# 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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