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Let $X$ be a normed space, and $d$ is the metric induced by the norm. Suppose $C$ is non-empty closed convex subset of $X$. Let $x_a$ be a point outside $C$, i.e., $x_a \in X \backslash C$. Define $$d(x_a,C) = \inf \left\{d(x_a,x_c) | x_c \in C\right\}.$$

Suppose $x_d$ is element of $C$ such that $d(x_a,C) = d(x_a,x_d)$.

Show that for any $\lambda \in (0,1)$, $d(\lambda x_a + (1-\lambda)x_d,C) = \lambda d(x_a,C)$. First of all, one direction is obvious: $$d(\lambda x_a + (1-\lambda)x_d,C) \leq \lambda d(x_a,C),$$ however I have a hard time proving the reverse part.

Muhammad Fuady
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  • I think the problem is not well formulated: probably $X$ is a normed spaced and $d$ is the metric induced by the norm, i.e. $d(x,y) = |x-y|$, isn't it? – Rigel Jul 09 '17 at 10:25
  • Yes, I have fixed the question. I hope the question now appear better. – Muhammad Fuady Jul 09 '17 at 10:31

1 Answers1

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Let $x_\lambda := \lambda x_a + (1-\lambda) x_d$, $\lambda\in (0,1)$. Let us fix $\lambda \in (0,1)$ and $\epsilon > 0$.

By definition of distance from $C$, there exists a point $y_\lambda \in C$ such that $$ d(x_\lambda, y_\lambda) \leq d(x_\lambda, C) + \epsilon. $$ Hence $$ d(x_a, C) \leq \|x_a - y_\lambda\| \leq \|x_a - x_\lambda\| + \|x_\lambda - y_\lambda\| \leq (1-\lambda) d(x_a, C) + d(x_\lambda, C) + \epsilon, $$ i.e. $$ d(x_\lambda, C) \geq \lambda d(x_a, C) - \epsilon. $$ Since this inequality holds for every $\epsilon > 0$, the conclusion follows.

Rigel
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