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.

  • $\begingroup$ Please add a reference for the third proof. $\endgroup$ Nov 27, 2022 at 11:14
  • $\begingroup$ Nash, J. F. (1950): Non-cooperative Games. Dissertation, Princeton University, Department of Mathematics. $\endgroup$
    – VARulle
    Nov 27, 2022 at 23:44
  • $\begingroup$ @VARulle I took a look; Nash used a slightly different construction in his thesis. $\endgroup$ Nov 28, 2022 at 14:03
  • 1
    $\begingroup$ @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 $\endgroup$
    – VARulle
    Nov 29, 2022 at 8:55
  • $\begingroup$ @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. $\endgroup$ Nov 29, 2022 at 8:57

1 Answer 1


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.


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