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Is there any equivalent criteria to show that subset $A$ in the Lebesgue space $L_p(X)$ is dense? In particular, I am interested in $L_2((0,1))$.

ersh
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    For $X=L_2(0,1)$ and a set $A$ within, consider the orthogonal of A, namely $A^{\perp}$, and the Hilbert space structure of X. $A$ is dense in $X$ iff $A^{\perp}={0}$. – Marko Karbevski Jan 02 '19 at 16:44
  • @MarkoKarbevski Where can I find the proof or how can I prove it? – ersh Jan 02 '19 at 19:48
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    I forgot to mention one thing, if $A$ is dense then its orthogonal is trivial for any set $A$. However if $A^\perp={0}$ we need an additional criteria in order to conclude: we want $A$ to be a vector subspace (otherwise choose the unit sphere, its orthogonal is trivial but it is not dense). – Marko Karbevski Jan 02 '19 at 20:45
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    For the first part, let $A$ be dense, $a_0 \in A^\perp$. There exists a sequence $(a_n)$ of elements in $A$ such that $a_n \to a$.
    1. What is the value of $\langle a_n,a \rangle$ ? ; 2) What's its limit in terms of $|a| $? ; 3) Conclude. You should find the proofs here: https://math.stackexchange.com/questions/1315321/if-m-is-a-non-empty-subset-of-a-hilbert-space-h-the-span-of-m-is-dense-in
     – Marko Karbevski Jan 02 '19 at 20:45
  • Thanks, I just needed this part "$A$ is dense then its orthogonal is trivial for any set $A$". – ersh Jan 02 '19 at 21:32

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Copying the comments in order to have this question answered:

We have the following proposition:

For $H=L_2(0,1)$ and a set $A$ within, consider the orthogonal of $A$, namely $A^⊥$, and the Hilbert space structure of $H$. If $A$ is dense in $H$ then $A^⊥=0.$

Moreover, if $A$ is a vector subspace of $H$ and $A^⊥=0$ then $A$ is dense in $H$.

Showing why $A$ needs to be a vector subspace of $H$ in order to have the second implication can be achieved (via contradiction) by considering the unit sphere $\mathbb{S}$ in $H$. $\mathbb{S}^⊥={0}$ yet $\mathbb{S}$ is not dense in $H$.

In order to prove the first implication, one can try the following steps: Let $A$ be dense in $H$, $a_0 \in A^⊥$. We know (why?) that there exists a sequence $(a_n)_{n\ge 1}$ of elements in $A$ such that $a_n \to a_0$.

  1. Calculate the terms of the sequence $(\langle a_n, a_0 \rangle)_{n\ge 1}.$
  2. Express the limit of the sequence in terms of $||a_0||$.
  3. Conclude.

The full proof can be find here: If $M$ is a non-empty subset of a Hilbert space $H$, the span of $M$ is dense in $H$ iff $M^{\perp} = \{0\}$.

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