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Yes. Your argument is valid. In a nutshell, your argument is this: Take any $f \in X^*$. Since $X$ is finite dimensional, $f$ is bounded. So, $f \in X'$. So, $X^* \subseteq X'$, as desired. When $X$ is infinite dimensional, we can always construct an unbounded linear function $f$ on $X$. Finding an explicit construction of such an $X$ is tricky. ...


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Hint: A basis of neighbourhoods of $x_0\in X$ for the weak topology is obtained by varying $\epsilon$, $k$, and the $f_i$'s in $E^\ast$ in the following expression: $$V(f_1,\cdots,f_k,\epsilon)=\{x \in E:\forall i=1,\cdots,k:|f_i(x-x_0)|<\epsilon\}$$ (see Proposition 3.4 of Brezis' "Functional Analysis, Sobolev Spaces and Partial Differential Equations")...


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Such a topological vector space is called "locally bounded". Note that any subset of a bounded set is bounded, so actually it suffices for there to exist just a single bounded neighborhood of $0$.


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Use the following lemma in topology: Let $\mathcal{B}$ and $\mathcal{B}'$ be bases for the topologies $\tau$ and $\tau'$, respectively, on a set $X$. Then $\tau=\tau'$ if and only if the following this true For each $x\in X$ and each basis element $B\in\mathcal{B}$, such that $x\in B$, there is a basis element $B'\in\mathcal{B}'$ such that $x\in ...



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