I am trying to understand this question: Show that this hyperplane is parallel to the null space of this linear functional. This is Exercise 2.8.12 from Kreyszig's Introductory Functional Analysis.
For one of the answers, the answerer misunderstood the question. The other by Pp.., I'm not able to follow past the second paragraph.
Specifically, I don't get why $f(f(x)^{-1}x) = f(x)^{-1}f(x)$. Actually, I don't think this even makes sense. $f(x)^{-1} \in \text{Im}(f)$, so we can't even apply $f$ to it. If Pp.. meant $f^{-1}(x)$, then I don't think we know $f^{-1}$ even exists. From my understanding, $f^{-1}$ exists iff $f(x)=0 \Rightarrow x=0$, which I don't believe follows from the assumptions.
I understand the goal: show that $H_1 \in X/\mathcal{N}(f)$, meaning $H_1 = x_1 + \mathcal{N}(f)$ for some $x_1 \in X$, but I don't know how to get there from the assumptions:
(1) $f\not = 0$ on $X$ is any linear functional.
(2) $\mathcal{N}(f)$ is a subspace of $X$
(3) codim$\mathcal{N}(f)=1$ (Equivalently dim$(X/\mathcal{N}(f))=1$)
Please help me understand what I'm missing!
First of all, your questions about $f(x)^{-1}$ are based on wrong interpretation of the symbol. It was supposed to mean $\frac 1 {f(x)}$ and not the inverse! Coming to the proof I will use (3) only indirectly, so no mention of it is made in this proof. Suppose $H_1=\{x:f(x)=1\}$. By (1) there exists u such that $f(u) \neq 0$. I will take $x_1 =\frac 1 {f(u)} u$. We have to show that $H_1=x_1+N(f)$. Take $x \in H_1$. Then $f(x-\frac {f(x)} {f(u)} u)=0$ by linearity of f. Let $v=x-\frac {f(x)} {f(u)} u$ so $v \in N(f)$. We have $x=\frac {f(x)} {f(u)} u+v =x_1 +v$ since $f(x)=1$. We have proved that $x \in x_1 +N(f)$. Thus $H_1 \subset x_1 +N(f)$. The reverse inclusion is much simpler and I will leave that to you. Hence $H_1 = x_1 +N(f)$.