# Problem

The ground field is $K$, $\operatorname{char}K\neq2$. Suppose $W$ is a (maybe infinite dimensional) subspace of a vector space $V$ with a symmetric/symplectic form $\langle\cdot,\cdot\rangle$. The orthogonal space of $W$, denoted as $W^\perp$, is defined as $W^\perp=\{\,v\in V\,\colon\,\langle v,w\rangle=0,\forall w\in W\,\}$. If the form is nondegenerate on $W$, say $W\cap W^\perp=\{0\}$, is it true that $V=W\oplus W^\perp$? What if the form is nondegenerate on $V$?

# Background

The statement is true when $W$ is finite dimensional. One can choose an orthogonal basis for $W$, then apply the projection formula to determine a orthogonal projection. However, such a process couldn't be applied in the infinite dimensional case. I'm interested in how far we can generalize our results of psuedo-Euclidean or symplectic forms on a finite dimensional vector space to the infinite-dimensional case.

• For finite dimensions, we always have $S\oplus S^\perp=V$. For infinite dimensions, it may fail.
– Pedro
Jan 13, 2014 at 3:42
• @PedroTamaroff Well, it seems more a stupid analysis problem, on which I have no experience for now. Have our mathematicians generalized the psuedo-Euclidean/symplectic case in functional analysis? They're apparently not Banach spaces. Jan 13, 2014 at 4:19

Counter-example

$K=\Bbb R$ and $V=\mathcal C([0,1]),\Bbb R)$. If $f,g\in E$, we put $\langle f,g\rangle=\int_0^1 f(t)g(t)dt$ and let $V=\{f \in W: f(0)=0\}$. We have $V^\bot=\{0\}$

Proof: Suppose $f \in V^\bot$. Let $g\in E$ such that $g:t\mapsto t f(t)$. Then $g \in V$. By definition of $f \in V^\bot$ we have $\langle f,g\rangle=0$ which gives $\int_0^1 tf^2(t)dt=0$

Since $tf^2(t)$ is continuous and positive we have $\forall t \in [0,1]\quad tf^2(t)=0$ then $f=0$ on $]0,1]$ and on $[0,1]$ by continuity at point $0$.

• Really elegant - thank you so much!
– user261214
Nov 17, 2016 at 17:24

Let $V = C[a,b]$ be the vector space of continuous functions on $[a,b]$ with inner product $\displaystyle \langle f,g \rangle = \int_a^b f(x)g(x)dx$.

Let $W =\{f \in V; f(a)=0\}$. Then $W^{\perp} = \{0\}$. To see why: if $h\in W^{\perp}$ then $\langle x-a,h(x) \rangle=0$ we must also have, for example, $\langle(x-a)g(x),h(x)\rangle=0$ for all $g\in V$. So $h\equiv 0$.

So $V\neq W \oplus W^{\perp}$.

edit: Mohamed beat me to it