2

Here is a problem I'm working on:

If $M$ is a closed subspace of an inner product space $V$, is it always true that $V=M\oplus M^\perp$?

I am asked to consider the following example: Let $V\subset\ell^2$ be the set of all sequences with only finitely many nonzero terms, and let $M=\{y\in V:\langle y,x\rangle=0\text{ in }\ell^2\}$, where $x=(1/n)_{n=1}^\infty$.

I imagine that to show $V\neq M\oplus M^\perp$ I must find $v\in V$ such that $v$ is not the sum of any $u\in M,w\in M^\perp$. Any hints on how to proceed?

buffle
  • 623
  • 1
    The orthogonal complement of $M$ in $V$ is ${0}$ (and $V \neq M$). If you are taking the orthogonal complement in $\ell^{2}$ then $M^{\perp}$ is not even subset of $V$. – Kavi Rama Murthy Oct 13 '20 at 07:55

0 Answers0