$\newcommand{\RR}{\mathbb R}\newcommand{\diff}{\, \mathrm d}$ We fix $T \in (0, \infty)$ and let$p, q \in [1, \infty)$. Let $\mathbb T$ be the interval $[0, T]$. Let $p, q \in [1, \infty)$.
Let $E$ be the space of all real-valued Lebesgue measurable functions on $\mathbb T \times \RR^d$ with the finite norm $$ \| g \|_E := \sup_{x \in \RR^d} \bigg ( \int_{t_0}^{t_1} \bigg ( \int_{B(x, 1)} | g(s, y) |^p \diff y \bigg )^{\frac{q}{p}} \diff s \bigg )^{\frac{1}{q}}. $$$$ \| g \|_E := \sup_{x \in \RR^d} \bigg ( \int_{\mathbb T} \bigg ( \int_{B(x, 1)} | g(s, y) |^p \diff y \bigg )^{\frac{q}{p}} \diff s \bigg )^{\frac{1}{q}}. $$
Let $F$ be the space of all real-valued Lebesgue measurable functions on $\mathbb T \times \RR^d$ with the finite norm $$ \| g \|_F := \bigg ( \int_{t_0}^{t_1} \bigg ( \sup_{x \in \RR^d} \int_{B(x, 1)} | g(s, y) |^p \diff y \bigg )^{\frac{q}{p}} \diff s \bigg )^{\frac{1}{q}}. $$$$ \| g \|_F := \bigg ( \int_{\mathbb T} \bigg ( \sup_{x \in \RR^d} \int_{B(x, 1)} | g(s, y) |^p \diff y \bigg )^{\frac{q}{p}} \diff s \bigg )^{\frac{1}{q}}. $$
Above, $B(x, 1)$ is the open ball centered at $x$ with radius $1$. Clearly, $\| g \|_E \le \| g \|_F$ and thus $F \subset E$.
Is there a constant $c >0$ such that $\| g \|_F \le c \| g \|_E$ for all $g \in E$?
Thank you so much for your elaboration!