Is $H^{1}(R;H^1(R^n))=H^1(R^{n+1})$?

Question

Is $H^{1}(R;H^1(R^n))=H^1(R^{n+1})$,

where

$H^1(R;H^1(R^n))=\{f\colon R\to H^1(R^n)\colon \int \|f(t,\cdot)\|_{H^1}^2\,dt <\infty \quad \text{and} \int \|f'(t,\cdot)\|_{H^1}^2\,dt <\infty \}$

(and $g=f'\Leftrightarrow \int f(t,\cdot)\phi'(t)\,dt=-\int g(t,\cdot)\phi(t)\,dt \quad \forall \phi \in C_c^\infty(R,R)$)

Answer 1

This is not true. Indeed, $f\in H^1(R^{n+1})\Leftrightarrow f$ and all of its first derivatives (in the sense of distributions) are in $L^2$. But if $f\in H^1(R,H^1(R^n))$, then necessarily $\frac{\partial^2 f}{\partial t\partial x_j}$ also lie in $L^2$, which is not the case for a general function in $H^1(R^{n+1})$.