Let $\mathcal{H} = L^p_{n}(0,a)$, where $p \in \mathbb{R}^+$, $p \geq 1$ and $n \in \mathbb{N}$, be the Hilbert space of vector-valued functions defined on the finite interval $(0,a) \subset \mathbb{R}$, $f = (f_1,\ldots,f_n)$, with norm $$ ||f||_{L^p_n(0,a)} = \sum_{i=1}^n ||f_i||_{L^p(0,a)}, $$ and similarly we define, for matrix-valued functions $K \in L^p_{n\times n}(-a,a)$, $$ ||K||_{L^p_{n\times n}(-a,a)} = \sum_{i,j=1}^n ||K_{ij}||_{L^p(-a,a)}. $$
Consider a convolution-type integral operator $T_K : \mathcal{H} \to \mathcal{H}$, $$ (T_K f)(t) = \int_0^a K(t-s)f(s) ds. $$
Suppose that the kernel is absolutely summable, i.e. $K \in L^1_{n\times n(-a,a)}$. Prove that $$ ||T_K||_{L^p_{n}(0,a)} \leq ||K||_{L^1_{n\times n}(-a,a)}. $$
Remarks: (i) sources; (ii) what I've tried:
(i) This is an intermediate step (see Lemma 1.1 following this link) to prove that under these assumptions, $T_K$ is compact.
(ii) To compute \begin{align} ||T_K f||_{L^p_n(0,a)} &= \sum_{i=1}^n ||(Kf)_i||_{L^p_n(0,a)} \\ &= \sum_{i=1}^n \left( \int_0^a |(Kf)_i(t)|^p dt \right)^{1/p} \\ &= \sum_{i=1}^n \left( \int_0^a \left| \sum_{j=1}^n \int_0^a K_{ij}(t-s) f_j(s) ds \right|^p dt \right)^{1/p} \\ &\leq \sum_{i=1}^n \left( \int_0^a \left( \sum_{j=1}^n \int_0^a |K_{ij}(t-s)| |f_j(s)| ds \right)^p dt \right)^{1/p}. \end{align}
Edit 1:
Clearly for $p=1$ it holds, since then by a change of variable,
\begin{align}
||T_K f||_{L^1_n(0,a)} &\leq \sum_{i=1}^n \sum_{j=1}^n \int_0^a \left( \int_{-s}^{a-s} |K_{ij}(t)| dt \right) |f_j(s)| ds \\
&\leq \sum_{i,j = 1}^n ||K_{ij}||_{L^1(-a,a)} ||f_j||_{L^1(0,a)}.
\end{align}
How to proceed with $p>1$?
Edit 2: Using (for $p\geq 1$) Holder's inequality with conjugate indices $q$ and $p$, I arrive at \begin{align} ||T_K f||_{L^p_n(0,a)} &\leq a^{1/p} \sum_{i,j = 1}^n ||K_{ij}||_{L^q(-a,a)} ||f_j||_{L^p(0,a)}. \end{align} Which does not solve the problem because in general we have $||K_{ij}||_{L^1(-a,a)} \leq ||K_{ij}||_{L^q(-a,a)}$ for $q>1$.