$\newcommand{\RR}{\mathbb R}\newcommand{\diff}{\, \mathrm d}$ We fix $T \in (0, \infty)$ and $p, q \in [1, \infty)$. Let $\mathbb T$ be the interval $[0, T]$.
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Let $E$ be the space of all real-valued Lebesgue measurable functions on $\mathbb T \times \RR^d$ with the finite norm
$$
\| g \|_E := \sup_{x \in \RR^d} \bigg ( \int_{\mathbb T} \bigg ( \int_{B(x, 1)} | g(s, y) |^p \diff y \bigg )^{\frac{q}{p}} \diff s \bigg )^{\frac{1}{q}}.
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Let $F$ be the space of all real-valued Lebesgue measurable functions on $\mathbb T \times \RR^d$ with the finite norm
$$
\| g \|_F := \bigg ( \int_{\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!
Best Answer
The opposite inequality cannot be true. If that were true, then consider a positive function $g$ with the property such that for all $s\in \mathbb{T}$ it holds that $g(s,x) \leq C g(s,y)$ whenever $|x-y|<1$ then it would hold that \begin{equation} \int_\mathbb{T} \sup_{x\in\mathbb{R}^d} g(x,s)ds \leq C_1 \sup_{x\in \mathbb{R}^d} \int_{\mathbb{T}}g(s,x)ds, \end{equation} for some $C_1>0$. This cannot be true consider for example $g(s,x)= \frac{\psi'(s)}{((|x|-\psi(s))^2+1)^\frac{d+1}{2}}$, where $\psi:(0,T)\to\mathbb{R}$ is a strictly increasing smooth function such that $\psi(0^+)=-\infty, \psi(T^-)=+\infty)$. Then the left hand side becomes \begin{equation*} \text{L.H.S.}=\int_0^T\psi'(s)ds = \int_\mathbb{R}dt=+\infty. \end{equation*} While the right hand side is \begin{equation*} \text{R.H.S.}= \sup_{x\in \mathbb{R^d}} \int_\mathbb{R} \frac{dt}{((|x|-t)^2+1)^{\frac{d+1}{2}}} = \int_\mathbb{R} \frac{dt}{(t^2+1)^{\frac{d+1}{2}}}<+\infty. \end{equation*}