Let $V$ be a normed vector space, $D\subseteq V$ and $\{f_n\}_{n\in\mathbb{N}}$ a sequence in $V^*$ such that for all $x\in D$, the sequence $\{f_n(x)\}_{n\in\mathbb{N}}$ is bounded. Is it true that for all $y\in\overline{D}$, the sequence $\{f_n(y)\}_{n\in\mathbb{N}}$ is bounded?
I don't know if this is true or if there is a counterexample. If there is a counterexample with a non Banach space, does the result holds if $V$ is a Banach space?
I tried taking $y\in\overline{D}$, $\{x_n\}_{n\in\mathbb{N}}$ in D that converges to $y$, then for all $n,k\in\mathbb{N}$ $$|f_n(y)|\leq |f_n(x_k-y)|+|f_n(x_k)| .$$ I have no idea how to bound the term $|f_n(x_k)-y|$.
Also, I have that the result holds if $D$ has no empty interior, as there exists $x\in D$ and $\varepsilon>0$ such that $B(x,0)\subseteq D$, so $x+\varepsilon y/2|y|$ is in the ball, and therefore in $D$, so there exists $M>0$ such that for all $n$ $$\left|f_n\left(x+\frac{\varepsilon y}{2|y|}\right)\right|<M .$$ Using the triangle inequality, $$\frac{\epsilon}{2|y|}|f_n(y)|-|f_n(x)|\leq \left|\frac{\epsilon}{2|y|}|f_n(y)|-|f_n(x)|\right|\leq\left|f_n\left(x+\frac{\varepsilon y}{2|y|}\right)\right|<M .$$ Then, because $x\in D$, there exists $N>0$ such that $\{f_n(x)\}_{n\in\mathbb{N}}$ is bounded by $N$ and $$|f_n(y)|<\frac{2|y|}{\epsilon}(M+N) .$$
I don't know how to deal with the empty interior scenario.
