I was trying to understand the proof of RRT from my lecture notes, but I have trouble in understanding the following step.
$$T : L^{q} \rightarrow (L^{p})'$$
For $1 \leq p < \infty$. I understand the proof for showing that $T$ is linear, bounded and is an isometric mapping between the two spaces. I have trouble understanding the part in which $R(T)=(L^{p})'$. The proof proceeds by showing that $R(T)$ is dense in $(L^{p})'$.
We use the following result:
Let X be a n.s. Let $G \subset X$ be a subspace such that $ \overline{G} \neq X$. Then $\exists f \in X', f \neq 0$ such that $$f(x)=0 \forall x \in G$$.
By contradiction, assume that $R(T)$ is not dense in $(L^{p})'$. Then by the result above, we get that $\exists f \in (L^{p})'', f \neq 0$ such that $$f(x)=0 \forall x \in R(T)$$
$$ f(Tu)=0, \forall u \in L^{q}$$
Then, we use the fact that $L^{p}$ is reflexive, thus there exists a canonical mapping from $L^{p}$ to $(L^{p})''$ given by $$C_{x}(f)=f(x)\hspace{2mm} f \in (L^{p})'$$
From this, the conclusion is that there exists $f \in L^{p}$ such that $Tu(f)=0 \forall u \in L^{q}$. I cannot understand this last conclusion. Can someone clarify?