2
$\begingroup$

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:

S&S vol 4-Page 43

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.

$\endgroup$
1
  • 2
    $\begingroup$ 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 $\endgroup$ Jun 18, 2017 at 11:21

1 Answer 1

8
$\begingroup$

Exercise 34 is true for all normed vector spaces and has a direct proof.

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.

$\endgroup$
1
  • $\begingroup$ I understand. However, I was thinking of trying to produce a proof using Zorn's lemma as an exercise. Is there any way out? $\endgroup$
    – Manan
    Aug 9, 2017 at 18:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.