Disintegration theorem: how to obtain $\|\pi_\sharp f\|_{L^\infty(Y)}\leq \|f\|_{C(X)}$ for all $f\in C(X)$

Question

Theorem $4$ of this blog entry of Terrence Tao states the following:

Let

• $X$ be a compact metric space, $\mathcal X$ its Borel $\sigma$-algebra, and $\mu$ a Borel probability measure on $X$.
• $(Y, \mathcal Y)$ a measurable space and $\pi:X\to Y$ a measurable map.
• $\nu := f_\sharp \mu$.

Then there is a collection $(\mu_y)_{y\in Y} \subset \mathscr P(X)$, where $\mathscr P(X)$ is the set of all Borel probability measures on $X$, such that
$$
\int_X f\cdot (g\circ \pi) \mathrm d\mu = \int_Y \left(\int_X f\mathrm d\mu_y\right)g(y)\mathrm d\nu(y)
$$
for all bounded measurable maps $f:X\to \mathbb C$ and $g:Y\to \mathbb C$.
Further, we have
$$g\circ \pi=g(y),\ \mu_y\text{-a.e. in } X$$
for y $\nu$-a.e. in $Y$.

The proof: We have the pullback map
$$
\pi^\sharp:L^2(Y, \mathcal Y, \nu)\to L^2(X, \mathcal X, \mu), g \mapsto g \circ \pi.
$$

We take its adjoint $\pi_\sharp:L^2(X, \mathcal X, \mu)\to L^2(Y, \mathcal Y, \nu)$, and have the duality
$$
\int_X f(\pi^\sharp g) \mathrm d \mu = \int_Y\left(\pi_{\sharp} f\right) g \mathrm d \nu \quad \forall f \in L^2(X, \mathcal X, \mu), \forall g \in L^2(Y, \mathcal Y, \nu).
$$

The author then said that

It is easy to see from duality that we have $\|\pi_\sharp f\|_{L^\infty(Y)}\leq \|f\|_{C(X)}$ for all $f\in C(X)$.

Can somebody explain how to obtain such inequality from the duality?

Answer 1

@Anne has answered my question in a comment. I formalize her ideas as below.

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We have $L^\infty(Y)=(L^1(Y))^*$, so $$ \|\pi_\sharp f\|_{L^\infty(Y)} = \sup \{\langle \pi_\sharp f, g \rangle_{L^1(Y)} ; \|g\|_{L^1(Y)} = 1\}. $$

We have the duality $$ \langle \pi_\sharp f, g \rangle_{L^1(Y)} = \langle f, \pi^\sharp g \rangle_{L^1(X)}. $$

So $$ \|\pi_\sharp f\|_{L^\infty(Y)} = \sup \{\langle f, \pi^\sharp g \rangle_{L^1(X)} ; \|g\|_{L^1(Y)} = 1\}. $$

Notice that $$ \begin{align} \langle f, \pi^\sharp g \rangle_{L^1(X)} &\le \|f\|_{C(X)} \langle \pmb 1, \pi^\sharp g \rangle_{L^1(X)} \\ &= \|f\|_{C(X)} \int_X \pi^\sharp g \mathrm d \mu \\ &= \|f\|_{C(X)} \int_X g \circ \pi \mathrm d \mu \\ &= \|f\|_{C(X)} \int_Y g \mathrm d \nu \quad (\star)\\ &\le \|f\|_{C(X)} \|g\|_{L^1(Y)}. \end{align} $$

Here $\pmb 1$ is the constant map $x \mapsto 1$. Also, $(\star)$ follows from change-of-variables formula. This completes the proof.

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Update 1: @Anne has pointed out a mistake in above proof, i.e., the duality holds for $g \in L^2(Y)$ but not necessarily for $g \in L^1(Y)$. However, it seems we can obtain the same inequality with a tweak on the definition of $\pi^\sharp$. We define a map $$ \pi^\sharp: L_1(Y) \to L_1(X), g \mapsto g \circ \pi. $$

Then $\pi^\sharp$ is a linear operator. Moreover, $$ \|\pi^\sharp\| = \sup_{g \in L_1(Y)} \frac{\|g \circ \pi\|_{L_1(X)}}{\|g\|_{L_1(Y)}} = \sup_{g \in L_1(Y)} \frac{\int_X |g| \circ \pi \mathrm d \mu}{\int_Y |g| \mathrm d \nu} = \sup_{g \in L_1(Y)} \frac{\int_Y |g| \mathrm d \nu}{\int_Y |g| \mathrm d \nu} = 1. $$

It follows that $\pi^\sharp$ is bounded. We denote its adjoint by $$ \pi_\sharp : L_\infty(X) \to L_\infty(Y). $$

We have the duality $$ \int_Y (\pi_\sharp f) g \mathrm d \nu = \int_X f (\pi^\sharp g) \mathrm d \mu \quad \forall f \in L_\infty(X), \forall g \in L_1(Y). \quad (\star) $$

For $f \in C(X)$, we have $f \in L_\infty(X)$, and $$ \begin{align} \|\pi_\sharp f\|_{L_\infty(Y)} &= \sup \left \{ \int_Y (\pi_\sharp f) g \mathrm d \nu \,\middle\vert\, \|g\|_{L_1(Y)} = 1 \right\} \\ &= \sup \left \{ \int_X f(\pi^\sharp g) \mathrm d \mu \,\middle\vert\, \|g\|_{L^1(Y)} = 1 \right\} \quad \text{by } (\star)\\ &\le \|f\|_\infty \sup \left \{ \int_X \pi^\sharp g \mathrm d \mu \,\middle\vert\, \|g\|_{L^1(Y)} = 1 \right\} \\ &= \|f\|_\infty \sup \left \{ \int_X g \circ \pi \mathrm d \mu \,\middle\vert\, \|g\|_{L^1(Y)} = 1 \right\} \\ &= \|f\|_\infty \sup \left \{ \int_Y g \mathrm d \nu \,\middle\vert\, \|g\|_{L^1(Y)} = 1 \right\} \quad (\star\star)\\ &\le \|f\|_\infty \sup \left \{ \int_Y |g| \mathrm d \nu \,\middle\vert\, \|g\|_{L^1(Y)} = 1 \right\} \\ &= \|f\|_\infty. \end{align} $$

Here $(\star\star)$ follows from change-of-variables formula. This completes the proof.

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Update 2: I show here why Tao's function $\pi_\sharp f\in L_2(Y)$ is equal $\nu$-a.e. to my function $\pi'_\sharp f\in L_\infty(Y)$. We have

• $$ \int_X f(\pi^\sharp g) \mathrm d \mu = \int_Y\left(\pi_{\sharp} f\right) g \mathrm d \nu \quad \forall g \in L_2 (Y). $$

• $$ \int_X f(\pi^\sharp g) \mathrm d \mu = \int_Y \left(\pi'_{\sharp} f\right) g \mathrm d \nu \quad \forall g \in L_1 (Y). $$

Because $\nu$ is finite, we get $L_\infty(Y) \subset L_2(Y) \subset L_1(Y)$. So

• $\pi_{\sharp} f - \pi'_{\sharp} f \in L_2(Y)$.
• $$ \int_Y (\pi_{\sharp} f - \pi'_{\sharp} f) g \mathrm d \nu =0 \quad \forall g \in L_2 (Y). $$

Pick $g:= \pi_{\sharp} f - \pi'_{\sharp} f$, we get $$ \int_Y (\pi_{\sharp} f - \pi'_{\sharp} f)^2 \mathrm d \nu =0 $$

It follows that $\pi_{\sharp} f - \pi'_{\sharp} f=0$ $\nu$-a.e.