Is it true: $\int ( \int f(u,z) d\mu(u))^2 dz \leq \|\mu \|_{TV} \int ( \int f(u,z) du)^2 dz$?

Question

I am trying to understand if for a measurable continuous nonnegative function $f$ on $\mathbb{R}^2$, $\sigma$-finite measure $\mu$ on $\mathbb{R}$ such that $\mu(A)>0$ for any measurable set on the Borel sigma-algebra on $\mathbb{R}$ and $ \| \cdot \|_{TV}$ total variation distance, it holds:

$$\int \bigg( \int f(u,z) d\mu(u) \bigg)^2 dz \leq \| \mu \|_{TV} \int \bigg( \int f(u,z) du \bigg)^2 dz$$

If you consider that the total variation distance as the supremum metric for measures this seems to me just Holder inequality.

Answer 1

Claim. Let $\mu$ be a finite non-negative Borel measure on $\mathbb{R}$. Then the followings are equivalent:

1. For any compactly supported continuous $f : \mathbb{R}^2 \to [0, \infty)$, $$ \int_{\mathbb{R}} \left( \int_{\mathbb{R}} f(x, z) \, \mu(\mathrm{d}x)\right)^2 \, \mathrm{d}z \leq \mu(\mathbb{R}) \int_{\mathbb{R}} \left( \int_{\mathbb{R}} f(x, z) \, \mathrm{d}x\right)^2 \, \mathrm{d}z. \tag{1}\label{eq:main} $$

2. There exists a Borel-measurable $g : \mathbb{R} \to \mathbb{R}$ such that $$ 0 \leq g(x) \leq \sqrt{\smash[b]{\mu(\mathbb{R})}} \qquad\text{and}\qquad \mu(\mathrm{d}x) = g(x) \, \mathrm{d}x . $$

Proof. The direction $(\impliedby)$ is easy, so we only prove the direction $(\implies)$.

First, consider $f(x, z) = \varphi(x)\eta(z)$ where $\varphi, \eta \in C_c(\mathbb{R})$ are non-negative and $\int_{\mathbb{R}} \eta(z)^2 \, \mathrm{d}z = 1$. Then $\eqref{eq:main}$ is recast as

$$ \int_{\mathbb{R}} \varphi(x) \, \mu(\mathrm{d}x) \leq \sqrt{\smash[b]{\mu(\mathbb{R})}} \int_{\mathbb{R}} \varphi(x) \, \mathrm{d}x . \tag{2}\label{eq:2} $$

Since $\eqref{eq:2}$ holds for any non-negative $\varphi \in C_c(\mathbb{R})$, it immediately follows that $\mu$ is absolutely continuous with respect to $\mathrm{d}x$, and the corresponding Radon–Nikodym derivative satisfies

$$ 0 \leq \frac{\mathrm{d}\mu}{\mathrm{d}x} \leq \sqrt{\smash[b]{\mu(\mathbb{R})}}. $$

Now the desired claim follows by setting $g = \frac{\mathrm{d}\mu}{\mathrm{d}x}$.