If $f : [0,1] \to H$ is weakly continuous, is it bounded with respect to the norm of $H$ on $[0,1]$?

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Let $H$ be a separable Hilbert space and $f: [0,1] \to H$ be a mapping that is weakly continuous in the sense that \begin{equation} t \to \langle f(t), v \rangle_H \end{equation} is continuous in the ordinary sense for all $v \in H$.

Then, by the fact that any weakly convergent sequence is norm-bounded, I suspect that $\lVert f(t) \rVert$ is bounded for all $t \in [0,1]$.

However, since I have to deal with the continuum $[0,1]$, I do not see how to justify my guess, using the uniform boundedness principle.

Could anyone please help me?