Question #272576

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

with Y β‰  X ,if 0 < π‘Ÿ < 1 prove that there exist 

𝑋r element of X such that ||X|| = 1And π‘Ÿ < 𝑑 π‘‹π‘Ÿ

, π‘Œ ≀ 1


1
Expert's answer
2022-01-31T16:07:06-0500

First of all, let us remark that

d(xr,Y)=inf⁑y∈Yd(xr,y)β‰€βˆ£βˆ£xr∣∣d(x_r, Y) = \inf_{y\in Y} d(x_r, y) \leq ||x_r||, as 0∈Y0\in Y and d(0,xr)=∣∣xr∣∣d(0,x_r)=||x_r||. Therefore, the right-hand side of the inequality is automatically satisfied if ∣∣xr∣∣=1||x_r||=1.

Secondly, let us remark the following facts :

  1. For a constant λ∈R\lambda\in \mathbb{R} and any x∈Xx\in X we have d(λx,Y)=∣λ∣d(x,Y)d(\lambda x, Y)=|\lambda|d(x,Y)
  2. For a vector y∈Yy\in Y and any x∈Xx\in X we have d(xβˆ’y,Y)=d(x,Y)d(x-y, Y)= d(x,Y)


We know that x↦d(x,Y)x\mapsto d(x, Y) is not a function that is constantly zero (as YY is closed). Therefore there exists x0x_0 such that d(x0,Y)=a>0d(x_0, Y) = a>0. By inf⁑\inf property for any Ο΅>0\epsilon>0, there exists yϡ∈Yy_\epsilon \in Y such that aβ‰€βˆ£βˆ£x0βˆ’yΟ΅βˆ£βˆ£β‰€a+Ο΅a\leq||x_0-y_\epsilon||\leq a+\epsilon. Now let us take the vector xΟ΅=x0βˆ’yϡ∣∣x0βˆ’yϡ∣∣x_\epsilon = \frac{x_0-y_\epsilon}{||x_0-y_\epsilon||}, by using the properties 1 and 2 we have

1βˆ’Ο΅a+ϡ≀d(xΟ΅,Y)≀11-\frac{\epsilon}{a+\epsilon}\leq d(x_\epsilon, Y) \leq 1

Let us fix 0<r<10<r<1. Now by taking Ο΅>0\epsilon>0 small enough, taking into the account that a>0a>0 (and so lim⁑ϡ→0Ο΅a+Ο΅=0\lim_{\epsilon\to 0} \frac{\epsilon}{a+\epsilon} = 0) , we have 1βˆ’Ο΅a+Ο΅>r1-\frac{\epsilon}{a+\epsilon}>r and thus r<d(xΞ΅,Y)≀1r<d(x_\varepsilon,Y)\leq 1 with ∣∣xϡ∣∣=1||x_\epsilon||=1


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