Assume that $p_n$ is the $n$-th step distribution of a random walk with state space $\mathbb{Z}^d$, i.e. $p_n(x,y)=\mathbb{P}(S_{n+1}=y\mid S_0=x)$, where $S_n=S_0+\sum_{i=1}^nX_i$ with $X_i$'s i.i.d. and independent of $S_0$. Define the operator $\nabla_yp_n(x):=p_n(x+y)-p_n(x)$.
Now writing $y=\sum_{i=1}^d{a_ie_i}$ (where $(e_i)_i$ is a basis of $\mathbb{Z}^d$) is it somehow true that $\nabla_yp_n(x)\leq\sum_{i=1}^d\nabla_{e_i}p_n(x)$? I do not see how to manage with the $a_i$'s. That's an argument often used by proving statement about $\nabla_yp_n(x)$ saying that by triangular inequality it is enough to prove the statement for $\nabla_{e_i}p_n(x)$.