by eurocoder
Last Updated July 17, 2017 10:20 AM

I know that a linear operator is bounded if and only if it is continuous. But what about a bounded linear functional? Is this just a special case of a linear operator, and hence it is also bounded if and only if it is continuous?

But if that is the case, it would seem to make the definition of the weak topology redundant:

The weak topology on $X$ is the weakest topology $U\subset2^X$ with respect to which every bounded linear functional $\Lambda:X\to \mathbb{R}$ is continuous.

If every bounded linear functional is continuous due to being bounded, then why would continuity even be required in the definition of the weak topology?

You know that bounded linear functions are continuous with respect to the norm topology. However, when we're on a topological space in general and we have some continuous functions, removing open sets may break continuity (e.g. if you reduce to the trivial topology, the only continuous functions will be the constant ones).

This is where the weak topology comes in. We're removing as many open sets as possible while still keeping the bounded linear operators continuous.

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