak continuity of a map $f: (0,T) \to X$ for a Banach space $X$ can only mean one thing: $f$ is continuous with respect to the standard topology on $(0,T)$ and the weak topology on $X$.
Since $(0,T)$ is a metric space, the weak continuity of $f$ is equivalent to the sequential weak continuity of $f$ (see here): For any convergent sequence $x_n\to x$ in $(0,T)$ $f(x_n) \to f(x)$ weakly.