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kamil9876:
For continous one to one functions from reals to reals there are some simple ways of showing the inverse is continous. One is to do - what is graphically intuitive - to use IVT and monotonicity: Simply suppose . To show is continous at just take consider and . Supposing is monotonically increasing (it's either strict increasing of decreasing to be one to one, a consequence of IVT), now for all we have .
So you can just set

edit: have to divide by two

/0:
Going over a proof of Bessel's inequality, I'm confused about the following bit:
Let and let be a countable set of numbers

Suppose that (where the sum is taken over an arbitrary finite collection of s). How do we know that ?

humph:
Write

We are given that

By definition,

This is lesser than or equal to

which is clearly lesser than or equal to by the definition of the sequence .

/0:
Thanks humph!

With bounded linear operators, , and

In Shakarchi's proof that , I understand the direction .

But in proving , the proof method used is simply to show:

.

How does this imply that ?

humph:
By definition, . So if we show that there exists such that , it must be true that by the definition of infimum. Thus if , then .

So really it shouldn't be , but rather .

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