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Let $(X,\mathcal{B},\mu)$ be a probability space, and let $P: L^2(X,\mu) \to L^2(X,\mu)$ denote the orthogonal projection onto some closed subspace of $L^2(X,\mu)$ that contains the (almost everywhere) constant functions. Is it then true that for any $A \in \mathcal{B}$ we have that $$\langle \left(P(\mathbb{1}_A)\right)^2, \mathbb{1}_A\ \rangle \geq \langle \mathbb{1}_A , 1 \rangle ^2 = \mu(A)^2,$$ or even that the LHS above is positive when $\mu(A)>0$ ? I know that $\langle P(\mathbb{1}_A), \mathbb{1}_A\ \rangle \geq \mu(A)^2 ,$ because of Cauchy-Schwartz, since $P$ is a projection and $P(1)=1$, but can't work out the other one.

Of course, by $\langle \cdot, \cdot \rangle$ I mean the inner product in $L^2(X,\mu)$.

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Update: It turns out that $$\langle \left( P\left( \mathbb{1}_A \right)\right)^2\ , \mathbb{1}_A \rangle \geq \mu(A)^4.$$ Indeed, just notice that $\mathbb{1}_A=(\mathbb{1}_A)^2$ and then use Cauchy-Schwartz to obtain that $$\langle \left( P\left( \mathbb{1}_A \right)\right)^2\ , (\mathbb{1}_A)^2 \rangle \geq \langle P\left( \mathbb{1}_A \right)\ , \mathbb{1}_A \rangle^2,$$ which in turn is at least $\mu(A)^4$ as I state in the question.

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    $\begingroup$ Since the closed subspace contains the constant functions, it follows $ \langle 1_A - P(1_A), 1 \rangle =0. $ Not sure what to make out of that. $\endgroup$
    – daw
    Commented Jun 28, 2023 at 10:05

2 Answers 2

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Let $P$ be the projection onto the closed subspace $V$. The projection satisfies $$ \langle 1_A - P(1_A),\ v \rangle =0 \quad \forall c\in V. $$ Since that subspace contains all constant functions, it follows that $$ \langle 1_A - P(1_A), 1 \rangle =0. $$ In particular, if $V$ is equal to the subspace of constant functions, then $P(1_A) = 1 \cdot \mu(A)$ and $$ \langle (P(1_A))^2, 1_A \rangle = \mu(A)^3. $$

Setting $v = P(1_A)$ in the above equality gives $$ \| P(1_A)\|_{L^2}^2 = \langle 1_A, P(1_A)\rangle, $$ but there is nothing that suggests a lower bound of $\langle ( P(1_A))^2, 1_A \rangle$.

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  • $\begingroup$ I didn't claim equality, but that the quantity is greater than or equal to $\mu(A)^4$. I included details in the update. Do you agree with my claim? $\endgroup$
    – User
    Commented Jun 28, 2023 at 16:03
  • $\begingroup$ yes. It is an usual way of writing Cauchy-Schwarz, but seems to be true. Note that my calculation with $V=$ space of constant functions shows that the lower bound is at most $\mu(A)^3$ not $\mu(A)^2$. $\endgroup$
    – daw
    Commented Jun 28, 2023 at 16:05
  • $\begingroup$ Indeed, but the question is more general and I think the bound $\mu(A)^3$ might not always hold. However, $\mu(A)^4$ is always a lower bound. $\endgroup$
    – User
    Commented Jun 28, 2023 at 17:34
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We have that $$\langle \left( P\left( \mathbb{1}_A \right)\right)^2\ , \mathbb{1}_A \rangle = \langle \left( P\left( \mathbb{1}_A \right)\right)^2\ , \left( \mathbb{1}_A \right)^2 \rangle \geq \langle P\left( \mathbb{1}_A \right)\ , \mathbb{1}_A \rangle ^2$$ by Cauchy Schwarz, but then $\langle P\left( \mathbb{1}_A \right)\ , \mathbb{1}_A \rangle ^2 \geq \mu(A)^4$, as was mentioned in the question, so we have that the LHS is positive and a slightly different lower bound.

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