I am curious as to whether there are any conditions on a vector-valued linear operator that imply injectivity of its adjoint. This is in part a continuation of my previous question here. For reference, let $T:X \rightarrow Y_{1} \times \ldots \times Y_{n} $, with $T \in L(X,Y_{1} \times \ldots \times Y_{n})$ and $X,Y_{i}$ Banach spaces. The adjoint of $T$ is given by \begin{equation} \begin{aligned} &T^{*}: Y_{1}^* \times \ldots \times Y_{n}^* \rightarrow X^* \\ &T^{*}(y_{1}^*,\ldots,y_{n}^*)(x) = \sum_{i=1}^{n} y_{i}^{*} \circ T_{i}x. \end{aligned} \end{equation} Setting the above summation equal to $0$, this $\textbf{looks}$ similar to the definition of linear independence. However, for common Banach spaces such as $L^{p}(\mathbb{R})$ spaces, $p \neq 2$, the expression $y_{i}^{*} \circ T_{i}x$ becomes the integration of the product of some $L^{p}(\mathbb{R})$ and $L^{q}(\mathbb{R})$ functions. To the best of my knowledge, sums of integrals are not necessarily linearly independent, even if their integrands are.
This leads to my question. Are there any properties of $T_{i}$ (perhaps linear independence in combination with something else) that can allow us to conclude injectivity of the adjoint? I have been reading Rudin's Functional Analysis. Chapter 4 (specifically Theorem 4.15) gives conditions for the injectivity of $T^*$ but this requires the strong assumption that $T$ is surjective.