Answer to Question #272576 in Functional Analysis for Prathibha Rose

Question

Let X be a normed linear space and Y a closed subspace of X 

with Y ≠ X ,if 0 < 𝑟 < 1 prove that there exist 

𝑋r element of X such that ||X|| = 1And 𝑟 < 𝑑 𝑋𝑟

, 𝑌 ≤ 1

Answer 1

First of all, let us remark that

$d(x_r, Y) = \inf_{y\in Y} d(x_r, y) \leq ||x_r||$, as $0\in Y$ and $d(0,x_r)=||x_r||$. Therefore, the right-hand side of the inequality is automatically satisfied if $||x_r||=1$.

Secondly, let us remark the following facts :

1. For a constant $\lambda\in \mathbb{R}$ and any $x\in X$ we have $d(\lambda x, Y)=|\lambda|d(x,Y)$
2. For a vector $y\in Y$ and any $x\in X$ we have $d(x-y, Y)= d(x,Y)$

We know that $x\mapsto d(x, Y)$ is not a function that is constantly zero (as $Y$ is closed). Therefore there exists $x_0$ such that $d(x_0, Y) = a>0$. By $\inf$ property for any $\epsilon>0$, there exists $y_\epsilon \in Y$ such that $a\leq||x_0-y_\epsilon||\leq a+\epsilon$. Now let us take the vector $x_\epsilon = \frac{x_0-y_\epsilon}{||x_0-y_\epsilon||}$, by using the properties 1 and 2 we have

$1-\frac{\epsilon}{a+\epsilon}\leq d(x_\epsilon, Y) \leq 1$

Let us fix $0<r<1$. Now by taking $\epsilon>0$ small enough, taking into the account that $a>0$ (and so $\lim_{\epsilon\to 0} \frac{\epsilon}{a+\epsilon} = 0$) , we have $1-\frac{\epsilon}{a+\epsilon}>r$ and thus $r<d(x_\varepsilon,Y)\leq 1$ with $||x_\epsilon||=1$