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.

Published by ayushkhaitan3437

Hello! My name is Ayush Khaitan, and I'm a graduate student in Mathematics. I am always excited about talking to people about their research. Please please set up a meeting with me if you feel that I might have an interesting perspective to offer- https://calendly.com/ayushkhaitan/meeting-with-ayush View more posts

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