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Let $V$ be an inner product space and $W$ be a dense subspace of $V$ .
Claim: $W^{\perp }={0}$.
Let non-zero $\ x\in W^{\perp} \implies (\forall w \in W,\ \langle x ,w\rangle=0)\ \implies W \subseteq x^{\perp}$ and $x^{\perp}$ is closed subspace.
So $W \subseteq x^{\perp} \implies \overline{W} \subseteq \overline{x^{\perp}} \implies V \subseteq x^{\perp}$. This is a contradiction because this is possible only when $x=0$.
So $W^{\perp}=0.$
Is it okay and also, is there any example of an inner product space $V$ and subspace $W \leq V$, such that $W$ is not dense in $V$, but $W^{\perp}=0$?

LSpice
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RUPAM
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2 Answers2

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Your proof seems fine to me. Quick alternative proof: ${W}$ dense in ${V}$ means that every vector in ${V}$ can be written as the limit of ${w_n}$ for some sequence in ${W}$. Take some ${x \in W^{\perp}}$, and take some ${w \in V}$. Then take the sequence ${w_n \in W\ |\ w_n\to w}$. We have $$ \langle w_n,x\rangle = 0\ \forall\ n \in \mathbb{N} $$ by the continuity of the inner product, $$ \Rightarrow \langle w,x\rangle = 0 $$ but the choice of $w$ was arbitrary. In particular, if we had ${w=x}$ we have that $$ \langle x,x\rangle = 0 $$ which of course implies ${x=0}$. As required.....................

Riemann'sPointyNose
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As a rule, I don't like writing proofs that go: "suppose $x \neq 0$ then ... and $x = 0$" if I don't have to. So I would just write: "let $x \in W^\perp$" then all that stuff you said so $x = 0$.

Is there any example of inner product space $V$ and subspace $W ≤ V$ , such that $W$ is not dense in $V$, but $W^⊥ = 0$?

Assuming the Hahn-Banach theorem (which is slightly weaker than the Axiom of Choice but still independent of ZF) then no.

Suppose $W$ is a closed, proper subspace of $V$ and $x \notin W$. Define a linear function $f : W \oplus \mathbb{R}x \to \mathbb{R}$ by $f(x) = 1$ and $f(w) = 0$ for all $w \in W$. Then $f$ is continuous and by the Hahn-Banach theorem, admits a continuous extension to all of $V$. By construction, this extension is in $W^\perp \setminus \{0\}$.

Trevor Gunn
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  • Your answer assumes that $W$ is closed, but of course this assumption can be removed easily (as we should, since the original question clearly has in mind non-closed subspaces when it speaks of density): the orthocomplements of $W$ and its closure in $V$ are the same. – LSpice Jan 16 '21 at 20:22
  • I'll add this as a comment since I'm not sure what I'm saying is 100% correct. I believe if you take the axiom of choice to be false, then there are vector spaces $V$ for which $V^* = {0}$ (the algebraic dual) whether something like this is true for normed/inner product spaces I'm not sure. – Trevor Gunn Jan 16 '21 at 20:25
  • I wonder whether the dependence on Hahn–Banach can be removed when we deal with an inner-product space. Certainly it can be if $V$ is complete, in which case we have that $V = \overline W \oplus W^\perp$. – LSpice Jan 16 '21 at 20:27
  • (Ha, beaten to the punch! :-) You are right that there are non-trivial vector spaces with trivial algebraic dual. I don't think that they can be taken to be inner-product spaces, but, in the absence of completeness, I'm not sure.) – LSpice Jan 16 '21 at 20:29