0

I am reading a proof of Theorem 2.4.1 in "Nonlinear Programming" by Bazaraa, Sherali, and Shetty, and I am wondering what the reasoning is on a step of the proof. I am almost certain it is something very simple I am blanking on.

The Theorem Says:

Let $S$ be a nonempty, closed convex set in $R^n$ and $y \notin S$. Then there exists a unique point $\bar{x} \in S$ with minimum distance from $y$.

The part of the proof I am reading (uniqueness):

Suppose $\bar{x}$ is the minimizing point in $S$. To show uniqueness, suppose that there is an $\bar{x}' \in S$ such that $||y - \bar{x}|| = ||y - \bar{x}'|| = \gamma$. By the convexity of $S$, $(\bar{x} + \bar{x}')/2 \in S$. By the triangle inequality we get

\begin{align*} || y - \frac{\bar{x}+\bar{x}'}{2} || \leq \frac{1}{2}||y-\bar{x}|| + \frac{1}{2}||y-\bar{x}'|| = \gamma. \end{align*}

If strict inequality holds, we have a contradiction to $\bar{x}$ being the closest point to $y$. Therefore equality holds, and we must have $y-\bar{x} = \lambda (y-\bar{x}')$ for some $\lambda$.

My Issue:

I don't understand why we must have $y-\bar{x} = \lambda (y-\bar{x}')$ for some $\lambda$. Can someone please explain why this makes sense?

user316720
  • 57

1 Answers1

1

It helps to think about "When does strict inequality hold for the triangle inequality in a vector space $V$"?

In general, if I draw a triangle on a 2d plane, the triangle inequality is an equality only when one vertex is on the opposite side of the triangle. When we now place one vertex at the origin, this means that the other two vertices lie on the same line through the origin, i.e. one is a scalar multiple of the other.

student91
  • 2,913