Suppose that $T \in L(V,W)$. Prove that if Img$(T)$ is dense in $W$ then $T^*$ is one-to-one.
Answer
Suppose $T^*$ is not one-to-one. Then there exists $f_1,f_2 \in W^*$ with $f_1\neq f_2$ such that \[ T^*(f_1)=T^*(f_2). \] Thus \[ f_1(Tv)=f_2(Tv), \forall v\in V. \] and hence $(f_1-f_2)(Tv)=0$, $\forall v \in V$, i.e. \[f_1-f_2=0, \forall w\in Img (T).\] Since $f_1-f_2$ is continuous, it also vanishes on $\overline{Img(T)}=W$, and therefore $f_1-f_2=0$ on $W$, which contradicts the assumption $f_1 \neq f_2$. The proof is complete.

574
The answer is accepted.
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- answered
- 2265 views
- $18.00
Related Questions
- Finding a unique structure of the domain of a function that gives a unique intuitive average?
- Compute $$\oint_{|z-2|=2} \frac{\cos e^z}{z^2-4}dz$$
- What is the Lebesgue density of $A$ and $B$ which answers a previous question?
- $\textbf{I would like a proof in detail of the following question.}$
- Reflexive Banach Space and Duality
- Prove the uniqueness of a sequence using a norm inequality.
- A lower bound on infinite sum of exponential functions (corrected version)
- Analyzing the Domain and Range of the Function $f(x) = \frac{1}{1 - \sin x}$