I am proving the closest point property of a subset of a Hilbert space $H$: given $h\in H$ and a closed, nonempty and convex subset $M\subset H$, consider $$d=\inf_{m\in M} \|m-h\|$$ I am trying to show the existence of a sequence $m_{n}\subset M$ such that $\lim_{n\to\infty}\|m_{n}-h\| \to d$.

I'll have no problem doing it, as long as I can prove the existence of a finite infimum. I have read some papers where they say that the infimum exists and is finite, if $M$ is nonempty but they don't say why. Can anyone help me?

  • $\begingroup$ You have a non-empty set of nonnegative quantities ($\Vert\cdot\Vert$ is nonegative). $\endgroup$ – David Mitra May 16 '13 at 11:04
  • $\begingroup$ Since each term is nonnegative, and at least one term is finite (choose say $m_1$ as any element of $M$, so the distance $||m_1,h||$ is some finite real), and since you're taking an infimum, the infimum will exist. Basically you have a nonempty set of reals bounded below by $0$. $\endgroup$ – coffeemath May 16 '13 at 11:06
  • $\begingroup$ I'm sorry, I misread the question. $\endgroup$ – yohBS May 16 '13 at 11:14

As the commenters said, the set $\{\|m-h\|:m\in M\}$ has a lower bound, namely $0$. It is also nonempty, since $M$ is nonempty. Therefore, it has a finite infimum.

Just for completeness: the rest of the proof involves taking a minimizing sequence $m_n$ as in the question, then observing it has a weakly convergent subsequence, then arguing that the limit of this subsequence is in $M$ and is the closest point to $h$.

  • $\begingroup$ The parallelogram identity comes in handy for obtaining a Cauchy sequence from a minimizing sequence. $\endgroup$ – J. David Taylor Oct 25 '14 at 20:58

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