First of all, let us remark that
d(xrβ,Y)=infyβYβd(xrβ,y)β€β£β£xrββ£β£, as 0βY and d(0,xrβ)=β£β£xrββ£β£. Therefore, the right-hand side of the inequality is automatically satisfied if β£β£xrββ£β£=1.
Secondly, let us remark the following facts :
- For a constant Ξ»βR and any xβX we have d(Ξ»x,Y)=β£Ξ»β£d(x,Y)
- For a vector yβY and any xβX we have d(xβy,Y)=d(x,Y)
We know that xβ¦d(x,Y) is not a function that is constantly zero (as Y is closed). Therefore there exists x0β such that d(x0β,Y)=a>0. By inf property for any Ο΅>0, there exists yΟ΅ββY such that aβ€β£β£x0ββyΟ΅ββ£β£β€a+Ο΅. Now let us take the vector xΟ΅β=β£β£x0ββyΟ΅ββ£β£x0ββyΟ΅ββ, by using the properties 1 and 2 we have
1βa+ϡϡββ€d(xΟ΅β,Y)β€1
Let us fix 0<r<1. Now by taking Ο΅>0 small enough, taking into the account that a>0 (and so limΟ΅β0βa+ϡϡβ=0) , we have 1βa+ϡϡβ>r and thus r<d(xΞ΅β,Y)β€1 with β£β£xΟ΅ββ£β£=1
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