Upper bound on Lipschitz continuous hessian Lipschitz continuous hessian for the twice differentiable function $f$ is defined as, for any $x, y$,
$$\|\nabla^2 f(x) - \nabla^2 f(y)\| \leq M\|x-y\|$$ for some postive $M$, how to derive the upper bound of
$|f(y) - f(x) - \nabla f(x)^T(y-x) - \frac{1}{2}(y - x) \nabla^2 f(x)^T (y-x)| \leq \frac{M}{6}\|y-x\|^3$ 
 A: $\newcommand{\d}{\mathrm{d}}$
We will here denote the inner product of $x$ and $y$ by $\langle x, y\rangle.
Since $f$ is twice continuously differentiable, we may use the fact that
$$
\begin{aligned}
f(y) = f(x) + \langle \nabla f(x), y-x\rangle + \tfrac{1}{2} \langle \nabla^2 f(x+\lambda (y-x))(y-x), y-x \rangle
\end{aligned}\tag{1}\label{1}
$$
for some $\lambda\in [0,1]$. We may write this equivalently as (and this is maybe more convenient)
$$
\begin{aligned}
f(y) = f(x) + \langle \nabla f(x), y-x\rangle + \int_0^1 \int_0^\tau \left\langle \nabla^2 f(x+\alpha (y-x))(y-x), y-x\right\rangle \d \alpha\ \d \tau.
\end{aligned}\tag{2}\label{2}
$$
Since $\nabla^2 f$ is $M$-Lipschitz, 
$$
\begin{aligned}
&\|\nabla^2 f(x+\alpha(y-x)) - \nabla^2 f(x)\| \leq M\alpha \|x-y\|, \\
\Leftrightarrow& \nabla^2 f(x+\alpha(y-x)) = \nabla^2 f(x) + H,
\end{aligned}\tag{3}
$$
where $\|H\|\leq M\alpha \|x-y\|$. We may now easily plug (3) into (2) and prove the original inequality. In particular, because of \eqref{2}:
$$
\begin{aligned}
f(y) &= f(x) + \langle \nabla f(x), y-x\rangle + \int_0^1 \int_0^\tau \left\langle \nabla f^2(x)(y-x), y-x\right\rangle \d \alpha\ \d \tau\\
&+ \int_0^1 \int_0^\tau \left\langle H(y-x), y-x\right\rangle \d \alpha\ \d \tau,
\end{aligned}\tag{4}\label{4}
$$
where the first integral in \eqref{4} is
$$
\begin{aligned}
\int_0^1 \int_0^\tau \left\langle \nabla f^2(x)(y-x), y-x\right\rangle \d \alpha\ \d \tau = \tfrac{1}{2}\left\langle \nabla f^2(x)(y-x), y-x\right\rangle
\end{aligned}
$$
and the second integral's absolute value is
$$
\begin{aligned}
\left|\int_0^1 \int_0^\tau \left\langle H(y-x), y-x\right\rangle \d \alpha\ \d \tau\right| &\leq \int_0^1 \int_0^\tau \|H\|\cdot \|y-x\|^2 \d \alpha\ \d \tau\\
&\leq \int_0^1 \int_0^\tau M \alpha \|y-x\|^3 \d \alpha\ \d \tau\\
&=\tfrac{M}{6}\|y-x\|^3.
\end{aligned}
$$
