I came across these steps when I study my math lectures:
$$x_k = \operatorname*{argmin}_x \langle \nabla f(x_{k-1}),x\rangle + \frac{\eta_k} 2 \| x - x_{k-1} \|_2^2 $$ $$x_k = \operatorname*{argmin}_x \frac{\eta_k} 2 \left\|x - x_{k-1} + \frac 1 {\eta_k} \nabla f(x_{k-1})\right\|_2^2$$ I cannot find how the first step reduced to the second step. I tried using the relation: $$\|x\|_2^2 = \langle x,x\rangle$$
But I still cannot find the relation. Any help would be appreciated.
I don't know what point of the second formula is as you can compute a solution to the first explicitly.
If you have $\min_x \langle g,x\rangle + {\alpha \over 2} \|x-y\|^2$ (Euclidean norm) then the solution is $g+ \alpha(x-y)=0$ and hence $x= y- {1\over \alpha} g$.
Substituting values for the above, we see that $x_k = x_{k-1}+{1 \over \eta_k} \nabla f(x_{k-1})$.
Hence it is true that $x_k = \arg\min_x \| x-x_{k-1}+{1 \over \eta_k} \nabla f(x_{k-1}) \|^2 $, but it is a strange leap, unless the author is using this to justify some subsequent intuition or step.