Consider a function $f\in\mathcal{C}^2$ with Lipschitz continuous gradient (with constant $L$)- we also assume the function is lowerbounded and has at least one minimum. Let $\{x^k\}_k$ be the sequence generated by Gradient Descent algorithm with initial point $x^0$ and step-size $0<\alpha<2/L$: \begin{equation} x^{k+1}=x^k-\alpha\nabla f (x^k). \end{equation} We know that the sequence will converge to a critical point.
Now consider the new function $\tilde{f}(x)=f(x)+x'Ax$ with some $A\succeq\mathbf{0}$. Let $\{\tilde{x}^k\}_k$ be the sequence generated by Gradient Descent algorithm with the same initial point $\tilde{x}^0=x^0$ and step-size $0<\alpha<2/\tilde{L}$ where $\tilde{L}$ is the Lipschitz constant of $\nabla \tilde{f}$: \begin{equation} \tilde{x}^{k+1}=\tilde{x}^k-\alpha\nabla \tilde{f} (\tilde{x}^k). \end{equation}
Can we prove that $\mathrm{dist}\left(\{\tilde{x}^k\}_k,\{{x}^k\}_k\right)$ is uniformly bounded, for any $A$ and step-size $0<\alpha<2/\tilde{L}$?
I tried to prove it by assuming existence of a compact sublevel set $\mathcal{L}=\{x:f(x)\leq f(x^0)\}$ and using the fact that Gradient Descent generates a decreasing sequence of objective values (implying that the sequence remains in the compact sublevel set). However I cannot prove existence of a set independent of both $A$ and $\alpha$.