Is the shift operator continuous on the real Hardy spaces?

Is the shift operator (or the translation operator) continuous on Hardy spaces $H^p(\mathbb{R}^n)$ (with $0<p\leq 1$)? i.e. given $f\in H^p(\mathbb{R}^n)$, is the following map \begin{align} \tau:&\mathbb{R}^n\rightarrow H^p(\mathbb{R}^n)\\ x&\mapsto \tau_x(f), \quad \text{where} \quad (\tau_x(f))(y):=f(y-x) \end{align} continuous?

First of all I want to answer this on $H^1$, because it is a Banach space of functions (not distributions).

I guess the answer is "Yes".

I'm trying to use the proof on $L^p(\mathbb{R}^n)$ for $1\leq p<\infty$ which uses the density of $C_c\cap L^p$ on $L^p$. I know that $C_c\cap H^1$ is dense on $H^1$ (also $C_c^\infty\cap H^1$ is dense on $H^1$).

Remember that $H^1$ is a subspace of $L^1$ but their norms are not equivalent ($H^1$ is not closed in $L^1$). The norm in $H^1$ is given by: Let $\phi$ a Schwartz function with $\int \phi=1$ and, for every $f\in H^1$, consider the maximal function $M_\phi(f)(x):=sup_{t>0}|(f*\phi_t)(x)|$ (where $\phi_t(x):={t^{-n}}f(x/t)$). The norm of $f\in H^1$ is defined to be the $L^1$-norm of $M_\phi(f)$.