Riesz Lemma from Hahn Banach Theorem

Question

A note by Jacob Manaker provides a proof of the Riesz lemma via Hahn Banach. I am unable to follow a key construction in this proof. I note the proof down below for reproducibility, along with a screenshot:

For $X$ a normed vector space, $Y$ a closed proper subspace and $\epsilon \in [0, 1]$, there exists $x \in X$ of unit norm such that for all $y \in Y$, $|x – y| \geq 1 – \epsilon$.

Proof:

Fix any point $z \not in Y$ such that $d(z, Y) \geq 0$. Note that $d(\cdot, Y)$ is:

• a sublinear function on $X$ that
• vanishes on $Y$ and
• equals "projection onto $\mathbb Rz \simeq \mathbb R$ on $Y + \mathbb Rz$.

By Hahn-Banach, the linear portion of $d(\cdot, Y)$ extends to a linear functional on all of $X$ that is also dominated by $d(\cdot, Y)$. Thus $Y$ is contained in the kernel of some nonzero functional $f$.

WLOG, we may rescale $f$ to have unit norm, and take some "approximate normer" $x \in X$ such that $|f(x)| \geq 1 – \epsilon$. But now, for any $y \in Y$,

$$
(1 – \epsilon) – 0 = |f(x) – f(y)| \leq |f| |x – y| = |x – y|
$$

My Question

• What is this notion of "approximate normer"? Does it just mean to pick some element $x_0 \in X$ such that $|f(x_0)| \geq 1 – \epsilon$? Why does such an element $x_0$ exist?
• Even given such an element, how does one find the estimate that $|f(x) – f(y)| \leq |f| |x – y|$? I guess this follows immediate from the definition of $|f| \equiv \sup_{x \in X} |f(x)|/|x|$?
• Where did we use the fact that $f$ was constructed from a linear functional that acts as a projector on the subspace $\mathbb Rz$? If we don't need this fact, then why state it? I'm guessing all we need is $d(\cdot, Y)$ is non-trivial.
• Does the last inequality $|f||x – y| = |x – y|$ hold because we normalized the linear functional such that $|f| = 1$?

Answer 1

1. "Approximate normer" of functional $f$ is element $x$ with unit norm s.t. $|f(x)| \approx \|f\|$. Such element exists by definition of norm: $\|f\| = \sup\limits_{x: \|x\| = 1} |f(x)|$.

Indeed, from definition of $\sup$: $$\forall \epsilon > 0 \exists y: \|y\| = 1 \wedge (|f(y)| > \sup\limits_{x: \|x\| = 1} |f(x)| - \epsilon)$$ $$\forall y: \|y\| = 1 \rightarrow |f(y)| \leq \sup\limits_{x: \|x\| = 1} |f(x)|$$ From the first part, substituting $\sup\limits_{x: \|x\| = 1} |f(x)| = 1$, we get $\exists y: \|y\| = 1 \wedge (|f(y)| > 1 - \epsilon)$ - such $y$ is "approximate normer".

2. Yes, it is from definition and linearity: $|f(x) - f(y)| = |f(x - y)| \leq \|f\| \cdot \|x - y\|$.
3. We need $d(\cdot, Y)$ to be not just non-trivial, but also to dominate some non-trivial linear functional.
4. Yes.