Let $T$ be a convolution operator on $L^2(\mathbb{R}^n)$, suppose $K$ is a tempered distribution in $\mathbb{R}^n$ that coincides with a locally integrable function on $\mathbb{R}^n\setminus \{0\}$. Suppose that $K$ satisfies $$ \int_{|x|>2|y|}|K(x-y)-K(x)|dx\leq B,~y\in\mathbb{R}^n $$ for a constant $B$ (this is called the Hörmander condition and is a smoothness condition). Then for $1<r<\infty$ and $1<p<\infty$, we have the boundedness estimate $$ \left|\left| \left(\sum_{j}|Tf_j|^r\right)^{\frac{1}{r}} \right|\right|_p \leq C_{p,r} \left|\left| \left(\sum_j|f_j|^r\right)^{\frac{1}{r}} \right|\right|_p. $$ For $p=1$, we have the estimate $$ \left| \{ x\in\mathbb{R}^n:\left(\sum_j|Tf_j(x)|^r\right)^{\frac{1}{r}}>\lambda \} \right| \leq \frac{C_r}{\lambda} \left|\left| \left( \sum_j|f_j|^r \right)^{\frac{1}{r}} \right|\right|_1. $$ (This is a weak-type $(1,1)$ inequality).
My question is why do we not have a similar inequality for $p=\infty?$
A similar example is that we do not have $L^{\infty}-$boundedness for the Hilbert transform and a counterexample can be provided via step functions.
Does anyone know of a counterexample why $L^{\infty}-$boundedness does not exist? What is the intuitive reason behind it?