# Prove: For any sequence of linearly independent elements $y_j \in X$ and $a_j \in \mathbb R$ there exists an element $f \in X^*$ s.t. $f(y_j)=a_j$

I’m trying to solve the following problem but I have no clue how to do it.

Let $(X,||.||)$ be a normed $\mathbb C$-vector space. Prove: For any sequence of linearly independent elements $y_j, 1 \leq j \leq N$, in $X$ and any sequence $(a_j)_{1 \leq j \leq N}$ in $\mathbb R$ there exists an element $f \in X^*$ s.t. $f(y_j)=a_j$ for any $1 \leq j \leq N$.

The only thing I know is that I need at some point the Hahn-Banach Theorem. I would be grateful if someone could help me to prove this statement.
Thanks!

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I’m assuming you learned about finite-dimensional spaces along the way. If you have a linearly independent set of vectors $\{ y_1,y_2,\cdots,y_N \}$, then the linear space $Y_N$ spanned by these is a finite-dimensional space. So there are linear functionals $f_j$ on $Y_N$ such that $f_j(y_k)=\delta_{j,k}$. Using these you can find a functional $f$ on $Y_N$ such that $f(y_j)=a_j$ (in fact $f=\sum_{j=1}^{N}a_jf_j$ is such a functional.)

The space $Y_N$ is a finite-dimensional normed space, inheriting its norm from $X$. All norms on a finite-dimensional space are equivalent. So the following norm is equivalent to the the induced one:
$$\|x\|_{1}=\sum_{j=1}^{N}|f_j(x)|$$
Therefore, there is a constant $C$ such that $\|x\|_1 \le C\|x\|$ for all $x\in Y_N$, which forces the $f_j$ to be continuous with
$$|f_j(x)| \le \|x\|_{1} \le C\|x\|.$$
So $f$ is continuous on $Y_N$ with $|f(x)| \le (C\sum_{j=1}^{n}|a_j|)\|x\|$ for all $x\in Y_N$. By the Hahn-Banach theorem you can extend $f$ to all of $X$ in such a way that $\|f\|_{X^*} \le C\sum_{j=1}^{n}|a_j|$. Any such extension $\tilde{f}\in X^*$ satisfies $\tilde{f}(y_j)=a_j$ for $1 \le j \le N$.