 - w_0}_2^2}\\
        \leq& \E{\lrp{\lrn{w_{k\delta} - w_0 - \delta \lrp{\nabla U(w_{k\delta}) - \nabla U(w_0)}}_2 + \delta \lrn{\nabla U(w_0)}_2}^2}\\
        \leq& \lrp{1 + \frac{1}{n}}\E{\lrn{w_{k\delta} - w_0 - \delta \lrp{\nabla U(w_{k\delta}) - \nabla U(w_0)}}_2^2 }\\
        &\quad + (1+n)\delta^2 \E{\lrn{\nabla U(w_0)}_2^2}\\
        \leq& \lrp{1 + \frac{1}{n}} \lrp{1 + \delta L}^2 \E{\lrn{w_{k\delta} - w_0}_2^2} + 2n\delta^2 L^2 \E{\lrn{w_0}_2^2}\\
        \leq& e^{1/n + 2\delta L }\E{\lrn{w_{k\delta} - w_0}_2^2} + 2n\delta^2 L^2 \E{\lrn{w_0}_2^2}
        \end{aligned}$$ where the first inequality is by triangle inequality, the second inequality is by Young’s inequality, the third inequality is by item 1 of Assumption \[ass:U\_properties\].

Inserting the above into gives $$\begin{aligned}
        \E{\lrn{w_{(k+1)\delta} - w_0}_2^2} \leq e^{1/n + 2\delta L }\E{\lrn{w_{k\delta} - w_0}_2^2} + 2n\delta^2 L^2 \E{\lrn{w_0}_2^2} + \delta \beta^2
        \end{aligned}$$

Applying the above recursively for $k=1...n$, we see that $$\begin{aligned}
        & \E{\lrn{w_{n\delta} - w_0}_2^2}\\
        \leq& \sum_{k=0}^{n-1} e^{(n-k) \cdot (1/n + 2\delta L)} \cdot \lrp{2n\delta^2 L^2 \E{\lrn{w_0}_2^2} + \delta \beta^2}\\
        \leq& 16 \lrp{n^2 \delta^2 L^2 \E{\lrn{w_0}_2^2} + n\delta \beta^2}\\
        =& 16 \lrp{T^2 L^2 \E{\lrn{w_0}_2^2} + T \beta^2}
        \end{aligned}$$

[Discretization Bounds]{} \[ss:discretization\_bonuds\]

\[l:vt-wt\] Let $v_{k\delta}$ and $w_{k\delta}$ be as defined in and . Then for any $\delta,n$, such that $T:= n\delta \leq \frac{1}{16L}$, $$\begin{aligned}
            \E{\lrn{v_{T} - w_{T}}_2^2} \leq 8 \lrp{2T^2 L^2 \lrp{T^2L^2 \E{\lrn{v_0}_2^2} + T\beta^2} + T L_\xi^2 \lrp{16 \lrp{T^2 L^2 \E{\lrn{w_0}_2^2} + T \beta^2}}}
            \end{aligned}$$ If we additionally assume that $\E{\lrn{v_0}_2^2} \leq 8 \lrp{R^2 + \beta^2/m}$, $\E{\lrn{w_0}_2^2} \leq 8 \lrp{R^2 + \beta^2/m}$ and $T \leq \frac{\beta^2}{8L^2\lrp{R^2 + \beta^2/m}}$, then $$\begin{aligned}
            \E{\lrn{v_T - 