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Let $E$ be a normed vector space and $F$ a vector subspace of $E$. If $y \in F$, $x \in E$ and $0 < a \in \mathbb R$, prove that $d(y + ax, F) = ad(x,F)$.

I've tried to write down the definitions explicitly, but I don't see how to continue from $$d(y+ax,F) = \text{inf}\{|y+ax-t| : t \in F\}$$ and $$ad(x,F) = \text{inf}\{a|x-r|: r \in F\}$$ to $d(y+ax,F) = ad(x,F)$. I tried to get from $$|y + ax - t|^2 = (a|x-r|)^2$$ to a nice result, but this didn't prove useful. I also tried to use the triangle inequality, but the two different metrics confused me. Any hints? (this is not a homework question, but I would still prefer hints to complete answers)

hampster
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1 Answers1

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Only definitions and the fact that $F$ is a linear subspace are needed here. No need for any inequality. If $r \in F$ take $t=y-ar$. Then $y \in F$ and $a|x-r| =|ax-ar|=|y+ax -t|$. On the other hand if $t \in F$ then $r=\frac 1 a (t-y) \in F$ and $|y+ax -t|=a|x-r|$. Thus the two infimums are infimums of the same set of real numbers!.

Kavi Rama Murthy
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  • Just to make sure I understood; the substitutions $t=y-ar$ and $r = (1/a)(t-y)$ are allowed because $F$ is linear i.e since $y \in F$, if $r \in F$ then $y-ar = t \in F$ etc. – hampster Jun 11 '19 at 12:00
  • Exactly. Linearity of $F$ is the only basic information you need. – Kavi Rama Murthy Jun 11 '19 at 12:01