- #1

Vulture1991

- 7

- 0

\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 $0<\alpha<2/L$:

\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, independent of $A$ and step-size $\alpha$?**

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 $\alpha$.