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Let $(\phi_n)_n$ be a weak* convergent sequence in the dual of some normed space $X$ with (weak*-)limit $\phi$.

If $X$ is Banach then it follows from the uniform boundedness principle that $\sup_n \lVert \phi_n \rVert < \infty$ since $\sup_n |\phi_n (x)| < \infty$ for every $x \in X$.

But what if $X$ is not Banach? Does this still hold true?

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Generally, without completeness, you can't deduce that a weak$^\ast$ convergent sequence is bounded.

Let $X = c_{00}$ be the space of sequences with only finitely many nonzero terms, endowed with the $\lVert\,\cdot\,\rVert_\infty$-norm. Its completion is $c_0$, the space of sequences converging to $0$, and its dual therefore isometrically isomorphic to $\ell^1(\mathbb{N})$. The sequence given by $x_n = n\cdot e_n$ is convergent to $0$ in $\bigl(\ell^1(\mathbb{N}),\sigma(\ell^1(\mathbb{N}),c_{00})\bigr)$ - since $\langle x_n, y\rangle = n\cdot y_n = 0$ for all sufficiently large $n$ - and unbounded.

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