Given $f : \mathbb{R}^n \rightarrow \mathbb{R}$ and a sequence $\{ x_k\} \in \mathbb{R}^n$, a sequence $\{ v_k\} \in \mathbb{R}^n$ is said to be gradient related ([1]) if for any subsequence $\{ x_k\}_{k \in \mathcal{K}} $ of $\{ x_k\}$ which converges to a non-critical point (a non-critical point is a point $p \in \mathbb{R}^n$ where $\text{grad} f(p) \ne 0$) of $f$,
$$\limsup_{k \in \mathcal{K}}\langle\text{grad} f(x_k),v_k\rangle < 0$$
Let me present my understanding of this definition, and I'm requesting an evaluation of my understanding.
For my understanding of $\limsup$ I shall rely heavily on the answers to Can someone clearly explain about the lim sup and lim inf?
After choosing $\mathcal{K}$, define $d_i := \langle\text{grad} f(x_i),v_i\rangle$ for $i \in \mathcal{K}$.
$\limsup_i d_i < 0$ which means that after a finite $N$ for all $i>N$, $d_i<0$. This is because $$ \limsup_i d_i = \inf _{n=1,2,3,...}\big(\sup\{d_{n},d_{n+1},d_{n+2},... \}\big)$$ Now if the said condition does not hold after finite $N$ then the inner $\sup$ will always return a positive value and the outer $\inf$ will be at least $0$. (refer https://math.stackexchange.com/a/493529/361497).
What I am confused about is that whether $d_i<0$ for all $i \in \mathcal{K}$?
[1] Definition 4.2.1, Absil, P.-A.; Mahony, R.; Sepulchre, R., Optimization algorithms on matrix manifolds., Princeton, NJ: Princeton University Press
It seems to me that your question is more concerned with the properties of $\limsup$ of a given sequence of real values $\{ d_i\}_{i\in\mathcal{K}}$ than on the specific optimization by gradient descent problem, therefore I focus my answer on this topic. Let's first analyze the properties of the supremum of subsets of the real line $\mathbb{R}$: the starting point is the following, very basic, lemma, stated without its (however very elementary) proof.
Lemma 1 ([1], §I.9, p. 26, Problem 8.3a) Let $S\subset\mathbb{R}$ be a set bounded from above, i.e. it exists a number $\sup S\in \mathbb{R}\iff \sup S\neq\infty$ : then, for each $\epsilon >0$ there exists $x_0\in S$ such that $$ \sup S-\epsilon<x_0\le\sup S.\label{1}\tag{1} $$ By using this lemma, we are able to prove the following important property of supremums of subsets of real numbers:
Lemma 2 ([1], §I.12, p. 38, Problem 12.1a) If $S\subseteq T\subseteq \mathbb{R}$ and $T$ is bounded from above, then $\sup S$ exists and $$ \sup S\le \sup T\label{2}\tag{2} $$ Proof Since $T$ is bounded, there exists at least one real number $l_\mathrm{ub}$ such that $x\le l_\mathrm{ub}$ for all $x\in T$: but since $x\in T$ for each $x\in S$ since $S\subseteq T$, then $x\le l_\mathrm{ub}$ for all $x\in S$ th $S$ is upper bounded and $\sup S\in \mathbb{R}$. To prove \eqref{2} we proceed by contradiction: suppose that $\sup S> \sup T$. By \eqref{1} there exists $x_0\in S$ such that $$ \sup S-\epsilon<x_0\le\sup S\quad\forall \epsilon >0 $$ Now choosing $\epsilon=\sup S-\sup T>0$, we see that there exists $x_0\in S$ and therefore $x_0\in T$ such that $$ \sup S-\epsilon=\sup T < x_0\in T $$ and thus we have reached the sought for contradiction.
We are now ready to apply these results to your problem: define the family of sets $\{D_{\mathcal{K}_n}\}_{n\in\mathbb{N}}$ as $$ D_{\mathcal{K}_n}=\{d_k=\langle\text{grad} f(x_k),v_k\rangle|k\in\mathcal{K}, \;k\ge n\} $$ It is immediate to see that $$ D_{\mathcal{K}_0}\supseteq D_{\mathcal{K}_1}\supseteq D_{\mathcal{K}_2}\supseteq \ldots \supseteq D_{\mathcal{K}_n}\supseteq\dots $$ and this, by lemma 2, implies $$ \sup D_{\mathcal{K}_0}\ge \sup D_{\mathcal{K}_1}\ge \sup D_{\mathcal{K}_2}\ge \ldots \ge \sup D_{\mathcal{K}_n}\ge\dots $$ This implies that $\{\sup D_{\mathcal{K}_n}\}_{n\in\mathbb{N}}$ is a monotonically decreasing sequence of real numbers thus, since $$ \limsup_i d_i=\inf_{n\in\mathbb{N}} \sup D_{\mathcal{K}_n} <0, $$ the supremum of each member of the nested family of set will never return positive if its index is greater than $N$, thus $$ d_i\le \sup D_{\mathcal{K}_n} <0\quad \forall i>N \iff d_i <0 \quad \forall i\in \mathcal{K}, i>N $$
NEW EDIT. After the answer was accepted, having read the comments and the question again, I felt obliged to share some considerations on the concepts of $\limsup$, $\liminf$ and the way they are introduced and studied in several university courses on mathematical analysis.
It is costumary to introduce $\limsup$ and $\liminf$ as a byproduct of the $\lim$ definition, usually given in terms of the $\epsilon$-$\delta$ construction. In my opinion, this trend overshades the real importance of such concepts, because
The book [1] of Fischer, despite being affected by many (however non disturbing) typos, analyzes througly the properties of $\limsup$ and $\liminf$ (see chapter 3:"Real sequences and their limits), including the fundamental theorems on their structure that we have implicitly used in this question and in the answer: you can find them in chapter 3, §3.6 pages 110-119. These theorem are so fundamental that they hold in every complete lattice as the algebras of subset of a given set: I found them proved and used also in a fundamental, scarce text on functional analysis authored by Gaetano Fichera (only in Italian, it was never translated).
[1] Emanuel Fischer (1983), "Intermediate Real Analysis", Undergraduate Texts in Mathematics, Berlin-Heidelberg-New York: Springer-Verlag, ISBN 0-387-90721-1, xiv+770, MR681692 (84e:26004), 0506.26002.