Let $E$ be a topological vector space and $E^*$ its topological dual. Let $f_1, f_2 \in E^*$ such that $\{f_1, f_2\}$ is linearly independent. Clearly, $f_1 \neq 0 \neq f_2$. We define $$ F:E \to \mathbb R^2, x \mapsto (f_1(x), f_2(x)). $$
Clearly, $E$ is infinite-dimensional and $F$ linear continuous. Are there some other special properties of $F$? Is it injective or surjective?
Thank you so much for your elaboration!
The range of any linear map is a linear subspace. In this case, if the range is one-dimensional then the range is a straight line which makes $f_1$ and $f_2$ multiples of each other. Of course the range cannot be $\{0\}$ either. So the map is onto.
It need not be injective. For example, consider the first two projections on $\ell^{2}$. [ A linear map is injective if and only if its kernel is $\{0\}$. In this case $(0,0,1,0,...)$ is in the kernel so the map is not injective]. Also, the dimension of the co-domain of an injective linear map is atleast as much as the dimension of the domain. Hence, $F$ can never be injective.
I have figured a proof and posted it as below answer.
Let $E$ be a topological vector space and $E^*$ its topological dual. Let $f_1, \ldots, f_n \in E^*$ such that $\{f_1, \ldots, f_n\}$ is linearly independent. We define $$ F:E \to \mathbb R^n, x \mapsto (f_1(x), \ldots, f_n(x)). $$ Then $F$ is surjective.
Lemma Let $f, f_1, \ldots, f_n$ be linear functionals on a vector space $E$. If $\bigcap_{i=1}^n \ker f_i \subset \ker f$ then $f$ is linear dependent of $\{f_1, \ldots, f_n\}$.
By Lemma, there is $x_i \in (\bigcap_{j \neq i} \ker f_j) \setminus \ker f_i$. Then $F(x_i)$ is zero everywhere except at the $i$-th coordinate. Clearly, $\{F(x_1), \ldots, F(x_n)\}$ is a basis of $\mathbb R^n$, This completes the proof.