## § 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 $A : X \to Y$ whose norm is bounded
*below *: That is, there exists a $k$such that for all $x$, $k||x|| \leq ||Ax||$. - Intuitively, this forces $A$ 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 $A$ to be bounded! We will still get a bounded inverse operator.

#### § Step 1: $A$ is one to one

- Suppose $At = 0$. We will show that this implies $t = 0$.
- $k ||t || \leq ||At||$. That is, $k ||t|| \leq 0$. Since $k > 0$, this implies $||t|| = 0$ or $t = 0$.

#### § Step 2: $A^{-1}$ is bounded

- Define $A^{-1}(y) \equiv x$ when $Ax = y$.
- Since $k ||x|| \leq ||Ax||$, we write $Ax = y$, and thus $x = A^{-1} y$.
- This gives us $k || A^{-1} y || \leq y$.
- This means that $||A^{-1} y|| \leq (1/k) y$, thereby establishing the boundedness of $A$.
- Thus, $A$ is a bounded linear operator.

#### § Claim: This is in fact sufficient: Every invertible operator $A$ with bounded inverse has such a lower bound $k$.

- Reverse the proof: take the bound of $A^{-1}$ to be $k$ and show that this lower bounds $A$.

- We can thus define $A^{-1} : Range(A) \to X$