I have a convex function $f$ on $\mathbb{R}^n$ and a collection of $m\in\mathbb{N}$ convex weights $\{w_i\}_{i=1}^m\subset\left]0,1\right]$ satisfying $\sum_{i=1}^mw_i=1$. I have two vectors $(x_1,\ldots,x_m)$ and $(z_1,\ldots,z_m)$ in $\mathbb{R}^{nm}$ (each subcomponent $x_i, z_i$ is in $\mathbb{R}^n$). Given that there exists $k\in\{1,\ldots,m-1\}$ such that $$(\forall i\in\{1,\ldots,k\})\quad z_i=x_i,$$ I want to find an upper bound of the quantity $$f\left(\sum_{i=1}^mw_ix_i\right) - f\left(\sum_{i=1}^mw_iz_i\right),$$ where the upper bound does not depend on the indices $\{1,\ldots,k\}$ where $x_i=z_i$ (ideally on the difference of $f$ evaluated at the averages of $x$ and $z$ over only the indices $\{k+1,\ldots,m\}$). Does such an inequality exist, or is there a nice counterexample?
work thus far: Letting $W_t = \sum_{i=1}^kw_i$ and $W_t^*=\sum_{i=k+1}^mw_i$, I can use convexity (Jensen's inequality) to sort-of isolate the shared terms (underbraced below) $$ \begin{aligned}f\left(\sum_{i=1}^mw_ix_i\right)-\underbrace{W_tf\left(\sum_{i=1}^k\frac{w_i}{W_t}z_i\right)}-W_t^*f\left(\sum_{i=k+1}^m\frac{w_i}{W_t^*}z_i\right) &\leq f\left(\sum_{i=1}^mw_ix_i\right) - f\left(\sum_{i=1}^mw_iz_i\right)\\ &\leq \underbrace{W_tf\left(\sum_{i=1}^k\frac{w_i}{W_t}x_i\right)}+W_t^*f\left(\sum_{i=k+1}^m\frac{w_i}{W_t^*}x_i\right)- f\left(\sum_{i=1}^mw_iz_i\right). \end{aligned} $$ My issue is that these terms do not cancel!