We were shown the following proof for the existence of "Nash equilibrium" in a lecture in our "Game Theory" course. There are two equalities that I don't understand why they are correct, and would appreciate some help understanding them.
The Proof
Define $X=\prod\limits _{i=1}^{N}\Delta\left(A_{i}\right)$. We want to define a continuous function $f$ s.t. $f(x)=x$ iff $x$ is the best response to itself (a.k.a a nash equilibrium). To do that, we'll first define $$c_{i}^{k}\left(x_{i},x_{-i}\right)=\max\left\{ U_{i}\left(e_{i}^{k},x_{-i}\right)-U_{i}\left(x_{i},x_{-i}\right),0\right\} $$ and clearly this function is continuous (it's the max of two continuous functions).
We'll define $f$ in each coordinate separately by defining $f_{i}^{k}\left(x\right)$ for each player $i$ and action $k$: $$f_{i}^{k}\left(x_{i},x_{-i}\right)=\frac{x_{i}^{k}+c_{i}^{k}\left(x_{i},x_{-i}\right)}{1+\sum\limits _{a_{i}^{\ell}\in A_{i}}c_{i}^{\ell}\left(x_{i},x_{-i}\right)}$$ intuitively it gives player $i$ a new probability to choose action $k$. Now each coordinate in $f$ is $f_{i}^{k}$ for all $i,k$ values.
Clearly $f$ is continuous. Therefore, by Brouwer's fixed point theorem, there's some fixed point $x\in X$. We'll prove $x$ is the best response to itself and conclude it's a nash equilibrium.
Assume towards a contradiction that there exists some $i$ s.t. $x_{i}$ isn't the best response to $x_{-i}$ (where $x_i$ refers to the coordinates in $x$ that correspond to the actions of player $i$, and $x_{-i}$ refers to the rest of the coordinates - those of the other players). Therefore
$$U_{i}\left(f_{i}\left(x_{i},x_{-i}\right),x_{-i}\right)-U_{i}\left(x_{i},x_{-i}\right) =\sum_{k=1}^{\left|A_{i}\right|}\left[\frac{x_{i}^{k}+c_{i}^{k}\left(x_{i},x_{-i}\right)}{1+\sum\limits _{\ell=1}^{\left|A_{i}\right|}c_{i}^{\ell}\left(x_{i},x_{-i}\right)}-x_{i}^{k}\right]U_{i}\left(e_{i}^{k},x_{-i}\right) =\sum_{k=1}^{\left|A_{i}\right|}\left[\frac{c_{i}^{k}\left(x_{i},x_{-i}\right)-x_{i}^{k}\sum\limits _{\ell=1}^{\left|A_{i}\right|}c_{i}^{\ell}\left(x_{i},x_{-i}\right)}{1+\sum\limits _{\ell=1}^{\left|A_{i}\right|}c_{i}^{\ell}\left(x_{i},x_{-i}\right)}\right]U_{i}\left(e_{i}^{k},x_{-i}\right) \overset{?}{=}\sum_{k=1}^{\left|A_{i}\right|}\left[\frac{c_{i}^{k}\left(x_{i},x_{-i}\right)\left[U_{i}\left(e_{i}^{k},x_{-i}\right)-U_{i}\left(x_{i},x_{-i}\right)\right]}{1+\sum\limits _{\ell=1}^{\left|A_{i}\right|}c_{i}^{\ell}\left(x_{i},x_{-i}\right)}\right] \overset{??}{=}\sum_{k=1}^{\left|A_{i}\right|}\left[\frac{\left(c_{i}^{k}\left(x_{i},x_{-i}\right)\right)^{2}}{1+\sum\limits _{\ell=1}^{\left|A_{i}\right|}c_{i}^{\ell}\left(x_{i},x_{-i}\right)}\right]>0$$
which is a contradiction to the assumption that $x$ is a fixed point (if it was a fixed point, the difference between $U_{i}\left(f_{i}\left(x_{i},x_{-i}\right),x_{-i}\right),U_{i}\left(x_{i},x_{-i}\right)$ should've been zero (they should've been the same). This finished the proof.
My Questions
The part I don't understand is why are the two equalities, marked with $\overset{?}{=}$ and $\overset{??}{=}$ correct?
- In $\overset{?}{=}$ I don't understand how did the numerator become what it did.
- In $\overset{??}{=}$ I don't understand why $U_{i}\left(e_{i}^{k},x_{-i}\right)-U_{i}\left(x_{i},x_{-i}\right)=c_{i}^{k}\left(x_{i},x_{-i}\right)$, it would've been correct if we used $\max$ between that value and zero (by the definition of $c$), but without that $\max$ I don't see how it's correct