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?

  • $\begingroup$ 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

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 $||\Lambda||\le 1$).

  • $\begingroup$ 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

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.



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