§ Uniform Boundedness Principle / Banach Steinhauss
- Consider a set of bounded linear operators . If is pointwise bounded, that is, exists for all , then the family is norm-bounded: exists.
§ Proof 1: Based on an ingenious inequality
- Reference: A really simple elementary proof of the uniform boundedness theorem by Alan D Sokal
§ Ingenious Inequality, Version 1
- Let be a bounded linear operator. Then for any we have .
- Proof: recall that .
- Now see that .
- This can be rewritten as , but this .
§ Ingenious Inequality, Version 2
- Let be a bounded linear operator, let be any basepoint. Then for any we have .
- We rewrite the optimization problem as .
- First, consider: .
- The last inequality follows from , that is, triangle inequality.
- Now we see that:
- and thus we get the bound that .
§ Proof of theorem
- Suppose for contradiction that , which it is indeed pointwise bounded (for all is bounded).
- Then choose a sequence such that . This is possible since the set is unbounded.
- Next, create a sequence of points, such that , and (that is, is a radius ball around .
- See that this sequence is cauchy, and thus converges. In particular, let the limit be . Then we can show that .
- Also see that we have the bound .
- Thus, .
- But this contradicts the pointwise boundedness of at the point . Hence proved.
§ Proof 2 using Baire category
- Suppose that for every , .
- We want to show that .
- For every integer , we build the subset .
- Since for every , there is some such that (by assumption, is pointwise bounded), we know that the sets cover .
- Furthermore, each is closed: A cauchy sequence of points such that will converge to a limit such that .
- Thus, by the baire category theorem, there is a ball for some , .
- Now this means that the set is norm bounded as .
- But this is a linear space, once we trap one ball we trap them all. By rescaling and translation, we can move the norm boundedness of into the norm boundedness of at which point we have proven that .
- Now let and . Calculate:
- This bound of does not in any way depend on or , then , which establishes the bound.