I have a vector $R$ which from previous work I found it to be equal to $\frac{1}{2}f''(x+a(y-x))(y-x)^2b$ where $a\in(0,b)$ and $x,y\in\Re^d$.
I am also given that $\|R\|_2\le$$L\|x-y\|^2_2$ and I have to find this constant L. Therefore I decided to find an upper bound on R but I do not know how. I need some help
To answer this question, since $f$ is $C^2$ class, you can use development of Taylor and the mean value theorem for functions of several variables taking values in $\mathbb{R}^d$. But it would be too much work. I believe the solution below is easier.
Hint: The secret is to understand that the second derivative of $f:\mathbb{R}^d\to\mathbb{R}^d$ as a bililinear transformation $$ \mathbb{R}^d\times \mathbb{R}^d\ni(u.v)\longmapsto D^{(2)}f(p)\bullet(u,v) \in \mathbb{R}^d $$ and then applying an important property of the matrices: if $v$ is a column vector and $M$ a square matrix then $\|Mv\|\leq \sup_{z\in\mathbb{R}^d}\|Mz\|\cdot \|v\|$. Then \begin{align} \|bD^{(2)}f(p)\bullet (u,v)\| \leq & \left(\sup_{z\in\mathbb{R}^d}\frac{\|bD^{(2)}f(p)\bullet (z,v)\|}{\|z\|}\right)\|u\| \\ \leq & \left(\sup_{z\in\mathbb{R}^d}\sup_{w\in\mathbb{R}^d}\frac{\|bD^{(2)}f(p)\bullet (z,w)\|}{\|z\|\|w\|}\right)\|u\|\|v\| \\ \leq & \left(\sup_{z\in\mathbb{R}^d}\sup_{w\in\mathbb{R}^d}\frac{\|D^{(2)}f(p)\bullet (z,w)\|}{\|z\|\|w\|}\right)b\|u\|\|v\| \end{align} Set $p= x+a(y-x)$ and $u=v=(y-x)$ to obtain the desired result.