Determine $R(A)^{\perp},R(A^*)^\perp$ of an operator $A$ related to a discontinuous linear functional

by weiyi sun   Last Updated October 10, 2019 05:20 AM

There is a functional analysis exercise:

Let $E$ be an infinite-dimensional Banach space. Fix an element $a \in E$, and a discontinuous linear functional $f: E \to \mathbb{R}$. Consider the operator $A: E \to E$ defined by$$ D(A)=E, \quad A x=x-f(x) a. $$ Determine $N(A), R(A), N(A^*), R(A^*),R(A)^{\perp},R(A^*)^\perp$, where $A^*$ is the adjoint of $A$.

I have determined $N(A), R(A),N(A^*), R(A^*)$ successfully:

$$N(A)=\{x\in E:x=f(x)a\}\subset\mathbb{R}a,$$ where $\mathbb{R}a=\{ka:k\in\mathbb{R}\}$. $$E\supset R(A)=\{x-f(x)a:x\in E\}\supset N(f)(\text{since } \forall x\in N(f),\ Ax=x).$$ But $\forall x=ka \in \mathbb{R}a$, $Ax=ka-f(a)ka=k(1-f(a))a=0$ iff $f(a)=1$.

Therefore if $f(a)=1$, $N(A)=\mathbb R a$ and $R(A) \subset N(f)$ therefore $R(A)=N(f)$.

If $f(a)\neq 1$, $N(A)=\{0\}$ and $R(A)=E$ since $$\forall x\in E, x=A\left(\dfrac{f(x)}{1-f(a)}a+x\right)\implies E \subset R(A).$$

To determine $N(A^*), R(A^*)$, we have to first computed the domain $D(A^*)$ of the adjoint of $A^*$ and $A^*$. $$\begin{aligned} D\left(A^{\star}\right)&=\left\{v \in E^{\star} ; \exists c \geq 0 \text { such that }|\langle v, A u\rangle| \leq c\|u\| \quad \forall u \in D(A)\right\}\\ &=\left\{v \in E^{\star} ; \exists c \geq 0 \text { such that }|\langle v, u\rangle-f(u)\langle v,a\rangle| \leq c\|u\| \quad \forall u \in E\right\}\\ &=\{v\in E^*:\langle v, a\rangle=0\}. \end{aligned}$$ The last "=" is because if $\langle v, a\rangle\neq 0$, since $f(x)$ is discontinuous, $|\langle v, u\rangle-f(u)\langle v,a\rangle|$ can not be bounded. Therefore $\forall u\in E, v\in D(A^*)$, we have $$ \begin{aligned} &\langle A^*v, u\rangle=\langle v, Au\rangle\\ \implies&\langle A^*v, u\rangle=\langle v, u\rangle-f(u)\langle v,a\rangle\\ \implies&\langle A^*v, u\rangle=\langle v, u\rangle\\ \implies&A^*v=v \end{aligned} $$ Therefore it's clear that $N(A^*)=\{0\}$ and $R(A^*)=D(A^*)$.

However, when it comes to $R(A)^{\perp},R(A^*)^\perp$, I encountered some trouble:

For $R(A)^\perp$, I know that from the orthogonality $R(A)^\perp=N(A^*)=\{0\}$ but when I compute $R(A)^\perp$ directly, if $f(a)=1$, $R(A)^\perp=N(f)^\perp$. How to verify that $N(f)^\perp=\{0\}$?

For $R(A^*)^\perp$, $R(A^*)^\perp=D(A^*)^\perp \supset \mathbb R a$ clearly, but how to verify that $D(A^*)^\perp \subset \mathbb R a$?

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