This is a proposition in the Optimization by vector space methods, 7.3 Frechet derivatives, proposition 3:
Let $T$ be a twice Frechet differentiable function on an open domain $D$. Let $x\in D$ and suppose $x+\alpha h \in D$ for all $0\leq \alpha \leq 1$, then show that $$\|T(x+h)-T(x)-T'(x)h\| \leq \frac{1}{2}\|h\|^2 \textsf{sup}_{\alpha}\|T''(x+\alpha h)\| $$
The book says the proof is similar to that of the previous proposition 2 in the book, which states that if $T$ is Frechet differentiable, then $\|T(x+h)-T(x)\| \leq \|h\|\textsf{sup}_{\alpha}\|T'(x+\alpha h)\|$. But I still couldn't figure out the proof after studying the proof of this proposition.
Let $y^* \in Y^*$ be such $\|y^*\| = 1$ and $\langle y^*, T(x+h) - T(x) - T'(x)h\rangle = \|T(x+h) - T(x) - T'(x)h\|$. The existence of such $y^*$ is guaranteed by Hanh-Banach Theorem. Now let $\varphi(t) = \langle y^*, T(x+th) \rangle$. Then $\varphi: \mathbb R \to \mathbb R$ is twice Frechet differentiable. Now by Taylor's Thoerem, we have \begin{align*} \varphi(1) = \varphi(0) + \varphi'(0) + \frac{1}{2}\int_{0}^1 \varphi''(\tau)d \tau. \end{align*} We only need to observe \begin{align*} \varphi''(\tau) = \langle y^*, T''(x+\tau h) [h, h]\rangle. \end{align*} Note here since $T: X \to Y$, $T'(x+\tau h) \in L(X,Y)$ and $T''(x+\tau h) \in L (X, L(X,Y) )$ where $L(X,Y)$ denotes bounded linear maps from $X$ to $Y$.