Let $X$ be a normed space. Suppose $E$ is a subset of $ X^*$ (The space of continuous linear functionals). For every $\phi\in E$, define seminorm $p_\phi: X\to [0,\infty)$ such that $p_\phi (x)= |\phi(x)|$. If $\tau$ is the topology generated by these seminorms and $(X,\tau)$ is a locally convex space then $(X,\tau)^* = {\rm span}(E)$.
I do not have any idea about it. Please help me. Thank so much.
Here is an answer to the question in a more general setting. Note that the topology generated by the collection of seminorms $\{p_{f} : f\in E\}$ as in the question coincides with the topology generated by the family of linear functionals $E$. This is because the collection \begin{equation} \{ \{x\in X : |f(x - x_{0})| < \varepsilon \; \text{for all} \; f\in F \} : x_{0} \in X, F\subseteq E \; \text{finite} \; \text{and} \; \varepsilon > 0 \} \end{equation} is a subbase for both topologies.
Proposition. Let $X$ be a vector space (over a real or complex field $\mathbb{K}$) and let $X^{\#}$ be the algebraic dual of $X$. Suppose $E \subseteq X^{\#}$ is nonempty and $\tau$ is the topology on $X$ generated by $E$, that is, the coarsest topology on $X$ making each member of $E$ continuous. Let $f\in X^{\#}$. Then the following are equivalent.
The linear functional $f$ belongs to ${\rm span} \, E$.
The linear functional $f$ is $\tau$-continuous.
The linear functional $f$ is bounded as a function on some $\tau$-open neighbourhood of $0$.
Proof. Suppose $f \in {\rm span} \, E$. Then there are $\{c_{1}, \ldots , c_{d} \} \subseteq \mathbb{K}^{d}$ and $\{f_{1}, \ldots , f_{d} \} \subseteq E$ such that $f = \sum_{k=1}^{d} c_{k} f_{k}$. Since $f$ is the composition of the $\tau$-continuous map $x\mapsto (f_{k}(x))_{k=1}^{d}$ from $X$ into $\mathbb{K}^{d}$ and the continuous map $(\xi_{k})_{k=1}^{d} \mapsto \sum_{k=1}^{d} c_{k} \xi_{k}$ from $\mathbb{K}^{d}$ into $\mathbb{K}$, it follows that $f$ is also $\tau$-continuous.
Next, suppose $f$ is $\tau$-continuous. Then $\{x\in X : |f(x)| < 1 \}$ is a $\tau$-open neighbourhood of $0$ and $f$ is bounded on this neighbourhood.
Finally, suppose $f$ is bounded on some $\tau$-open neighbourhood $U$ of $0$. Then there is some finite $F\subseteq E$ and some $\varepsilon > 0$ such that \begin{equation} \bigcap_{g \in F} \{x\in X : |g(x)| < \varepsilon \} \subseteq U. \tag{1} \end{equation} It is observed from ($1$) that $f$ is bounded as a function on the vector subspace $\bigcap_{g \in F} {\rm ker} \; g$. It follows from a scaling argument that $\bigcap_{g \in F} {\rm ker} \; g \subseteq {\rm ker} \, f$. Then by this result we have $f \in {\rm span} \, F \subseteq {\rm span} \, E$ as desired. This completes the proof.
Corollary. Let $X$ be a vector space (over a real or complex field $\mathbb{K}$) and let $X^{\#}$ be the algebraic dual of $X$. Suppose $E \subseteq X^{\#}$ is nonempty. Let $\tau$ be the topology on $X$ generated by $E$ and let $\tilde{\tau}$ be the topology on $X$ generated by ${\rm span} \, E$. Then $\tau = \tilde{\tau}$ and $(X, \tau )^{*} = {\rm span} \, E$ as sets.
Proof. As $\tilde{\tau}$ is a topology on $X$ making each member of $E$ continuous it follows that $\tau \subseteq \tilde{\tau}$. Conversely, by the previous proposition $\tau$ is a topology which makes each member of ${\rm span} \, E$ continuous and so $\tilde{\tau} \subseteq \tau$. Furthermore, by the proposition we also have that the $\tau$-continuous linear functionals on $X$ are precisely the linear functionals on $X$ which belong to ${\rm span} \, E$. This completes the proof.