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This came up in a proof I was reading. Define $$\inf_{z \in K} \|x-z\| = d$$

Let $y_n\in K$ be a minimizing sequence How do we know that such a minimizing sequence exists? Here K is a closed convex subset of a Hilbert space H. Is this some axiom for from the reel numbers?

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  • $\begingroup$ Just remember what the infimum of a set of real numbers is. $\endgroup$ – Thomas Dec 17 '12 at 10:24
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As pointed out by Thomas, it follows from the definition of infimum.

A subset $A$ of the real numbers is said to be bounded below if there is $b \in \mathbb{R}$ such that $b \leq a$ for all $a \in A$. For such a set, the infimum of $A$ is the greatest lower bound for $A$ which is guaranteed to exist due to the completeness of $\mathbb{R}$.

As $\inf_{z\in K}\|x - z\| = d$ we know that $d \leq \|x - z\|$ for every $z \in K$. Now for each $n \in \mathbb{N}\setminus\{0\}$ let $y_n \in K$ be such that $d \leq \|x - y_n\| \leq d + \frac{1}{n}$. For each $n$, such a $y_n$ exists; if no such $y_n$ existed for a particular $n$, then we would have $d + \frac{1}{n} \leq \|x - z\|$ for all $z \in K$ which contradicts the fact that $\inf_{z\in K}\|x-z\|=d$.

So we have a sequence $(y_n)$ such that $d \leq \|x-y_n\| \leq d + \frac{1}{n}$. Taking limits we obtain $d \leq \displaystyle\lim_{n\to\infty}\|x-y_n\|\leq d$ and hence $\displaystyle\lim_{n\to\infty}\|x-y_n\| = d$.

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  • $\begingroup$ When defining inf$_{z \in K} \lVert x-z \rVert = d$, how do we know that such a $d$ exists? $\endgroup$ – A Slow Learner Mar 21 at 18:23
  • $\begingroup$ @ASlowLearner: The set $\{\|x - z\| \mid z \in K\}$ is bounded below by zero. $\endgroup$ – Michael Albanese Mar 21 at 18:24

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