Linear Analysis - Riesz's lemma

In Linear Analysis, during the chapter discussing finite-dimensional normed spaces, we proved Riesz's lemma:

Let $Y$ be a proper, closed subspace of a normed space $X$. Then $\forall \varepsilon > 0$, $\exists x \in X$, $\|x\|=1$, such that $d(x,Y) > 1 - \varepsilon$.

I find the proof very interesting.

In school, we learn Pythagorus' theorem in all of its (potentially tautological) majesty. From that moment onwards, the name Pythagorus is endowed with a reverent, mystical quality that we revere because as small, impressionable children we were taught to. In France, there's another name that is given equal attention that as far as I know is less widespread in the UK - Thales (of Miletus). Apparently, he used the properties of similar triangles to calculate the height of the pyramids (although I've also heard said that this is another instance of misplaced mathematical credit, and he had nothing to do with the pyramids). Anyway, besides Pythagorus, we are patiently force-fed the théorème de Thalès, i.e. that the ratios of the sides of similar triangles are equal (image courtesy of Wiki)

Going through the proof of Riesz's theorem, it strikes me as very similar to this basic theorem of Euclidean geometry.

Proof of Riesz's lemma:

• We assumed $Y$ proper, so pick $z \in X\backslash Y$. 
• We assumed $Y$ closed, so $d(z,Y)=\inf\{\|x-y\| \mid y \in Y\} > 0$. By definition of infimum, we can pick $y \in Y$ such that $d(z, y)(1 - \varepsilon) < d(z,Y)$.
• Set $x = \displaystyle \frac{z-y}{\|z-y\|}$. Thus $\|x\|=1$. This is the point that we are looking for.
• Write $\lambda =\displaystyle \frac{1}{\|z-y\|}$ for clarity. $d(x,Y)=d(\lambda (z-y), Y) = \inf\{ \| \lambda (z - y) - y^\prime\| \mid y^\prime \in Y\}$. 
• Using the fact that subspaces are closed under scalar multiplication, $d(x,Y) = \inf \{\lambda \| (z-y) - y^\prime \| \mid y^\prime \in Y \}$.
• Using the fact that subspaces are closed under vector addition, $d(x,Y) = \inf\{\lambda\| z - y^\prime \| \mid y^\prime \in Y\}$.
• We conclude that $d(x,Y)=\lambda d(z,Y) > 1 - \varepsilon$, by choice of $y$. $\square$

(Perceived) link with similar triangles

I'm interested in one specific step of the proof: the assertion that for $Y$ a closed vector subspace, $d(\lambda x, Y) = \lambda d(x,Y)$, for $\lambda \in \mathbb{R}$.

 

Look familiar? The above argument with the infimum holds for any subspace. But how analagous is it really? There are a few traps for our intuition here. We haven't assumed that $X$ is a Hilbert space, so we can't assume that the shortest distance from $x$ to $Y$ is along a perpendicular line. That's fine - similar triangles don't have to be right-angled. More subtly, for the shortest distance to be along a line, we're assuming that lines are geodesics in normed spaces. Apparently, they are, but apparently there can be other geodesics too. 

I was sort of hoping we might be able to argue that the similar triangles property holds in this case without using the analytic definition of $d(x,Y)$, but the potential non-uniqueness of geodesics messes that up. There are other complications - recall that there may be multiple closest points in $Y$, since $X$ is not a Hilbert space, and we don't yet have any reason to believe that the closest points to $x$ and $\lambda x$ respectively are colinear in $Y$. Thus it seems that studying this question is out of my reach at the moment (my knowledge of geometry and geodesics is currently rather sketchy).

Normed spaces are sort of paradoxical - it feels to us like they have an extremely nice, rigid structure, yet things can apparently fail in mysterious ways. I certainly catch myself thinking of them intuitively as if they necessarily admitted an inner product (and thus had an orthonormal basis). This section is unfortunately rather inconclusive. 

Important corollary of Riesz's lemma

The unit ball of any infinite dimensional normed space is not compact.

Related problem: Example sheet 1Q12

Show that in the unit ball of every infinite-dimensional normed space, there is a sequence $(x_n)$ with $\|x_n - x_m \| \geq 1$, $\forall n \neq m$. Can $\geq$ be replaced with $>$?

In finite dimensions the unit ball is compact. If Riesz's lemma works for any $\varepsilon > 0$, then it has to work for $\varepsilon = 0$ by compactness. This approach might have been very interesting in infinite-dimensional normed spaces with compact unit balls. However, Riesz's lemma shoots itself in the foot by preventing any such objects from existing. Damn.

 Can $\geq$ be replaced with $>$?

Yes. Our supervisor came up with a proof that doesn't use Hahn-Banach. [more details to come]