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

Continuity implies: for any for

We want the following result: , where is a constant, and is any vector in .

What constants can be construed from and , knowing that they are prone to change? As is a linear operator, is constant. We need to use this knowledge.

We want , or .

We have .

Hence, .

.

We have just deconstructed the proof given on pg.97of Kreyszig’s book on Functional Analysis. The substitution 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. is also constant! Let us assume . Multiplying on both sides by , we get . This shows . Does this substitution also prove boundedness?

We have to show . only if . Hence, this is conditionally true.

Similar investigations taking to be constant can also be conducted.

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