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

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 $$L^2(X,\mu)$$.

$$\\$$

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.

• 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.
– daw
Commented Jun 28, 2023 at 10:05

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

• 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?
– User
Commented Jun 28, 2023 at 16:03
• 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 $\mu(A)^2$.
– daw
Commented Jun 28, 2023 at 16:05
• 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.
– User
Commented Jun 28, 2023 at 17:34

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.