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.
 A: 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 $V=E$. 
$\square$
A: 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 $V=E$.
