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I will first give some context for my question to be as defined as possible.

Prove that a projection onto the set K: $P_K(v) = \text{argmin}_{w \in K}||v-w||$ exists and is unique. Without lose of generality assume that the vector $v$ is always $0$ (Because translation does not modify the convergence of K).

Assumptions of the proof:

  • Hilbert space.
  • K is convex subset.

The proof started with the parallelogram law and reached this formula (I skipped some steps but assume that this formula is correct. I can expand it if requested):

  • Let $w \in K, d = min||v- w||= min ||0-w|| = min ||w||$ $$u, w \in K, ||u-w||^2 \le 2||u||^2 + 2||w||^2-4d^2$$

Then the proof moved to showing that such projection $P_K(v)$ exists. In order to do so they defined the following Cauchy sequence of elements in K:

  • Let $\{V_n\} \subset K $ be a minimizing sequence.
  • $||V_n|| \to d$ when $n \to \infty$

At this point is where my question starts. Their next step is to show that the sequence indeed converges to $d$. But in a previous exercise I had to prove that something was a Cauchy sequence (CS) to begin with, how can we already asume that a definition of a sequence that we made up is a CS without proving it?

Thank you for reading my question.

Bill Dubuque
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1 Answers1

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They do prove it's a CS although it may not be obvious: they show it converges to $d$. So in particular it converges - which implies that it's a Cauchy sequence.

Edit: Here's a discussion with proof that Convergent $\implies$ Cauchy.

SV-97
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