Let $X$ be an infinite dimensional, reflexive and separable real Banach space. Consider a function $f: X \to \mathbb{R}$, and assume $f$ is sequentially continuous with respect to the weak topology, that is, if $a_n \rightharpoonup a$ then $f(a_n) \to f(a)$.

What conditions are needed to show that $f$ is also continuous with respect to the weak topology?

Moreover, let $B_R = \{a \in X \mid \|a\|_X \leq R \}$ endowed with the weak topology, this is a Polish space, and after a metric is defined is a compact metric space. Convergence in the metric is weak convergence.

If $f$ is sequentially weak continuous on $X$, the restriction of $f$ to $B_1$ should be sequentially weak continuous as well, which is sequentially continuous with respect to the metric. In turn, this means that $f$ is continuous on the topology induced by the metric, which is the inherited weak topology on $B_1$. Hence, if $f$ is sequentially weak continuous, the restriction of $f$ to $B_1$ is weak-continuous.

Is the argument right? This is enough for my application, but I am afraid I may be overlooking something.

I guess there are functions that are weak continuous when restricted to closed balls of any radius, but only sequentially continuous in the whole space $X$. Can you think of some particular easy example of this?

Thanks