I need to bound the trace of a matrix.
Can anyone help me see where the following bound comes from, I have spent ages trying to do the calculation but cant see it ! I will provide my working so far :
Let $0<\epsilon <1$ and $c>0$ be constants, and $\|\cdot\|$ the usual Euclidean norm.
Let $\phi:\mathbb{R}^d\to\mathbb{R}$, $\phi(y)=(c+\|y\|^2)^{1/\epsilon}$, denote its matrix of second derivatives be $\nabla^2\phi(y)\in \mathbb{R}^{d\times d}$. Note we know the derivatives :
$$\nabla \phi(y)=\frac{2\phi(y)}{\epsilon(c+\|y\|^2)}y$$.
$$\frac{\partial \phi(y)}{\partial y_i} = \frac{2}{\epsilon}\Big( \frac{\phi(y)}{(c+\|y\|^2)}+2(\frac{1}{\epsilon}-1)y_i^2\frac{\phi(y)}{(c+\|y\|^2)^{-2}} \Big) $$
$$\frac{\partial^2 \phi(y)}{\partial y_iy_j} =\frac{2}{\epsilon}2(\frac{1}{\epsilon}-1)y_iy_j\frac{\phi(y)}{(c+\|y\|^2)^2} $$
Now here comes $\textbf{the question :}$
Let $\sigma \in \mathbb{R}^{d'\times d}$ with the Frobenius norm $\|\sigma\|_F\leq M(c+\|y\|^2)^{1/2}$, how do we get the bound
$$ \frac{\epsilon}{2}\text{Trace}(\sigma \nabla^2\phi(y) \sigma')\leq \frac{M^2(d+2)}{\epsilon}\phi(y). $$
My work : $\text{Trace}(\sigma \nabla^2\phi(y) \sigma')=\text{Trace}(\sigma \sigma'\nabla^2\phi(y))\leq \|\sigma \|_F^2 \|\nabla^2 \phi(y)\|_F\leq M^2(c+\|y\|^2)\|\nabla^2 \phi(y)\|_F$. Now i'm stuck.