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This question may be quite related to the following link: but I am not sure. Sorry, if it is trivial.

https://math.stackexchange.com/questions/1277612/advantage-disadvantage-of-complete-incomplete-metric-space

In many application specially in engineering, we look for a function by minimizing or maximizing a cost function i.e. a given metric. The metric may mean square error (MSE) or sometime may be L1 norm. The questions are:

Is it mandatory that the space of function with the given metric must be complete to get a reasonable answer?

Is it possible that the space of function may be complete under L1 norm but not under MSE?

Any comments would be highly appreciated.

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Creator
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    If you're in a normed, finite dimensional vector space, it will be complete... see: http://math.stackexchange.com/questions/168275/proof-that-every-finite-dimensional-normed-vector-space-is-complete (so you may not have to worry). – TravisJ May 11 '15 at 21:19
  • Both $L_1(\mathbb{R}^n)$ and $L_2(\mathbb{R}^n)$ (MSE) are complete normed vector spaces. In many applications (e.g. optimization problems) we refine approximations of functions iteratively. It is nice to know that the limit of the iterates remains "in the class." – matt biesecker May 11 '15 at 21:23
  • @TravisJ Is it complete for atomic norm as well? – Creator May 11 '15 at 21:41
  • @Creator, I'm not familiar with the term atomic norm... my understanding (which is possibly wrong--hopefully someone will correct me if it is) is that all norms on $\mathbb{R}^{n}$ are equivalent in the sense that there exist $m$ and $M$ so that for any $x$ it follows that $m\Vert x\Vert \leq \Vert x\Vert^{\prime}\leq M\Vert x\Vert$. So it should be complete regardless of norm (assuming an atomic norm is a norm). – TravisJ May 11 '15 at 23:49
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    Are you talking about norms of functions? Or do you mean some finite dimensional space? If finite dimensional any norm will make it complete (which doesn't mean any metric, but I don't think metrics that don't arise from norms appear in engineering). If it's a function space, yes we must be careful that the space be complete, otherwise most theorems don't hold. For example an iterative proceedure that produces successive approximations must converge in the same space. – GPerez May 12 '15 at 00:00
  • @GPerez Yes, here I mean some finite dimension space not functions. I have just asked another question which is related to function. – Creator May 12 '15 at 00:02
  • Can you give an example or link to MSE? I only get statistics pages and none interpret it as a metric. – GPerez May 12 '15 at 00:10
  • i think you got this http://en.wikipedia.org/wiki/Mean_squared_error but i use this as a measure – Creator May 12 '15 at 00:11

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The completeness of the space under the given norm is essential in several aspects.

First, since you are talking about optimization: If the underlying space is not complete, there is the risk that the optimization problem has no solution in the given space. For example, if you consider an incomplete space of smooth functions, then it may happen that minimizers/maximizers are non-smooth.

Second, completeness means that every Cauchy sequence converges actually. That is, if you observe a sequence of iterates from a numerical method, such that their elements come arbitrarily close to each other, then you need completeness to be ensure that the limit exists.

Of course, both aspects are linked to each other.

daw
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