3
$\begingroup$

Let $Z$ be a subspace of normed linear space $X$ and that $y$ is an element of $X$ whose distance from $Z$ is $d$. Then there exists a $\Lambda \in X^* $ (the dual space of $X$) so that $\| \Lambda\| \leq 1$, $\Lambda(y) = d$ and $\Lambda(z) = 0$ for all $z \in Z$.

This corollary is left as an exercise to the reader in chapter 3 of Reed and Simon's Functional Analysis, Corollary 3, page 77. I assumed that this corollary would be short and have relatively simple ideas as the previous 2 corollaries which were proved, but after multiple fruitless attempts I tried to find some solutions online to similar problems.

First I found some variations of the problem, for instance if $Z$ is closed, then one can define a linear functional $\lambda$ from $\operatorname{span}\{ y, Z \}$ to $\mathbb{R}$ by noting that each $x \in \operatorname{span}\{ y, Z \}$ can be written as a unique linear combination $x = \alpha y + z$ where $\alpha \in \mathbb{R}$ and $z \in Z$. Now define $\lambda$ by $$ \lambda (\alpha y + z) = \alpha$$ This is a linear functional and the kernel of $\lambda$ is the closed subspace $Z$ and hence $\lambda$ is also continuous. The solution then refers to some theorem in which it was proved that $$\| \lambda \|_{\operatorname{span}\{ y, Z \}} = \frac{1}{d(0, \lambda^{-1} \{1 \})}$$ Since $\lambda^{-1} \{ 1 \} = y + Z$ it follows that $$ \| \lambda \|_{\operatorname{span}\{ y, Z \}} = \frac{1}{d(0, y + Z)} = \frac{1}{d(y,Z)} = \frac{1}{d}$$ Now we have $\lambda(y) = 1$, $\lambda(z) = 0$ for all $z \in Z$ and $\| \lambda \|_{\operatorname{span}\{ y, Z \}} = \frac{1}{d}$. By an earlier corollary we may now extend the functional $\lambda$ to a functional $\lambda'$ on the whole space $X$ and furthermore $\| \lambda'\|_X = \| \lambda \|_{\operatorname{span}\{ y, Z \}}$. Now if we define $$ \Lambda(x) = d \lambda' (x)$$ then $\Lambda$ has the desired properties.

This solution is fine as is, but the book I am using does not mention either of the two important steps (kernel is closed is equivalent to continuity, and the lemma for the relationship between the norm of $\lambda$ and preimage of $1$), so I am unsure if this was the method they had in mind. I am also unsure as to how to force a closed subspace. One method might be for instance to define the functional $\lambda$ from $\overline{\operatorname{span} \{ y, Z\}}$.

Does anyone have a completely alternate, perhaps simpler solution, or a way to dodge the fact that subspace $Z$ is not closed?

$\endgroup$
  • $\begingroup$ Is $X^*$ the dual space of $X$? $\endgroup$ – Xianjin Yang Jul 7 '16 at 13:59
  • $\begingroup$ Yes it is, I'll add that to the original question. $\endgroup$ – Kayle of the Creeks Jul 7 '16 at 13:59
  • $\begingroup$ Re "some theorem in which it was proved...": $\|\lambda\|_{span (y,Z)}$ $=\sup \{|a|/\|ay+z\| : z\in Z\land ay+z\ne 0\}$ $=\max (0, \sup \{1/\|y+z/a\| : z\in Z\land a\ne 0\})$ $=\sup \{1/\|y+z\|: z\in Z\}=$ $(\inf _{z\in Z}\|y+z\|)^{-1}=1/d.$ And $d(0,\lambda^{-1}\{1\})$ $=\inf \{\|ay+z\|: a=1\land z\in Z\})=$ $\inf \{\|y+z\|z\in Z\|)=d.$ $\endgroup$ – DanielWainfleet Jul 7 '16 at 16:20
  • $\begingroup$ Ran out of editing time. In my comment above, the last line should be $\inf\{\|y+z\| : z\in Z \}=d.$ $\endgroup$ – DanielWainfleet Jul 7 '16 at 16:29
2
$\begingroup$

Two answers. First, you can easily derive the general case from the case where $Z$ is closed. Second, and I think more important, $Z$ being closed is really irrelevant in the first place!

(i) You can derive the general case from the case where $Z$ is closed:

If $d=0$ you can just set $\Lambda=0$ and you're done.

Suppose $d>0$. Then $y\notin\overline Z$. And $d$ is also the distance from $y$ to $\overline Z$. Apply your argument to $y$ and $\overline Z$.

(ii) It really doesn't matter whether $Z$ is closed or not.

You start by defining $\lambda$ on the span of $Z$ and $y$ by $$\lambda(\alpha y+z)=\alpha d\quad(\alpha\in\Bbb R, z\in Z).$$

(You said $\alpha$, but it should be $\alpha d$ if you want $\lambda y=d$.)

As far as I can see, the only place you used the fact that $Z$ was closed was in showing that $\lambda$ is continuous. But $\lambda$ is clearly continuous, just from the definition. Saying $d$ is the distance from $y$ to $Z$ implies that $||y+z||\ge d$ for all $z\in Z$. Hence $||\alpha y+z||\ge|\alpha|d$. So $$|\lambda(\alpha y+z)|=|\alpha|d\le||\alpha y+z||.$$

In other words, $|\lambda x|\le||x||$ for every $x$ in the domain of $\lambda$. This says precisely that $\lambda $ is bounded, in fact $||\lambda||\le1$. Hence $\lambda$ is continuous (and in particular HB says $\lambda$ extends to $\Lambda$ with $||\Lambda||\le 1$).

$\endgroup$
  • $\begingroup$ Thank you! I had some problems proving the continuity of the mapping $\lambda$ so I assumed that there was something else going on, but your method makes it obvious why the mapping is continuous. $\endgroup$ – Kayle of the Creeks Jul 7 '16 at 14:45
1
$\begingroup$

Let $P(x)=\inf_{z\in Z}\|x-z\|$. It is easy to show that $P(ax)=aP(x),a>0$ and $P(x+y)\leq P(x)+P(y)$. Meanwhile, $P(z)=0, \forall z\in Z$ and $P(y)=d$. We can show that $P(x)\leq \|x\|$ by choosing $z=0$ in the definition.

Next, we define another function $L(x)$. $L(z)=0, \forall z\in Z$ and $L(y)=d$. Let $L$ be defined on the subspace $Y=\{by+z|z\in Z, b\in R\}$ and linear. It is easy to show that on $Y$, $L(x)\leq P(x)$. By the Hahn-Banah theorem, we can extend $L$ to the whole domain $X$ and $L(x)\leq P(x) \leq \|x\|, \forall x$. So, $\|L\|\leq 1$ and satisfy all other conditions.

Done!

$\endgroup$

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.