Is it true that $\langle \left(P(\mathbb{1}_A)\right)^2, \mathbb{1}_A\ \rangle \geq \mu(A)^2$?

Question

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.

Answer 1

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.

Answer 2

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 $\langle ( P(1_A))^2, 1_A \rangle$.