Are these two norms on localized versions of $L^p_q$ equivalent?

$$\newcommand{\RR}{\mathbb R}\newcommand{\diff}{\, \mathrm d}$$ We fix $$T \in (0, \infty)$$ and $$p, q \in [1, \infty)$$. Let $$\mathbb T$$ be the interval $$[0, T]$$.

• Let $$E$$ be the space of all real-valued Lebesgue measurable functions on $$\mathbb T \times \RR^d$$ with the finite norm $$\| g \|_E := \sup_{x \in \RR^d} \bigg ( \int_{\mathbb T} \bigg ( \int_{B(x, 1)} | g(s, y) |^p \diff y \bigg )^{\frac{q}{p}} \diff s \bigg )^{\frac{1}{q}}.$$

• 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!

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 $$$$\int_\mathbb{T} \sup_{x\in\mathbb{R}^d} g(x,s)ds \leq C_1 \sup_{x\in \mathbb{R}^d} \int_{\mathbb{T}}g(s,x)ds,$$$$ for some $$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*}$$