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### Continuous linear operators are bounded.: decoding the proof, and how the mathematician chances upon it

Here we try to prove that a linear operator, if continuous, is bounded.

Continuity implies: for any $\epsilon>0, \|Tx-Tx_0\|<\epsilon$ for $\|x-x_0\|<\delta$

We want the following result: $\frac{\|Ty\|}{\|y\|}\leq c$, where $c$ is a constant, and $y$ is any vector in $X$.

What constants can be construed from $\epsilon$ and $\delta$, knowing that they are prone to change? As $T$ is a linear operator, $\frac{\epsilon}{\delta}$ is constant. We need to use this knowledge.

We want $\frac{\|Ty\|}{\|y\|}\leq \frac{\epsilon}{\delta}$, or $\delta\frac{\|Ty\|}{\|y\|}\leq {\epsilon}$.

We have $\|Tx-Tx_0\|=\|T(x-x_0)\|<\epsilon$.

Hence, $x-x_0=\delta.\frac{y}{\|y\|}$.

$\|T(\delta.\frac{y}{\|y\|})\|=\frac{\delta}{\|y\|}\|Ty\|$.

We have just deconstructed the proof given on pg.97of Kreyszig’s book on Functional Analysis. The substitution $x-x_0=\delta.\frac{y}{\|y\|}$ did not just occur by magic to him. It was the result of thorough analysis. And probaby such investigation.

But hey! Let’s investigate this. $\frac{\delta}{\epsilon}$ is also constant! Let us assume $\epsilon\frac{\|Ty\|}{\|y\|}\leq \delta$. Multiplying on both sides by $\frac{\epsilon}{\delta}$, we get $\frac{\epsilon^2}{\delta}\frac{\|Ty\|}{\|y\|}\leq \epsilon$. This shows $x-x_0=\frac{\epsilon^2}{\delta}\frac{y}{\|y\|}$. Does this substitution also prove boundedness?

We have to show $\|x-x_0\|<\delta$. $\frac{\epsilon^2}{\delta}<\delta$ only if $\epsilon<\delta$. Hence, this is conditionally true.

Similar investigations taking $(\frac{\epsilon}{\delta})^n$ to be constant can also be conducted.

### Linear operators mapping finite dimensional vector spaces are bounded,

Theorem: Every linear operator $T:V\to W$, where $V$ is finite dimensional, is bounded.

Proof $\frac{\|Tx\|}{\|x\|}=\frac{\|T(a_1e_1+a_1e_2+\dots+a_ne_n)\|}{\|a_1e_1+a_1e_2+\dots+a_ne_n\|}\leq \frac{\|T(a_1e_1+a_1e_2+\dots+a_ne_n)\|}{c(|a_1|+|a_2|+\dots+|a_n|)}\leq \frac{\|T(e_i)\|}{c}$

where $\|T(e_i)\|=\max\{\|T(e_1)\|,\|T(e_2)\|,\dots\}$.

What we learn from here is

$\|e_i\|(|a_1|+|a_2|+\dots+|a_n|)\geq\|a_1e_1+a_1e_2+\dots+a_ne_n\|\geq c(|a_1|+|a_2|+\dots+|a_n|)$

where

$\|e_i\|=\max\{\|e_1\|,\|e_2\|,\dots,\|e_n\|\}$.

Similarly,

$\|e_i\|(|a_1|+|a_2|+\dots+|a_n|)\geq\|a_1e_1+a_1e_2+\dots+a_ne_n\|\geq \|e_k\|(|a_1|+|a_2|+\dots+|a_n|)$

where

$\|e_k\|=\min\{\|e_1\|,\|e_2\|,\dots,\|e_n\|\}$

Another proof of the assertion is

$\frac{\|Tx\|}{\|x\|}=\frac{\|T(a_1e_1+a_1e_2+\dots+a_ne_n)\|}{\|a_1e_1+a_1e_2+\dots+a_ne_n\|}\leq \frac{\|T(a_1e_1+a_1e_2+\dots+a_ne_n)\|}{\|e_k\|(|a_1|+|a_2|+\dots+|a_n|)}\leq \frac{\|T(e_i)\|}{\|e_k\|}$

which is a constant.

Note: why does this not work in infinite dimensional spaces? Because the max and min of $\|e_r\|$ and $\|Te_r\|$ might not exist.

### 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.