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

by ayushkhaitan3437

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.