Riesz’s lemma decoded

This is a rant on Riesz’s lemma.

Riesz’s lemma- Let there be a vector space $Z$ and a closed proper subspace $Y\subset Z$. Then $\forall y\in Y$, there exists a $z\in Z$ such that $|z-y|\geq \theta$, where $\theta\in (0,1)$, and $|z|=1$.

A proof is commonly available. What we will discuss here is the thought behind the proof.

For any random $z\in Z\setminus Y$ and $y\in Y$, write $\|z-y\|$. Let $a_{y\in Y}=\inf\|z-y\|$. Then $\|z-y\|\geq a$. Also, there exists a $y_0\in Y$ such that $\|z-y_0\|\leq\frac{a}{\theta}$. Then $\left\|\frac{z}{\|z-y_0\|}-\frac{y}{\|z-y_0\|}\right\|\geq\theta$. Because the vector space $Z$ is closed under scalar multiplication, we have effectively proved $\|z-y\|\geq\theta$ for any $\theta\in (0,1)$ and $y\in Y$.

If there is some other vector $v$ such that $\|v-v_0\|\leq\frac{a}{\theta}$, then $\|\frac{z}{\|v-v_0\|}-\frac{y}{\|v-v_0\|}\|\geq\theta$.

Hence, one part of Riesz’s lemma, that of exceeding $\theta$ is satisfied by every vector $z\in Z\setminus Y$. The thoughts to take away from this is dividing by $\theta$ or a number less than $1$ increases everything, even a small increase from the infimum exceeds terms of a sequence converging to the infimum, and every arbitrary term in the sequence is greater than the infimum. When we say $\theta$ can be any number in the interval $(0,1)$, we know we’re skirting with boundaries. We could aso have thought of a proof in this direction: let $b=\sup_{y\in Y} \|z-y\|$. Then $b\theta\leq\|z-y_0\|\leq b$. However, for an arbitrary $y\in Y$, $\left\|\frac{z}{\|z-y_0\|}-\frac{y}{\|z-y_0\|}\right\|\leq\frac{1}{\theta}$.

Hence, for every $\theta\in (0,1)$, $\theta\leq \|z-y\|\leq\frac{1}{\theta}$.

Now what about $\|z\|=1$? This condition is satisfied only when $z=z-y_0$ in the expression $\left\|\frac{z}{\|z-y_0\|}-\frac{y}{\|z-y_0\|}\right\|\geq\theta$.

Hence, over in all, for every vector $z\in Z-Y$, there are infinite vectors which satisfy the condition of Riesz’s lemma. Also, for every such $z$, there is AT LEAST one unit vector which satisfies Riesz’s lemma (there can be more than one). Hence, to think there can be only one unit vector in $Z-Y$ which satisfies Riesz’s lemma would be erroneous.

Graduate student