Show that this hyperplane is parallel to the null space of this linear functional.

Question

This is an exercise of my homework of "Functional Analysis".

If $Y$ is a subspace of a vector space $X$ and $\dim(X/Y)=1$, then every element of $X/Y$ is called a hyperplane parallel to Y. Show that for any linear functional $f\neq 0$ on X, the set $H_1=\{x\in X: f(x)=1\}$ is a hyperplane parallel to the null space $N(f)$ of $f$.

Answer 1

I didn't understand your attempt, so I propose the following:

Since $\;0\neq f\in X^*\;$ , we know $\;H:=\ker f\;$ is a hyperplane or maximal subspace of $\;X\;$ , or in other words: $\;\dim X/H=1\;$ , which means this last vector space is spanned (algebraically, not only topologically) by any non-zero vector $\;\overline v\in X/H\;$.

But clearly $\;\overline v:=f(x)+H\;,\;\;f(x)\neq 0\;$ fulfills the condition...

Answer 2

Since $f \neq 0$ on $X$, $H_1$ is not empty. Fix an $x_0 \in H_1$, and consider the coset $x_0 + N (f).$ Note that this is well-defined irrespective of elements in $H_1$. Indeed, for any $y \in H_1$, $y \neq x_0$, $y - x_0 \in N (f)$ since $f(y - x_0) = f(y)-f(x_0) =1 - 1 = 0;$ this shows that $x + N (f) = y + N (f)$ for any $x, y \in H_1.$

$1.$ For any $x \in x_0 + N (f)$, there exists an $y \in N (f)$ such that $x = x_0 + y.$ Since $f$ is linear, $f(x) = f(x_0 + y) = f(x_0) + f(y)=1 $ implies $ x \in H_1.$ This shows that $x_0 + N (f) \subset H_1.$

$2.$ For any $x \in H_1$, $x = x +0= x + x_0- x_0 = x_0 + (x - x_0) \in x_0 + N (f) $ since $f(x - x_0) = f(x)- f(x_0)=1 - 1 = 0$. This shows that $ H_1 \subset x_0 + N (f).$ Finally, combining $1.$ and $ 2.$ the inequality gives $H_1 = x_0+N (f)$ and so $H_1$ is one of the cosets in $X/N(f)$, i.e. it is a hyperplane parallel to $N(f)$.

Answer 3

According to the definition of "hyperplane parallel to $N(f)$", we just need to show that $H_1$ is an element of $X/N(f)$, i.e. it is of the form $H_1=x_1+N(f)$ for some vector $x_1\in X$.

First $H_1\neq\emptyset$. In fact, since $f\neq0$ there is $x\in X$ with $f(x)\neq0$. Therefore $f(x)^{-1}x$ satisfies $f(f(x)^{-1}x)=f(x)^{-1}f(x)=1$, i.e. $f(x)^{-1}x\in H_1$.

Let $x_1$ be any element of $H_1$, then we can show that $H_1=x_1+N(f)$. In fact, if $x\in H_1$ then $x=x_1+(x-x_1)$ where $f(x-x_1)=f(x)-f(x_1)=1-1=0$, i.e. $x-x_1\in N(f)$. Conversely, if $y=x_1+x\in x_{1}+N(f)$ then $f(y)=f(x_1+x)=f(x_1)+f(x)=1+0=1$, i.e. $y\in H_1$.

Therefore $H_1$ is one of the cosets in $X/N(f)$, i.e. it is a hyperplane parallel to $N(f)$.