Let $(X, \|\cdot\|)$ be a Banach space and $\mathbb{K}$ be its field of scalars ($\mathbb{R}$ or $\mathbb{C}$).
A function $f : [0,1] \to X$ is said to be strongly continuous in $x \in [0,1]$ if it is continuous in $x$ as a function $f : ([0,1], |\cdot|) \to (X, \|\cdot\|)$, while it is said to be weakly continuous in $x$ if, for all $\Lambda \in X^* := \mathcal{L}(X,\mathbb{K})$, $\Lambda \circ f : [0,1] \to \mathbb{K}$ is continuous in $x$.
Of course, strong continuity implies weak continuity, and in finite dimension weak continuity implies strong continuity too (it suffices to write $f =: (f_1, f_2, \dots, f_d)$ and to remember that $f$ is continuous iff the $f_k$s are), but there are functions that are weakly continuous everywhere but not strongly continuous in one point: this post for example features such a function (they chose to make the function go from a domain of $\mathbb{C}$ there but it's very much adaptable to an interval of $\mathbb{R}$). and I feel like it should not be hard to have countably many discontinuities by summing countably many functions that are each weakly continuous everywhere and strongly continuous everywhere except in one point each.
On the other hand, the existence of nowhere (strongly) continuous functions is guaranteed, for example by picking any non-linear solution of the Cauchy functional equation $g(x+y) = g(x) + g(y)$ and a vector $x$ in $X$, and then taking $f : t \in [0,1] \mapsto g(t)x \in X$, and you can probably take Hamel bases of $\mathbb{R}$ and $X$ and make an example with an infinite-dimensional range $f : \sum_i t_i e_i\mapsto \sum_i g(t_i)e'_i$, but it's hard to see whether those highly irregular functions can make place for a weakly continuous function...
As such this got me wondering how "badly" can weak continuous functions behave in regards to strong continuity, and as such here is my question:
Question: Does there exist $X$ an infinite-dimensional Banach space such that there exists an everywhere weakly continuous yet nowhere strongly continuous function $f : [0,1] \to X$?
Feel free to edit or re-tag if needed.
This is not what you are looking for, but here is an example of a weak-star continuous function that is nowhere strongly continuous:
Let $X= M([0,1])$ be the space of regular Borel measures, which is the dual of $C([0,1])$. Define $f:[0,1]\to M([0,1])$ by $f(t) = \delta_t$, i.e., $f(t)( \phi) = \phi(t)$ for $\phi\in C([0,1])$. Clearly $t\mapsto f(t) (\phi)$ is continuous for $\phi\in C([0,1])$. However, $\|f(s)-f(t)\|_M = 2$ for $s\ne t$.