be an $m$-strongly convex function, and let $x, y$ be two points in the domain of $f$, also let $g \in \partial f$ be a subgradient of $f$. We have $$\begin{aligned}
        f(y) \geq f(x) + g^T (y-x) + \frac{m}{2} || x - y ||^2 .
        \label{eq:functions_with_bounded_curvature}
    \end{aligned}$$ After substituting $f = b_t$, $x = w_t$, and $y = w_{t-1}$ in (\[eq:functions\_with\_bounded\_curvature\]), and noting that the zero vector is a subgradient of $b_t$ at $w_t$, $0 \in \partial b_t(w_t)$, we get: $$\begin{aligned}
        \frac{m}{2} \left( b_t(w_{t-1}) - b_t(w_t) \right) \geq || w_t - w_{t-1} ||^2 .
        \label{eq:bounds_with_bounded_curvature}
    \end{aligned}$$ We can replace $b_t(w_{t-1})$ by $v_{t-1}$ in (\[eq:bounds\_with\_bounded\_curvature\]) due to (\[eq:valid\_bounds\_eta\]). By adding the inequalities for $T$ iterations, similar to what we did in the proof of Theorem \[thrm:gmm\_bounds\_touch\_F\], we get: $$\begin{aligned}
        \frac{m}{2} \left( F(w_0) - F(w^*) \right) \geq \sum_{t=1}^T || w_t - w_{t-1} ||^2
        \label{eq:gmm_converges}\\
        \Rightarrow \exists w^\dagger \text{ {\emph{s.t.\ }}} \lim_{t \rightarrow \infty} w_t = w^\dagger.
    \end{aligned}$$ Similar to the proof of Theorem \[thrm:gmm\_bounds\_touch\_F\], if a global minimum does not exist, we can replace $F(w^*)$ with the value of the lower bound of $F$ in (\[eq:gmm\_converges\]).

Let $F$ be continuously differentiable, and $\mathcal{F}$ be a family of strongly convex and smooth functions. 