Problem 5

Exercise 10.5 (Injectivity and a coercive estimate)   Let $X$ and $Y$ be Banach spaces and $A: X\to Y$ be a bounded linear operator. Show that the following are equivalent:

1. $A$ is injective and the range of $A$ is closed
2. There exists a constant $M>0$ such that

   $$\Vert x\Vert \leq M\Vert Ax\Vert \quad \forall x\in X$$ (10.29)

Proof. If $A$ is injective and the range of $A$ is closed, consider the operator

$$A\vert _{A(X)} : A(X) \to A(X)$$ (10.30)

which is known to have a closed range. This means the domain is also closed, from which we can decide that the graph is closed, so that the operator and its inverse are continuous. Therefore, we take

$$M = \Vert(A\vert _{A(X)})^{-1}\Vert _{B(A(X))}$$ (10.31)

Now lets show $M$ is satisfactory. Let $x\in X$ be given, then $Ax\in A(X)$, and we can determine that

$$\Vert(A\vert _{A(X)})^{-1}(Ax)\Vert \leq M\Vert Ax\Vert$$ (10.32)

It remains to simply contract $(A\vert _{A(X)})^{-1}(Ax)=x$ to find the desired result.

Now suppose there exists such an $M>0$. To show that $A$ is injective, suppose $Ax=Ay$. By linearity we have the inequality

$$\Vert x-y\Vert \leq M\Vert Ax-Ay\Vert = 0$$ (10.33)

which implies $x=y$ by squeezing the difference down. We are left to show that $A(X)$ is closed. Suppose $Ax_n\to y$. By the estimate involving, we can determine that the sequence $\{x_n\}$ is Cauchy, from which we select its limit $x$. Then to show $y=Ax$, let us utilize the triangle inequality

$$\Vert Ax - y\Vert \leq \Vert Ax - Ax_n\Vert + \Vert Ax_n - y\Vert$$ (10.34)

The left summand vanishes because $A$ is bounded and the right summand vanishes by the assumption $Ax_n\to y$. Therefore, $A(X)$ is closed.

The above shows (a)$\iff$(b). $\qedsymbol$