# Closed kernel implies continuous linear functional : Zorn's Lemma

I know that this question has been asked before, but I am thinking of a solution using Zorn's lemma. Stein and Shakarchi's volume 4 wants me to prove the above statement(exercise 35) using exercise 34:

What I've been trying to do: Let $l$ be a discontinuous linear functional having a closed kernel $S$. Let $v\notin S$ and $\|v\|=1$. By the above exercise, we can get a continuous linear functional $l_1$ such that $v+S\subseteq ker(l-l_1)$ Hence we have furnished a discontinuous linear functional having $v+S$ in its kernel. Now, I have been trying to use Zorn's lemma in some way to show that we can find a discontinuous linear functional having the whole space as its kernel which would be contradictory. I am unable to choose a proper partial order and am stuck here.

• What you get is a continuous linear functional $l_1$ with $Ker l\subset Ker l_1$. But for linear functionals, this implies that $\alpha l= l_1$ for some $\alpha$. But since $l_1(v) = 1\neq 0$, this implies $\alpha\neq 0$, so that $l= \frac{1}{\alpha} l_1$ is also continuous Jun 18, 2017 at 11:21

Lemma. Let $(X, ||\cdot||)$ be a normed space over $\mathbf{C}$ and let $\ell : X \rightarrow \mathbf{C}$ be a linear map. Then $\ell$ is continuous if and only if $\ker{\ell}$ is closed.
Proof. If $\ell$ is continous, then $\ker{\ell}$ is the preimage of the closed set $\{0\} \subset \mathbf{C}$ and thus closed.
Conversely, assume that $W:= \ker{(\ell)}$ is closed. If $W = X$, then $\ell = 0$ and there is nothing to show. Otherwise, pick $x_0 \in X$ with $x_0 \notin W$. Then, as $W$ is closed, $\delta := \inf{\{||x_0 -w||\, :\, w \in W\}}$ is $> 0$. Let $x \in X$ be a vector of norm $||x|| = 1$. Write $x = \lambda x_0 + w$ for some $w \in W$. If $\lambda \neq 0$, then $\lambda^{-1}x = x_0 - (-\lambda^{-1}w)$ and by defintion of $\delta$, we get $|\lambda^{-1}| = ||\lambda^{-1}x|| \geq \delta$ or $|\lambda| \leq \delta^{-1}$. This last equaltiy holds of course for all $\lambda \in \mathbf{C}$. We conclude that $$|\ell(x)| = |\ell(\lambda x_0)| = |\lambda| |\ell(x_0)| \leq \delta^{-1}|\ell(x_0)|\,,$$ i.e. $\ell$ is bounded.