Let $0<q\leq p<\infty$. For $f:\mathbb{R}\to \mathbb{R}$, we define the norm
\begin{equation} \|f\|_{p,q}=\sup_{a\in \mathbb{R},r>0} r^{\frac{1}{p}} \left(\frac{1}{r} \int_{a-r}^{a+r} |f(x)|^q \ dx \right)^{1/q} \end{equation}
and the seminorm
\begin{equation} [[f ]]_{p,q}=\sup_{a\in \mathbb{R},r>0} r^{\frac{1}{p}-\frac{1}{q}} \sup_{\gamma>0} \gamma m\left(\{x\in(a-r,a+r): |f(x)|>\gamma\}\right) \end{equation} where $m$ is the Lebesgue measure on $\mathbb{R}$.
Observe that if $f$ is Dirac delta function, then $\|f\|_{p,q}=\infty$ and $[[f]]_{p,q}=0$. But $f$ is a distribution.
Could we find a function $f$ such that $\|f\|_{p,q}=\infty$ and $[[f]]_{p,q}<\infty$?
Thanks.
For every pair $(p,q)$ with $1< q\le p<\infty$ there is such a function.
Namely, $f(x)=|x|^{-\alpha}$ with $\alpha = 1+p^{-1}-q^{-1} \in (0,1]$.
Since $f$ is symmetrically decreasing away from $0$, it suffices to consider $a=0$ in the definition of seminorm above: shifting the interval to $0$ will increase the quantity under the supremum. The set where $|f|>\gamma$ is simply $|x|<\gamma^{-1/\alpha}$. So we are looking at $$2\sup_{r>0} r^{\frac{1}{p}-\frac{1}{q}} \sup_{\gamma>0} \min(\gamma r,\gamma^{1-1/\alpha}) $$ As a function of $\gamma$, $\min(\gamma r,\gamma^{1-1/\alpha})$ first goes up (when the smaller of two things is $\gamma r$) and then does not grow, because $1-1/\alpha \le 0$. So its maximum is attained when $\gamma r =\gamma^{1-1/\alpha}$, that is, $\gamma = r^{-\alpha}$. The seminorm is $$2\sup_{r>0} r^{\frac{1}{p}-\frac{1}{q} +1 - \alpha} =2 $$
On the other hand, $$\left(\frac{1}{r} \int_{-r}^{r} |x|^{-\alpha q} \ dx \right)^{1/q} \tag{1}$$ is either $\infty$ when $\alpha q\ge 1$, or is a constant multiple of $r^{-\alpha}$ otherwise. In the former case we are done, in the latter
$$ \|f\|_{p,q}= C \sup_{r>0} r^{\frac{1}{p}-\alpha} =C \sup_{r>0} r^{\frac{1}{q}-1} \tag{2} $$ Since $1/q-1<0$, the supremum is infinite.
More generally, the example works when
$$q>\frac{p}{p+1} \text{ and } q\ne 1$$ (the first ensures $\alpha>0$, the second that (2) is infinite). It also works when $p=q=1$, because then $\alpha=1$, making (1) infinite.