Show that a linear continuous functional can be expressed as finite linear combination of a collection of functionals separating points. 
Let $\{l_\alpha\}$ be a collection of linear functions in a linear space $X$ over $\mathbb{R}$ that separates points. Put $\tau$ as the weakest topology in which all $l_\alpha$ are continuous (Such topology is locally convex). Show that a linear functional $l$ is continuous in $(X,\tau)$ precisely when it's a finite combination of the $\{l_\alpha\}$.

I have been puzzled about this question. I have been thinking if we can separate $\mathrm{span}\{l_\alpha\}$ and a continuous linear functional $l \notin \mathrm{span}\{l_\alpha\}$ by a "point" $x$, viewed as a linear functional, yet unfortunately failed. Can anyone hint on the solution? Any help will be greatly appreciated :-)
 A: Say $l$ is continuous. By the definition of $\tau$ there exist finitely many  $l_\alpha$'s, which I'm going to relabel $l_1,\dots,l_n$, such that $$|lx|
\le c\sum_{j=1}^n|l_jx|$$for all $x$. (See Details below.) In particular $$\bigcap_{j=1}^n\ker(l_j)\subset\ker(l).$$This implies that $l$ is a linear combination of $l_1,\dots,l_n$ by "just linear algebra".
(Say the scalar field is $F$. Let $$V=\{(l_1x,\dots,l_nx):x\in X\}\subset F^n.$$ The map from $V$ to $F$ defined by $(l_1x,\dots,l_nx)\mapsto lx$ is well-defined by the above; it extends to a linear functional on $F^n$.)
Details, added on request: The definition of $\tau$ shows that there exist $l_1,\dots,l_n$ and $\delta>0$ such that $$\max_{1\le j\le n}|l_jx|<\delta\implies|lx|<1.$$Define $$\rho(x)=\sum_j|l_jx|.$$Then we certainly have $$\rho(x)<\delta\implies |lx|<1.$$And this implies $$|lx|\le c\rho(x),$$by homogeneity. Suppose first that $\rho(x)=0$. Then $\rho(\lambda x)=0$ for every $\lambda>0$, so $$\lambda|lx|=|l(\lambda x)|<1$$for every $\lambda>0$, hence $lx=0$.
Now suppose $\rho(x)>0$. Let $$x'=\frac{\delta}{2\rho(x)}x.$$Then $\rho(x')<\delta$, so $$\frac{\delta
}{2\rho(x)}|lx|=|lx'|<1;$$hence $$|lx|\le\frac{2}{\delta}\rho(x).$$
