# Existence of Nash Equilibrium

I know there are two proofs of the existence of Nash equilibrium: One by using Kakutani's fixed point theorem in Nash (1950), and the other by using Brouwer's fixed point theorem in Nash (1951).

I saw a third proof but lack of details. Could anyone help me fill in the details and understand the proof? I really appreciate it!

The background information is the following:

Let G denote a game with n players. Let $$S = \prod_n S_n$$ where $$S_n$$ is player n's finite pure strategy set. Let $$G_n: S \to \mathbb R$$ denote the payoff function. Let $$\Sigma_n$$ be player n's set of mixed strategies:

$$\Sigma_{n} = \{ \sigma_{n} \in \mathbb{R}^{S_{n}}_{+} \vert \sum_{s_n \in S_n} \sigma_{n,s_n} = 1 \}$$

Let $$\Sigma = \prod_n \Sigma_n$$. Then $$G_n(\sigma) = \sum_{s \in S} G_n(s)\sigma_{1,s_1} \dots \sigma_{n,s_n}$$.

$$\forall n$$, $$\Sigma_{-n} = \prod_{m \neq n} \Sigma_m$$. Then, $$G_n(\sigma) = \sum_{s_n \in S_n} G_n(s_n, \sigma_{-n})\sigma_{n,s_n}$$

Definition: $$\sigma^{\ast}$$ is a Nash equilibrium of G if $$\forall n$$, $$\tau_n \in \Sigma_n$$, $$G_n(\sigma^{\ast}) \ge G_n(\tau_n, \sigma_{-n})$$

Lemma: $$\sigma^{\ast}$$ is a Nash equilibrium if and only if $$\forall n, s_n, G_n(\sigma^{\ast}) \ge G_n(s_n, \sigma_{-n})$$

Now we have the existence theorem:

$$\mathbf {Theorem: \ G \ has \ a \ Nash \ equilibrium.}$$

The third $$\mathbf {proof}$$ goes like this:

$$\forall n$$, let $$h_n: \Sigma_n \to \mathbb R$$ be a strictly concave continuous function. $$\forall \epsilon \gt 0$$, define $$G^{\epsilon}_n:\Sigma \to \mathbb R$$ by $$G^{\epsilon}_n(\sigma) = G_n(\sigma)+\epsilon h_n(\sigma_n)$$ Then the best response function $${BR}^{\epsilon}_n(\sigma) = \underset{\tau_n \in \Sigma_n}{ArgMax} \ G^{\epsilon}_n(\sigma_{-n}, \tau_n)$$.

Also, $${BR}^{\epsilon} = {BR}^{\epsilon}_1 \times \dots \times {BR}^{\epsilon}_n$$ Apply Brouwer's fixed point theorem to get a fixed point $$\sigma^{\epsilon}$$ of $${BR}^{\epsilon}$$.

Now, take a convergent sequence {$$\sigma^{\epsilon}$$} such that $$\sigma^{\epsilon} \underset{\epsilon \to 0}{\to} \sigma$$ Then $$\sigma$$ is a Nash equilibrium of G.

Basically, I cannot see why Brouwer's fixed point theorem can be applied in that step. Also, in the last step, I cannot understand why is $$\sigma$$ is a Nash equilibrium of G.

• Please add a reference for the third proof. Commented Nov 27, 2022 at 11:14
• Nash, J. F. (1950): Non-cooperative Games. Dissertation, Princeton University, Department of Mathematics. Commented Nov 27, 2022 at 23:44
• @VARulle I took a look; Nash used a slightly different construction in his thesis. Commented Nov 28, 2022 at 14:03
• @MichaelGreinecker, true. Tbh, I didn't even look it up, just had a look at where I remembered having seen it---the "second proof" chapter in homepage.univie.ac.at/josef.hofbauer/00sel.pdf Commented Nov 29, 2022 at 8:55
• @VARulle That makes sense. It is also a nicer reference than the original thesis. Nash's handwriting, used for the formulas, makes me feel good about mine. Commented Nov 29, 2022 at 8:57

The first step is showing that $$BR^\epsilon$$ is a continuous function from the compact convex set $$\Sigma$$ to itself, which amounts to showing $$BR_n^\epsilon$$ is a continuous function for each $$n$$.
First, note that $$G_n$$ and is linear in $$\Sigma_n$$ and, therefore, concave. Since $$h_n$$ is strictly concave, the function $$G_n^\epsilon$$ is strictly concave as the sum of a concave function and a strictly concave function. Now, a strictly concave function can have at most one maximizer, otherwise, there would be a proper convex combination of two maximizers with a strictly higher value. Also, the function $$G_n^\epsilon$$ is continuous on $$\Sigma$$ and $$\Sigma_n$$, a compact set, so a maximum exists. To show the unique maximizer is a continuous function of $$\Sigma_{-n}$$, you can use that the best-reply correspondence is upper hemicontinuous by Berge's maximum theorem, which in the case of a function reduces to continuity; the problem als.
Second, to see that $$\sigma^{\epsilon} \underset{\epsilon \to 0}{\to} \sigma$$ implies that $$\sigma$$ is a Nash equilibrium, suppose it does not. Then there exists some $$n$$ and some $$\tau_n\in\Sigma_n$$ such that $$G_n(\tau_n,\sigma_{-n})-G_n(\sigma_n,\sigma_{-n})>0.$$ The function $$\sigma'\mapsto G_n(\tau_n,\sigma_{-n}')-G_n(\sigma_n',\sigma_{-n}')$$ is continuous, so for some $$\epsilon$$ we must have $$G_n(\tau_n,\sigma_{-n}^\epsilon)-G_n(\sigma_n^\epsilon,\sigma_{-n}^\epsilon)>0,$$ which gives us a contradiction.