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Among the principal properties of the orthogonal complement, we have the following:

$$A\subset(A^\perp)^\perp$$

Where $A$ is a subset of an inner product space $X$, and $A^\perp$ is the orthogonal complement of $A$.

The proof given in Rynne and Youngson, "Linear Functional Analysis", is as follows:

"Let $a\in A$. Then for all $x\in A^\perp$, $(a,x)=\overline{(x,a)}=0$, so $a\in(A^\perp)^\perp$. Thus, $A\subset(A^\perp)^\perp$."

How is it, exactly, that this shows that $a\in(A^\perp)^\perp$? And consequently that $A\subset(A^\perp)^\perp$? Has it to do with the sesquilinear nature of the inner product, and that the second arguement is conjugate linear?

Furthermore, why is it that we do not have equality, namely that $A=(A^\perp)^\perp$? Clearly it is not the case, but when I was reading it and thinking about it, I thought that they should be equal.

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2 Answers 2

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First, refresh some definitions:

Definition 1:

By definition, if $X$ is a subset of $V$, then $X^\bot$ is defined as $$X^\bot = \{v\in V: \forall x\in X:\langle x, v\rangle = 0\}$$

Definition 2:

By definition, $A$ is a subset of $B$ if and only if every element of $A$ is also an element of $B$, i.e. $$A\subset B\iff \forall x\in A: x\in B$$


Now, you want to prove that $A\subset (A^\bot)^\bot$. By definition 2, you must prove that every $a\in A$ is also an element of $(A^\bot)^\bot$.

So, you take an arbitrary $a\in A$. And you prove that for all $x\in A^\bot$, you have $\langle a, x\rangle = 0$. This, by definition 1, means that $a\in (A^\bot)^\bot$!


As far as your second question is concerned, it does sometimes happen that $A=(A^\bot)^bot$. For example, if $V$ is a vector space, then $(\{0\}^\bot)^\bot = \{0\}$ and $(V^\bot)^\bot = V$. Also, if $V$ is finite dimensional, you have equality. But in general, you do not.

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    $\begingroup$ How on earth does this answer deserve a downvote?! $\endgroup$
    – 5xum
    Aug 17, 2015 at 6:56
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    $\begingroup$ That was to your previous answer. $\endgroup$ Aug 17, 2015 at 7:02
  • $\begingroup$ That makes things more clear, thanks. I think I was over complicating it trying to think what the orthogonal complement of the orthogonal complement of A was, particularly when equality is not always a given. Cheers! $\endgroup$ Aug 17, 2015 at 7:07
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    $\begingroup$ @JeremyJeffreyJames If an answer is unclear, but correct, then ask the poster for a clarification. A downvote means the answer is wrong or bad. Merely unclear does not call for a downvote. $\endgroup$
    – 5xum
    Aug 17, 2015 at 7:08
  • $\begingroup$ Yes, although your first answer offered no clarification to the point in my question and for that reason, I thought it was bad. $\endgroup$ Aug 17, 2015 at 7:17
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To your second point:

In a Banach space it holds: $$A\subseteq E:\quad\overline{\langle A\rangle}=\left(A^\perp\right)^\perp$$

(This fails in incomplete spaces.)

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