Riesz's Lemma for finite-dimensional spaces

Question

Let $Z$ be any linear normed space and $Y$ a closed subspace of $Z$, then by Riesz's Lemma there exist an element $z_\theta \in Z$ such that $||z_\theta||=1$ and $dist(z_\theta,Y)<\theta$ where $\theta\in (0,1)$, i.e we can get away from the closed subspace arbitrarily close to $1$ and still be on the unit ball.

When however, $Z$ is a finite dimensional linear normed space, one can find an element $z\in Z$ such that $\operatorname{dist}(z,Y)=1$ with $\|z\|=1$. Could someone give me an idea of what happens in finite dimensional case?

I'm not quite clear what you are asking about. Perhaps about applying this Riesz Lemma in finite dimensional normed linear spaces? — hardmath, Jun 05 '15 at 19:26
In the case where $Y$ is a finite dimensional subspace of the normed space $X$, one can find $x\in S(X)$ with $\text{dist},(x,Y)=1$. This can be found in Lemma 1.4.22 of Robert Megginson's An Introduction to Banach Space Theory. — David Mitra, Jun 05 '15 at 19:37
I find this a legitimate question, once the wording is improved. — Tomasz Kania, Jun 07 '15 at 13:55

David Mitra · Accepted Answer · 2015-06-05T20:37:04.100

10

Since the following result doesn't seem to be well-known, I'll post its statement and an outline of its proof, both taken from the Lemma mentioned in the link in my comment above:

Theorem:

Let $Y$ be a finite dimensional subspace of the normed vector space $X$. Then there is an element $x\in X$ of norm one with $\text{dist}\,(x,Y)=1$.

Outline of proof:

Let $z\in X\setminus Y$. Then, as $Y$ is closed, $t=\text{dist}\,(z, Y)\ne0$. One can show, using the fact that $t^{-1}Y=Y$, that
$$\text{dist}\,(t^{-1}z,Y)=1.$$ Let $(y_n)$ be a sequence in $Y$ with $$\Vert y_n-t^{-1}z\Vert\rightarrow 1\tag1$$ Then $(y_n)$ is bounded and as $Y$ is finite dimensional, there is a subsequence $(y_{n_k})$ of $(y_n)$ that converges to some $y\in Y$.

Let $x=y-t^{-1}z$.

One can use $(1)$ to show that $\Vert x\Vert =1$.

Using the fact that $y-Y=Y$, one can show that $\text{dist}\,(x, Y)=1$.

Thus, $x$ is as advertised.

edited Jun 05 '15 at 20:37

answered Jun 05 '15 at 20:02

David Mitra

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1

Why $t^{-1} Y = Y$? – Alex Pozo Jun 05 '15 at 23:48
$Y$ is a subspace; so, $t^{-1}Y={ t^{-1} y\mid y\in Y}$ is just $Y$ again. – David Mitra Jun 06 '15 at 06:40
Merci beaucoup! – Alex Pozo Apr 23 '18 at 03:17
why does "$(y_n)$ is bounded and as $Y$ is finite dimensional" implies "there is a subsequence $(y_{n_k})$ of $(y_n)$ that converges to some $y∈Y$"? – John Mars Jun 17 '21 at 00:37
@JohnMars Finite dimensional spaces have compact unit balls. – David Mitra Jun 17 '21 at 01:29
@DavidMitra, know this post is several years old already, but just wondering how you know there is a sequence yn in Y that gets to distance 1 within $t^(-1)z$ Is it possible that $t^(-1)z$ is so far away from Y that no element of Y is within 1 distance away? – Bill Jul 28 '21 at 03:04
@William The distance of $t^{-1}z$ to $Y$ is 1; so it follows from the definition, no? – David Mitra Jul 28 '21 at 15:56

Tomasz Kania · Answer 2 · 2015-06-05T20:37:38.790

This is a comment to David Mitra's answer. There is a beautiful result by Krein, Milman and Krasnoselski based on the Borsuk–Ulam theorem which strengthens the Riesz lemma in finite dimensions.

Theorem. Let $M$ and $N$ be finite-dimensional subspaces of a normed space. If $\dim M > \dim N$, then there is $x_0\in M$ such that ${\rm dist}(x_0, N)=\|x_0\|$.

I am happy to provide a proof on request. I am very curious whether there exists a proof of this statement which would avoid using the Borsuk–Ulam theorem.

Reference:

M. Kreĭn, M. Krasnoselski, D. Milman, On defect numbers of operators on Banach spaces and related geometric problems, Trudy Inst. Mat. Akad. Nauk Ukrain. SSR, 11 (1948), 97–112.

Riesz's Lemma for finite-dimensional spaces

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