What is the meaning of the Riesz's lemma? I am reading about the Riesz's lemma but I am struggling to understand the real meaning of it. I have read different proofs of the lemma and even though I understood the proofs I am still not sure what the lemma means or what are its consequences or why its important. Is there a simple explanation of a graphical representation of the lemma?
Thanks. 
 A: Consider first a Hilbert space with a closed subspace $Y$. Then there exists the orthogonal complement $Y^{\bot}$. In particular, one can find a unit vector $x$ such that $x\bot Y$, or equivalently, $|x-y|\ge 1$ for all $y\in Y$.
The situation for Banach spaces (and normed spaces in general) is more complicated. One cannot find a unit "normal" vector to a closed subspace $Y$, i.e. to get $|x-y|\ge 1$ for all $y\in Y$ is impossible. However, it is possible to find an "almost normal" unit vector (in fact, arbitrary close to "being normal") in the sense that $|x-y|>r$, $\forall y\in Y$, where $r<1$ can be chosen arbitrary close to one.
Riesz's lemma says that for any closed subspace $Y$ one can find "nearly perpendicular" vector to the subspace.
A: Many familiar normed spaces (for example $\mathbb R^n$) are Hilbert spaces. That means it makes sense to find a vector perpendicular to a given subspace (plane). In the case of $\mathbb R^n$ it's easy to see that if $V$ is a proper subspace then we can choose a vector $u$ perpendicular to it on the unit sphere, and have $d(u,V) \ge 1$. Reisz's Lemma says this process can be almost be generalised to normed spaces where there exists no notion of a perpendicular vector.
A: Let $X$ be a normed space and $Y \subseteq X$ be a closed proper subspace. Let us denote by $U = \{ x \in X \, | \, |x| = 1 \}$ the unit sphere of $X$. Given $x \in U$, what can we say about $d(x,Y)$? We obviously have
$$ 0 \leq d(x,Y) \leq d(x,0) = |x| = 1. $$
Riesz lemma tells us that we can choose $x \in U$ such that $d(x,Y)$ is arbitrary close to $1$. 
If $X$ is a Hilbert space, then we have a geometric construction that maximizes $d(x,Y)$ and gives us a vector $x \in U$ with $d(x,Y) = 1$. To see this, let $x \in U$ and decompose it as $x = y + y^{\perp}$ with $y \in Y$ and $y^{\perp} \in Y^{\perp}$. Then
$$ d(x,Y)^2 = \inf_{y' \in Y} d(x,y')^2 = \inf_{y' \in Y} |y + y^{\perp} - y'|^2 = \inf_{y' \in Y} \left( |y - y'|^2 + |y^{\perp}|^2 \right) = |y^{\perp}|^2 $$
and so we see that if we want to maximize the expression, we need to choose $x \in Y^{\perp}$ with $|x| = 1$.
However, if $X$ is merely a normed space, we don't have a notion of an orthogonal projection and the Pytaghoras theorem so we can't perform the geometric construction above. Even so, Riesz lemma tells us that we can choose $x \in U$ (in a less obvious and non-constructive way) such that $d(x,Y)$ will be arbitrary close to $1$ (and we can't neccesarily make $d(x,Y) = 1$).
In a seperable Hilbert space $X$, you know that you can choose a complete orthonormal system $(e_i)$ which is a collection of linearly independent vectors such that $\overline{\operatorname{span} \{ e_i \}} = X$ and $\left< e_i, e_j \right> = \delta_{ij}$. If $X$ is merely a normed space, then Riesz' lemma allows us to construct a collection $(e_i)$ of linearly independent vectors with $\overline{\operatorname{span} \{ e_i \}} = X$, $|e_i| = 1$ and $|e_i - e_j| > 1 - \varepsilon$ for $i \neq j$. This can serve as a rough substitute for an orthonormal system in a general normed space and is good enough to generalize some arguments that work in the Hilbert space context trivially to the more general setting of normed vector spaces.
For example, if $X$ is a Hilbert space that is not finite dimensional, the existence of an orthonormal sequence in $X$ immediately shows that the unit sphere of $X$ is not compact. But the existence of an "Riesz-orthonormal" sequence in $X$ does the same with precisely the same proof.
