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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?

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    $\begingroup$ "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?" Because a map is always continuous with respect to a topology. Bounded maps are continuous with respect to norm topology, and now we are defining another topology on our space, if we don't do it carefully it may be that bounded maps are not continuous in this new topology. $\endgroup$ – s.harp Jul 17 '17 at 12:02
  • $\begingroup$ @s.harp Ok, now I get that we are moving from the norm topology to the weak topology. But I'm not sure what the difference is between these topologies..what sets are discarded when we move to the weak topology? Because it seems to me that the sets in both topologies are exactly the same?! $\endgroup$ – eurocoder Jul 18 '17 at 18:18
  • $\begingroup$ As an example: Unless the space is finite dimensional the ball $\{ x\in V\mid \|x\|<1\}$ is not open in the weak topology, indeed the interior of this set is actually empty in the weak topology. Generalise this to see that every bounded set in the weak topology must have empty interior. So you are throwing away quite a few open sets. $\endgroup$ – s.harp Jul 18 '17 at 18:27
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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 functionals continuous.

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  • $\begingroup$ Ok I think I have slightly better understanding. But one thing I'm not sure about. We have 'moved' from the norm topology to this new weak topology. But what sets have we discarded in the process? What sets are in the norm topology that won't be in the weak topology? It seems like both of these topologies contain the same open sets to me..am I missing something? $\endgroup$ – eurocoder Jul 18 '17 at 18:16
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    $\begingroup$ You lose all bounded open sets $\endgroup$ – Daminark Jul 18 '17 at 19:11

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