= Solution
{c}
Choose the common dominating <probability measure> $\nu=(\mu+\mu')/2$ and let $p=d\mu/d\nu$, $q=d\mu'/d\nu$ be the <Radon-Nikodym derivatives>. The assumed second <moments> imply first-moment integrability by the <Cauchy-Schwarz inequality>. Since $Y$ is a separable <Banach space>, the measurable $f$ is a <strongly measurable function>, and both <expected values> exist as <Bochner integrals>.
The difference of these <Bochner integrals> satisfies
$$
\begin{aligned}
\|\mathbb E^\mu f-\mathbb E^{\mu'}f\|_Y
&=\left\|\int_Xf(p-q)\,d\nu\right\|_Y\\
&\leq\int_X\|f\|_Y\,|\sqrt p-\sqrt q|(\sqrt p+\sqrt q)\,d\nu\\
&\leq\left(\int_X\|f\|_Y^2(\sqrt p+\sqrt q)^2\,d\nu\right)^{1/2}
\left(\int_X(\sqrt p-\sqrt q)^2\,d\nu\right)^{1/2}.
\end{aligned}
$$
The first inequality is the norm bound for a <Bochner integral>, and the second is the scalar <Cauchy-Schwarz inequality>. Since $(\sqrt p+\sqrt q)^2\leq2(p+q)$, the first factor is at most
$$
\sqrt{2}\left(\mathbb E^\mu\|f\|_Y^2+\mathbb E^{\mu'}\|f\|_Y^2\right)^{1/2}.
$$
With the <Hellinger distance> convention of the original PDF, the second factor is $\sqrt2\,d_{\mathrm{Hell}}(\mu,\mu')$. Multiplication proves the <Hellinger bound for differences of expectations>:
$$
\boxed{\|\mathbb E^\mu f-\mathbb E^{\mu'}f\|_Y
\leq2\left(\mathbb E^\mu\|f\|_Y^2+\mathbb E^{\mu'}\|f\|_Y^2\right)^{1/2}
d_{\mathrm{Hell}}(\mu,\mu').}
$$
The proof uses only the two laws and their common dominating measure; no <absolute continuity of measures> assumption between $\mu$ and $\mu'$ is needed.
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