Lipschitz constant of the convex function $f(x) - \frac{a}{2} |x|^2$ I was going through this blog post 
https://blogs.princeton.edu/imabandit/2013/04/04/orf523-strong-convexity/
It has been mentioned without proof that for a function $f:\mathbb{R}^n\rightarrow\mathbb{R}$ that is Lipschitz continuous with Lipschitz constant $\beta$ and strongly convex with coefficient $\alpha$, the function $g:\mathbb{R}^n\rightarrow\mathbb{R}$ defined as
\begin{equation}g(\mathbf{x})\triangleq f(\mathbf{x}) - \frac{\alpha}{2}\|\mathbf{x}\|^2\end{equation}
is Lipschitz continuous with coefficient $\beta-\alpha$.
I tried proving this as follows
\begin{align}
\nabla g(\mathbf{x}) =& \nabla f(\mathbf{x}) - \alpha \mathbf{x} \\
\|\nabla g(\mathbf{x}) - \nabla g(\mathbf{y})\|^2 = & \|\nabla f(\mathbf{x}) - \nabla f(\mathbf{y})\|^2 + \alpha^2 \|\mathbf{x} -\mathbf{y}\|^2 - 2 \alpha (\nabla f(\mathbf{x}) - \nabla f(\mathbf{y}))^T(\mathbf{x}-\mathbf{y}) \\
\leq & \beta^2 \|\mathbf{x} -\mathbf{y}\|^2 + \alpha^2 \|\mathbf{x} -\mathbf{y}\|^2 - 2 \alpha (\nabla f(\mathbf{x}) - \nabla f(\mathbf{y}))^T(\mathbf{x}-\mathbf{y})
\end{align}
where the last inequality was from the definition of Lipschitz continuity of $f$. I am stuck at the last inequality and am not sure on how to proceed. My hunch is that the last term should become $2\alpha\beta\|\mathbf{x} - \mathbf{y}\|^2$ but am not able to prove that. The problem seems fairly straightforward though.
Any help would be greatly appreciated.
 A: $f$ is $\beta$-smooth : $$ f(x) - f(y) \leq \nabla f(y)^\perp (x-y)
+ \beta/2|x-y|^2 $$
Hence $$ g(x)-g(y) \leq \nabla g(y)^\perp (x-y) +\alpha y^\perp
(x-y) + \beta/2|x-y|^2 - \alpha/2(|x|^2-|y|^2 ) $$
So we have a claim : $$ \alpha y^\perp (x-y) + \beta/2|x-y|^2 -
\alpha/2(|x|^2-|y|^2 ) = (\beta - \alpha )/2|x-y|^2 $$
which is a direct computation. 
A: From the blog, $f$ is $\beta$-smooth and $\alpha$ strongly convex. This means,
$$\
\|\nabla^2f(x)\|\leq \beta \quad \text{and} \quad \nabla^2f(x) \succeq\alpha I
$$
The maximum value of $|\lambda_i|\leq\beta$ from smoothness and property of spectral norm. $\lambda_i\geq\alpha \ \forall i$ from strong convexity.
$$\
\nabla^2g(x)=\nabla^2f(x)-\alpha I \implies \|\nabla^2g(x)\|=\max\text{abs eigenvalue of } \nabla^2f(x)-\alpha I
$$
We use the fact that for each eigenvalue $\lambda$ of $A$ corresponding eigenvalue of $A-\mu I$ is $\lambda-\mu$. Hence, we conclude,
$$\
\|\nabla^2g(x)\|\leq\beta-\alpha
$$
Note: The strong convexity of $f$ is crucial orelse we can only show $\|\nabla^2g(x)\|\leq\beta+\alpha$
