§ Reisez Lemma

• Let $M$ be a closed proper subspace of a normed linear space $X$. Then for all $0 < \alpha < 1$, there exists a $p \in X$ (dependent on $\alpha$), such that $d(M, p) \geq \alpha$. That is, $\forall m \in M, d(m, p) \geq \alpha$.
• This is easy to establish for say $\mathbb R^2$. pick a unit orthogonal vector, it will be at least $1$ unit apart (or more) by pythogoras.
• This lemma provides a convenient substitute for orthogonality.

§ Proof via Hahn Banach

• Hahn banach is also a substitute for orthogonality.
• Pick a point $z \not in M$. Thus, $d(z, M) > 0$. Note that $d(-, M)$ is:
• (a) a sublinear function on $X$.
• (b) vanishes on $M$.
• (c) equals the projection onto $\mathbb R z \simeq \mathbb R$ on $M + \mathbb Rz$.
• By Hahn Banach, the portion of it that is linear extends to a linear functional on all of $M + \mathbb Rz$, and is dominated above by $d(-, M)$.
• Now, normalize the bounded linear functional so obtained to get a functional $f$ such that $|f| = 1$. Note that noramalization does not change the fact that $f(M) = 0$.
• Next, we build an "approximate normer" $z'$. This is an element $z'$ of unit norm such that $|f(z')| \sim |f|$. Such an element exists by definition of norm: $||f|| = \sup_{||x|| = 1} |f(x)|$. See that since $|z'| = 1$, we must surely have that $|f(z)| \leq ||f||$, thus $|f(z') \leq 1$. We claim that $|f(z')| = 1 - \epsilon$. (This is clear by clear and distinct perception since $f$ "behaves differently" along $Y$). This must happen, for if not, then $f(z') = 1$ for all $|z'| = 1$. This is patently untrue since $f(Y) = 0$, thus the unit vector along $Y$ must vanish at the very least.
• Now, consider $f(z' - m) = f(z') - f(m) = (1 - \epsilon) - 0 = 1 - \epsilon$.
• Next, estimate $|f(z' - m)| = |1 - \epsilon| = 1 - \epsilon$.
• This gives $1 - \epsilon = |f(z' - m)| \geq |f| |z' - m| = |z' - m|$.
• The first inequality follows from the defn of norm $|f| = \sup_k |f(k)|/|k|$, and thus $|f| < |f(k)|/|k|$, or $|f(k)| > |f||k|$.
• The second inequality follows from the fact that $|f| = 1$.
• Reference

§ What about $\alpha > 1$?

• $\alpha > 1$ does not even hold in $\mathbb R^2$. If I pick $\alpha = 5$, there is no unit vector that is $5$ units away from the $x$-axis. A vector is at most $1$ unit away from the $x$ axis.

§ What about $\alpha = 1$?

• Apparently, this case holds for any reflexive space (double dual equals original space).
• To show counterexample, we need a non-reflexive space. consider $l_\infty$, space of sequences under max norm.
• Alternatively, pick $C[0, 1]$ under max norm.
• We begin by picking a subspace $X \equiv \{ f \in C[0, 1] : f(0) = 0 \}$. So $f$ is continuous and $f(0) = 0$.
• Let $M$ be the subspace of $X$ such that $\int_0^1 f(x) dx = 0$.
• We want to show that there exists no function $p \in X$ such that (a) $||p|| = 1$, (that is, $\sup_{x \in [0, 1]} p(x) = 1$ and (b) $d(p, m) \geq 1$ for all $m \in M$.

§ Pedestrian proof when $\alpha = 1$.

• If $d(p, m) \geq 1$, then we must have $d(p, 0) \geq 1$. This means that $\int_0^1 (p(x) - 0) \geq 1$, or that $\int_0^1 p(x) \geq 1$.
• Intuitively, since $p \in X$, we know that $p(0) = 0$, and since $p$ is continuous, it must "spend some time" around $0$, thereby losing some of the integral. Furthermore, since we know that $||p|| = 1$, the maximum integral any such function can attain is $1$.
• Since $p$ is continuous and $p(0) = 0$ (as $p \in X$), pick $\epsilon = 0.5$. Then there exists a $\delta$such that for all $0 \leq x < \delta$, we have that $p(x) < \epsilon = 0.5$. Thus, we can upper bound the integral of $p(x)$ over $[0, 1]$ by $\delta \times 0.5 + (1 - \delta) \times 1$. Ie, we surround $p$ by two rectangles, one of height $0.5$, one of height $1$, since we have the bounds. Since $\delta > 0$, we can see that $\int_0^1 p(x) dx < 1$ from the above estimate.
• Thus, this means that $d(p, m) < 1$, thereby violating the claim that we can find such a $p$ such that $d(0, p) \geq 1$. Hence proved!

§ Slightly more sophisticated proof when $\alpha = 1$.

• The same setup. We consider the integral operator $F: X \to \mathbb R$, defined as $F(f) \equiv \int_0^1 f(x) dx$.
• We note that $M = ker(f)$.
• We note that $||F|| \geq 1$. (In fact, $||F|| = 1$, since the function is maximized by being evaluated on $one(x) = 1$ which lies on the unit sphere).
• We note that $d(f, M) = |F(f)|/||F||$. That is, the distance of a point to the kernel of an operator $F$is the norm of $F(x)$ rescaled by the norm of $F$.
• We need an estimate on $|F(f)|$. By the above argument, we know that $|F(f)| < 1$ by continuity of $f$, $f(0) = 0$, and that $f(x) < 1$ for all $x$ (as $||f|| = 1$).
• Combine the two estimates to see tha $d(f, M) = (<1)/(>1)$, which is indeed less than $1$. Done.