Effect of bounding action space on the set of equilibria

Question

Suppose $N$ players play a game, where each player's action space is $[0,1]$.

Each player has an identical continuous utility function $u:[0,1]\times [0,1]^{N-1}\rightarrow\mathbb{R}$, where the first argument is their own action, and the other arguments are the actions of other players, with $u$ being invariant under re-orderings of the other players.

Suppose that the game has a unique pure, symmetric, Nash equilibrium with each player taking action $a\in (0,1)$.

Now suppose that the game's rules are changed, and players can now only choose actions in the interval $[b,1]$, where $b\in (0,a)$. By revealed preference, each player taking action $a$ is still a Nash equilibrium (if they did not have a profitable deviation before, they do not now).

Is it possible (for some $u$, hopefully not too weird!) that each player taking action $b$ is now an additional Nash equilibrium, even though it was not before? Is it possible that the $b$ equilibrium survives some refinement (e.g. trembling hand) that the $a$ equilibrium does not?

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Now this has been answered, I'll briefly note the particular circumstance I had in mind. The $N$ players are countries, choosing the level of their corporation tax rate. $a$ is the moderate level of corporation taxes (most) countries have set in the absence of international agreement. $b$ is the new minimum level now internationally agreed. Whether the $u$ in the answer looks anything like countries' pay-offs in setting their corporation tax rate is another question!

Answer 1

Let $N=2$ and for $(x,y)$ and $(p,q)$ in $[0,1]^2$ let $d_{p,q}(x,y)$ be the Euclidean distance between $(p,q)$ and $(x,y)$, i.e. $d_{p,q}(x,y)=[(p-x)^2+(q-y)^2]^{1/2}$.

Choose $k>0$ such that $k<a-b$ and $k<b$.

Now define $$u(p,q):=\max\{ -d_{p,q}(a,a),-d_{p,q}(b,b),k-d_{p,q}(b,0),k-d_{p,q}(0,b) \}.$$ This function is continuous as a maximum of continuous functions.

The construction makes $u(p,q)$ negative everywhere except at $(a,a)$ and at $(b,b)$, where it is $0$, and in the (small) $k$-discs around $(b,0)$ and $(0,b)$, where it reaches a maximum value of $k$ at these points.

In the unrestricted case, $(a,a)$ is a symmetric, pure NE, but $(b,b)$ is not, since each player would want to unilaterally deviate to playing $0$. In all other symmetric strategy profiles players would want to unilaterally deviate to either $a$, $b$, or $0$.

In the restricted case, $(b,b)$ is also a pure, symmetric NE, since a downward deviation is no longer possible.

This example can straightforwardly be generalized to $N\ge 3$ players. I leave the refinement part unanswered here but I think this could also be constructed in a similar way.