§ Coercive operator
- This is called as the lax milgram theorem, but in lawrence and narici, it's a fucking lemma (lol).
- Suppose there is an operator whose norm is bounded below : That is, there exists a such that for all , .
- Intuitively, this forces to "spread vectors" out, making it one-one.
- Intuitively, this makes the inverse bounded, because the inequality "flips direction" when we consider the inverse operator.
- See that we do not require to be bounded! We will still get a bounded inverse operator.
§ Step 1: is one to one
- Suppose . We will show that this implies .
- . That is, . Since , this implies or .
§ Step 2: is bounded
- Define when .
- Since , we write , and thus .
- This gives us .
- This means that , thereby establishing the boundedness of .
- Thus, is a bounded linear operator.
§ Claim: This is in fact sufficient: Every invertible operator with bounded inverse has such a lower bound .
- Reverse the proof: take the bound of to be and show that this lower bounds .
- We can thus define