# A problem involving a hyperplane and an affine subspace

Any ideas on how to solve this??

Let $E$ be a normed vector space and let $H \subset E$ be a hyperplane. Let $V \subset E$ be an affine subspace containing $H$. Prove that either $V=H$ or $V=E$.

If $E$ is an finite dimensional vector space then this is obvious. But I don't know how to go about the general case.

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What's you definition of affine subspace? – Davide Giraudo Jan 23 '12 at 18:07
By "hyperplane" do you mean a codimension-$1$ subspace? I'm not so clear what this means in the infinite-dimensional case. – Neal Jan 23 '12 at 18:07
I think this all relies ultimately on working out definitions, which you should post here. It would help you get on the right track. – Patrick Da Silva Jan 23 '12 at 18:40
OK, here are the definitions at the level of generality needed. A hyperplane is a set $H=\{x \in E: f(x)=\alpha\}$ for some linear functional $f$ (not necessarily continuous) and some real number $\alpha$. An affine subspace is a subset $V$ of $E$ that can be written as $V=A+a$ for some $a\in A$ and an (ordinary) vector subspace $A$ of $E$. – chango Jan 24 '12 at 12:16

I finally worked out the way to do it. Here it is for those of you interested. Thank you Davide for your attempt. Although it was wrong it inspired my proof.

Proof:

Suppose $\exists x \in V \setminus H$. To see that we must have $V=E$, choose $y \in E$ arbitrary. Also, write $V=U+a$.

We will have to distinguish two cases:

• $x-y \in Ker(f)$:

In this case $f(x-y)=0$. Let $y_0 \in E$ be s.t. $f(y_0) \neq 0$. Then

$x-y=\underbrace{x-y+ \alpha \frac{y_0}{f(y_0)}}_{\in H \subset V} - \underbrace{\alpha \frac{y_0}{f(y_0)}}_{\in H \subset V}$

So we can write

$x-y=u_0+a - (u_1+a) \implies y=\underbrace{x}_{\in V} + \underbrace{u_1-u_0}_{\in U} \in V$

• $x-y \notin Ker(f)$:

In this case $f(x-y) \neq 0$. Then $\exists \beta \in \mathbb{R}$ (take $\beta=\frac{\alpha-f(y)}{f(x)-f(y)}$) s.t.

$f( \beta x+(1-\beta)y)=\alpha$

i.e. $\beta x+(1-\beta)y \in H \subset V$

$\implies \beta (a+u_0) +(1-\beta)y=a+u_1$

$\implies (1-\beta)y= (1-\beta) a + u_1-\beta u_0$

$\implies y= a + \underbrace{\frac{u_1-\beta u_0}{1-\beta}}_{\in U} \in V$

Hence, we conclude that $y\in V$ and since $y$ was arbitrary this implies that $V=E$.

$\square$

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Denote $H=\{f=\alpha\}$ where $f$ is a linear functional and $\alpha$ a real number. If $f\equiv 0$, either $\alpha=0$, then $H=E$ and the result is clear, or $\alpha\neq 0$ and $H$ is empty, so the result can be false, for example if $V=\{0\}$ and $E$ has a dimension $\geq 2$.

So we assume $f\neq 0$. We can write $V=W+a$ where $a\in E$ and $W$ is a linear subspace of $E$. In fact, we have the inclusion $\{f=\alpha-f(a)\}\subset W\subset E$, since $H_0:=H+(-a)=\{f=\alpha-f(a)\}$. We put $\beta :=\alpha-f(a)$. If $\beta\neq 0$, and $y_0\in E$ is such that $f(y_0)\neq 0$ then $\beta \frac{y_0}{f(y_0)}\in H\subset W$ so $y_0\in W$. So for $x\in \ker f$, $x=\underbrace{x+\beta\frac{y_0}{f(y_0)}}_{\in H_0\subset W}-\underbrace{\beta\frac{y_0}{f(y_0)}}_{\in W}$. So in any case $\ker f\subset W\subset E$. If $\ker f=W$ we are done (we get $H=V$) and if $\ker \subsetneq W$ and $w_0\in W\setminus\ker f$ and $x \in E$, then
$$x=\underbrace{x+\frac{\beta w_0}{f(w_0)}}_{\in H_0\subset W}-\underbrace{\frac{\beta w_0}{f(w_0)}}_{\in W},$$ so $x\in W$ and $W=E$, hence $V=E$.

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I follow your argument up until you claim $x+\frac{w_0}{f(w_0)}\in H_0$. Why is this true? It should be very easy to prove since you just need to show $f(x+\frac{w_0}{f(w_0)})=\alpha-f(a)$ but I don't necessarily think that is the case for a general $x \in E$. – chango Jan 26 '12 at 12:36
If $x\in \ker f$ then $f\left(x+\beta\frac{y_0}{f(y_0)}\right)=f(x)+\beta=\beta$ (it's true only for $x\in\ker f$ indeed, not for a general $x$). – Davide Giraudo Jan 26 '12 at 14:31
Yes, that part is fine. But what I mean refers to what is below. So you agree that last part is not correct? – chango Jan 26 '12 at 15:30