a dense subspace of inner product space whose codimension is 1 Is there an example of an inner product space $V$ and a dense subspace $D$ whose codimension is $1$?
In other words, there exists $z\in V$ such that every element of $V$ can be written in the form $e+\lambda z$ for some $e\in E$ and some scalar $\lambda$.
 A: As Kavi Rama Murthy says in the comments, "Every infinte dimensional inner product space has a discontinuous linear functional f and the kernel of such a functional is dense and has codimension 1".
I would like to add a proof of the fact that every infinite dimensional normed space (in general) has an unbounded linear functional.
The kernel of a linear functional on a Banach space is either closed or dense: the kernel is closed if-f the functional is bounded. The kernel is dense if-f the functional is unbounded, see this post.
So let $X$ be a normed space with infinite dimension. Take $\{x_n\}$ a linearly independent subset of $X$ (which is possible, since $X$ has infinite dimension) and extend it to a Hamel basis of $X$, say $\{x_n\}_{n=1}^\infty\cup Z$, where $Z\subset X\setminus(\{x_n\}_{n=1}^\infty)$. We define a functional on the basis and we extend it linearly on all $X$. We simply set $f(x_n)=n\cdot\|x_n\|$ and $f(z)=0$ for all $z\in Z$. Note that $f$ cannot be bounded: if $|f(x)|\leq c\|x\|$ for all $x\in X$ for some $c>0$, then $n\|x_n\|\leq c\|x_n\|$ for all $n$, which is impossible since $\mathbb{N}$ has no upper bound.
