I am studying strongly convex functions and they mention if $f(x)$ is strongly convex with Lipschitz gradients $L$, which means $\parallel \nabla f(y) - \nabla f(x)\parallel \leq L\parallel x - y \parallel $ , then $ g(x):= f(x) - \frac{m}{2}\parallel{x}\parallel^2 $ has Lipschitz gradients $L-m$. I tried proving that but I am getting L+m instead:
$\parallel \nabla g(y) - \nabla g(x)\parallel =\parallel \nabla f(y) - \nabla f(x) + m(x-y)\parallel \leq \parallel\nabla f(y) - \nabla f(x)\parallel + m\parallel{x-y}\parallel\leq L\parallel x - y \parallel + m\parallel{x-y}\parallel = (L+m)\parallel{x-y}\parallel$
I have crossed checked with couple of sources and they all agree that $L-m$ is the correct bound. I cannot find anything wrong with my logic, can someone please help me see my fault?
To arrive to the answer $L-m$ one simple way is using the fact that: $\parallel \nabla^2f(x) \parallel \leq L$ thus $\parallel\nabla^2q(x)\parallel=\parallel\nabla^2f(x)\parallel- m \leq L-m $