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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|)




\|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|)



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.