= Solution
The defining <convolution> integral is absolutely convergent for every $x$ when $e\in L^1+L^2$. Indeed, writing $e=a+b$ with $a\in L^1$ and $b\in L^2$, the <Cauchy-Schwarz inequality> gives
$$
\int|K(x-y)e(y)|\,dy\leq\|K\|_\infty\|a\|_1+\|K\|_2\|b\|_2<\infty.
$$
Disjoint supports imply $\sum_n|e_n(y)|=|e(y)|$ almost everywhere. The <Tonelli theorem> therefore yields
$$
\sum_n\int|K(x-y)e_n(y)|\,dy=\int|K(x-y)e(y)|\,dy<\infty.
$$
We may exchange the sum and integral; the resulting numerical series is absolutely convergent. Its <triangle inequality> gives
$$
\boxed{|T(e)(x)|\leq\sum_n|T(e_n)(x)|.}
$$
Here $e$ belongs to the stated operator domain, as is implicit when writing $T(e)$. Disjoint supports also ensure that each summand belongs to that domain.
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