A function $f:\mathbb{R} \to \mathbb{R}$ with infinite norm but finite weak seminorm 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.
 A: 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.
