I'm supposed to show that for a normed space $X$ then we have that for a functional $f\in X^{*}$ ($X^{*}$ is the algebraic dual space of $X$) we have that
$$f\in X' \Leftrightarrow\ N(f)\text{ is closed}.$$
I've shown the $\Rightarrow$ part of the statement above but I'm having trouble showing the $\Leftarrow$ part. In the exercise it is hinted to show that if $f$ is unbounded then $N(f)$ is not closed.
The hint states that I should consider a sequence $\{x_k\}_{k=1}^{\infty}\subset X$ such that $||x_k||=1$ and $|f(x_k)|\geq k$. Then I should let $x\in X\setminus N(f)$ which just means $f(x)\neq 0$ and then construct a sequence $\{y_k\}_{k=1}^{\infty}\subset N(f)$ and show that $y_k\rightarrow x$.
I think I see the big picture behind the strategy of the hint but I have no clue as to how I would construct the sequence $\{y_k\}_{k=1}^{\infty}$ in such a way. If anyone can give any hints as to how to construct this sequence that would be highly appreciated!
Take $y_k=x-a_kx_k$ where $a_k=\frac {f(x)} {f(x_k)}$. Then $f(y_k)=0$ so $y_k \in N(f)$. Also $|a_k| \leq \frac {|f(x)|} k \to 0$ so $\|y_k -x\|=|a_k|\|x_k\|=|a_k| \to 0$.