### 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.