# Further questions about showing that a hyperplane is parallel to the null space of a linear functional

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$(X/\mathcal{N}(f))=1$)

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 $H_1 = x_1 +N(f)$.
• Okay, so the part with $f(x)^{-1}$ is like saying we have some scalar $\lambda$ applied to $u$, and $\lambda$ exists since $f(u) \not = 0$. So $x_1$ is just some arbitrary element of $X$ but we named it nicely. – llll Feb 21 '18 at 14:10
• And then the proof proceeds to show equivalence of $H_1$ and $x_1 + \mathcal{N}(f)$ by showing inclusion both ways. So then the next part should be: Let $x \in x_1 + \mathcal{N}(f)$. Then $x = x_1 + v$, for some $v \in \mathcal{N}(f)$. We see $f(x)=f(x_1)+f(v)=1+0$. Thus $x \in H_1$. – llll Feb 21 '18 at 14:17
• Could you explain where (3) is used, even indirectly? Also, I think I'm confused about what codim$\mathcal{N}(f)=1$ means. If dim$(X/\mathcal{N}(f))=1$, then is there only one coset, $X$? Which means $X/\mathcal{N}(f) = X/\{0\} = \{X\}$? Then $\mathcal{N}(f) = \{0\}$? – llll Feb 21 '18 at 14:23
• Actually 1) implies 3). Codimension of N(f)=1 means that if you take any two elements $x+N(f)$ and $y+N(f)$ in $X/N(f)$ then one is a multiple of the other; $y+N(f)=c(x+N(f)$ means $y-cx \in N(f)$ or $f(y-cx)=0$. If $x+N(f) \neq 0$ then x is not in $N(f)$, so $f(x) \neq 0$. We can always choose c such that $f(y-cx) =0$: just take $c=\frac {f(y)} {f(x)}$. That is why 1) implies 3) and there is no need to use 3) explicitly. – Kavi Rama Murthy Feb 23 '18 at 0:30