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Hi i have a little problem in understanding the proof of the following theorem

If $N$ is a finite-dimensional proper subspace of normed linear space $X$, there exists an element in the 1-sphere of $X$ whose distance from $N$ is 1.

The proof basically takes a point $z$ on $X$ but not in $N$ so there exists a sequence $(n_k)$ of points in $N$ such that $||z - n_k|| \rightarrow d(z,N)$ (why?) ssince $N$ is finite dimensional and $n_k$ is bounded (why?) there exists a subsequence $v_k$ of $n_k$ which converges to some $n \in N$. Hence

$||z - n|| = \lim_{k \rightarrow +\infty} ||z - v_k|| = d(z,N) = d(z-n,N)$ (why $d(z,N) = d(z - n,N)$?).

The remaining part it is easy to understand...

Martin Argerami
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user8469759
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  • When you say the "1-sphere" do you mean ${x\in X: ||x||=1 }$? This is properly called the unit sphere and generally a 1-sphere is just a circle. – PVAL-inactive Jul 28 '15 at 19:26
  • $N$ is locally compact. The intersection of some closed $\epsilon$-ball around a point $m_0 \in N$ with $N$ is a closed and bounded set in a Euclidean space, and compact. Other than that, what is the source for this???? Meanwhile, the closed unit ball in a normed vector space is compact if and only if the space is finite dimensional – Will Jagy Jul 28 '15 at 19:37
  • @PVAL it should be $\left{ x \in X : ||x|| = 1 \right}$ – user8469759 Jul 28 '15 at 19:39
  • @Will Jagy. Is taken from a book. – user8469759 Jul 28 '15 at 19:42
  • author and title, please – Will Jagy Jul 28 '15 at 19:47
  • https://books.google.it/books?id=wkBNXonCtPIC&printsec=frontcover&dq=unbounded+linear+operator&hl=en&sa=X&ved=0CCoQ6AEwAGoVChMI-6Hdtsz-xgIVZP5yCh3kFwNW#v=onepage&q=unbounded%20linear%20operator&f=false – user8469759 Jul 28 '15 at 19:50
  • evidently Lemma I.4.5 on page 14; in Seymour Goldberg, Unbounded Linear Operators, in case anyone wants to pursue this. – Will Jagy Jul 28 '15 at 20:03
  • Yeah but i still don't get some point in that proof, especially some statement about the distance. – user8469759 Jul 28 '15 at 20:05
  • The definition of $d(x,M)$ for a point $x$ and a subset $M$ is I.2.1 on page 6. You need that, and various notions on pages 6-14. – Will Jagy Jul 28 '15 at 20:09
  • Which notions? (i try to derive the equality for the distance) – user8469759 Jul 28 '15 at 20:13
  • for the equality... $ ||z - n|| = inf_{m \in N} ||z - n - m|| = d(z-n,N)$... i don't understand the existence and boundness of $n_k$. – user8469759 Jul 28 '15 at 20:51

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Since $z\not\in N$ and $N$ is closed, necessarily $\text{dist}(z,N)>0$ (if the distance is zero, it means that $z$ is in the closure of $N$).

By definition, $$ \text{dist}(z,N)=\inf\{\|z-n\|:\ n\in N\}. $$ So the numbers $\|z-n\|$ can be arbitrarily close to $\text{dist}(z,N)$, i.e. we are allowed to choose a sequence $\{n_k\}_k\subset N$ with $$ \lim_{k\to\infty}\|z-n_k\|=\text{dist}(z,N). $$ As the sequence of numbers $\{\|z-n_k\|\}_k$ is convergent, it is bounded; there exists $c>0$ with $\|z-n_k\|\leq c$ for all $k$. Then $$ \|n_k\|\leq\|z-n_k\|+\|z\|\leq c+\|z\|, $$ which shows that the sequence $\{n_k\}$ is bounded. Since $N$ is finite-dimensional, closed and bounded sets are compact. So the sequence $\{n_k\}_k$ admits a convergent subsequence. If $n=\lim n_k\in N$ (remember that $N$ is closed, since it is finite-dimensional), $$ \|z-n\|=\lim_k\|z-n_k\|=\text{dist}(z,N). $$ The first equality is simply the continuity of the norm: using the reverse triangle inequality, $$ \left|\,\|z-n\|-\|z-n_k\|\,\right|\leq\|(z-n)-(z-n_k)\|=\|n-n_k\|\to0. $$

Finally, I don't see why it is needed, but $\text{dist}(z,N)=\text{dist}(z-n,N)$ because $n\in N$ and $N$ is a subspace: $$ \text{dist}(z,N)=\inf\{\|z-r\|:\ r\in N\}=\inf\{\|z-n-r\|:\ r\in N\}=\text{dist}(z-n,N); $$ as $N$ is a subspace, the elements $n-r$, as $r$ goes through all of $N$, describe all of $N$.

Martin Argerami
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  • Thank you for your reply... it starts to get clear (especially that i should keep with me a book of real analysis with sequence properties while i'm study). First question you said that by defintion of $dist(\cdot,\cdot)$ we can choose the sequence $n_k$ with that properties what are the logical steps that allow you to say that? – user8469759 Jul 29 '15 at 08:47
  • It's just the very basic fact that in any set of real numbers with an infimum there is a sequence of elements of the set that converges to said infimum. – Martin Argerami Jul 29 '15 at 08:56
  • Is this the answer? http://math.stackexchange.com/questions/252756/existence-of-sequences-converging-to-sup-s-and-inf-s – user8469759 Jul 29 '15 at 09:01
  • Yes. Now, if you are not comfortable with sup, inf, and inequalities, you will have a hard time going on with functional analysis, since they are basic tools that are used a lot. – Martin Argerami Jul 29 '15 at 09:07
  • Slowly slowly... i will review all i need. – user8469759 Jul 29 '15 at 09:11