Analytic Proof of Browser Fixed Point theorem

Let $\displaystyle B^n=\{x=(x_1,\ldots,x_n)\in\mathbb R^n:|x|\le1\},$ and let $f:B^n\to B^n$ be a smooth map. We will prove that $f$ has a fixed point. The case of continuous functions can be obtained by reduction to smooth functions by mollification. Rather than trying to solve $f(x)=x$ directly, we deform the simple map $x\mapsto x$ into the displacement map $x\mapsto x-f(x)$ and construct an analytic signed count of their zeros. The essential fact will be that this signed count cannot change while the deformation avoids zero on the boundary. At the beginning the count is visibly $1$ ; at the end, a nonzero count forces a zero of $x-f(x)$ , hence a fixed point.

Set $\displaystyle F_s(x):=x-sf(x),\quad 0\le s\le1.$

At $s=0$ , the map is $F_0(x)=x$ , with its unique zero at the origin. At $s=1$ , we have $F_1(x)=x-f(x)$ , whose zeros are exactly the fixed points of $f$ . Assume, for contradiction, that $f$ has no fixed point on the boundary $\partial B^n$ ; otherwise there is nothing to prove. Then $F_s(x)$ never vanishes for $x\in\partial B^n$ . Indeed, $F_s(x)=0$ would give $x=sf(x)$ , and hence $1=|x|=s|f(x)|\le s\le1$ . Thus $s=1$ and $x=f(x)$ , contradicting the boundary assumption. Since $[0,1]\times\partial B^n$ is compact, there is therefore a uniform gap

$\displaystyle |F_s(x)|\ge a>0,\quad x\in\partial B^n,\quad 0\le s\le1.$

Choose $0<\varepsilon<\min\{a,1\}$ , and let $\rho_\varepsilon:\mathbb R^n\to[0,\infty)$ be a smooth bump function supported in $|y|<\varepsilon$ , with total integral $1$ . One should think of $\rho_\varepsilon$ as a smooth approximation to a point mass at the origin. The quantity

$\displaystyle Z_s(\varepsilon):=\int_{B^n}\rho_\varepsilon(F_s(x))\det DF_s(x),dx$

is therefore a smoothed signed count of the zeros of $F_s$ . The bump function sees only points for which $F_s(x)$ is close to zero, while the Jacobian determinant records orientation: locally orientation-preserving sheets contribute positively, and orientation-reversing sheets contribute negatively.

We now show directly that $Z_s(\varepsilon)$ is independent of $s$ . Write

$\displaystyle A_{ai}(x,s):=\frac{\partial F_s^a}{\partial x_i}(x),\qquad J_s:=\det A,\qquad V^a(x,s):=\frac{\partial F_s^a}{\partial s}(x)=-f^a(x),$

and let $C_{ai}$ be the cofactor of the entry $A_{ai}$ . The cofactors satisfy the two identities

$\displaystyle \sum_{i=1}^n A_{bi}C_{ai}=J_s\delta_{ab},\qquad \sum_{i=1}^n\frac{\partial C_{ai}}{\partial x_i}=0.$

The first is the standard cofactor identity for determinants. The second follows from cancellation of mixed partial derivatives: when one differentiates a cofactor, each resulting term involving $\partial_{x_i}\partial_{x_j}F_s^a$ is paired with the same term having $i$ and $j$ exchanged; equality of mixed derivatives leaves only the sign change caused by exchanging two determinant columns. For example, if $n=2$ and $F=(u,v)$ , then

$\displaystyle (C_{ai})=\begin{pmatrix}v_y&-v_x\\-u_y&u_x\end{pmatrix},$

and the divergence of each row is zero by $v_{xy}=v_{yx}$ and $u_{xy}=u_{yx}$ .

We now differentiate the signed density itself. Since $R_s(x)=\rho_\varepsilon(F_s(x))$ and $J_s(x)=\det DF_s(x)$ ,

$\displaystyle \frac{\partial}{\partial s}(R_sJ_s)=J_s\frac{\partial R_s}{\partial s}+R_s\frac{\partial J_s}{\partial s}.$

The two terms have different origins. The first records the motion of the bump function in the target as $F_s$ changes, while the second records the change in the local oriented volume factor. By the chain rule,

$\displaystyle \frac{\partial R_s}{\partial s}=\sum_{a=1}^n\frac{\partial\rho_\varepsilon}{\partial y_a}(F_s(x))V^a(x,s),$

where $V^a=\partial_sF_s^a$ . On the other hand, differentiating the determinant one column at a time gives

$\displaystyle \frac{\partial J_s}{\partial s}=\sum_{a,i=1}^n C_{ai}\frac{\partial V^a}{\partial x_i}.$

Thus

$\displaystyle \frac{\partial}{\partial s}(R_sJ_s)=J_s\sum_{a=1}^n\frac{\partial\rho_\varepsilon}{\partial y_a}(F_s)V^a+R_s\sum_{a,i=1}^nC_{ai}\frac{\partial V^a}{\partial x_i}.$

We now look for this expression as an $x$ -divergence. The cofactor identity suggests exactly what must be differentiated: insert the cofactors between the target velocity $V^a$ and a source coordinate direction. Indeed,

$\displaystyle \begin{aligned} \sum_{i=1}^n\frac{\partial}{\partial x_i}\left(R_s\sum_{a=1}^nC_{ai}V^a\right) &=\sum_{a,i=1}^n\frac{\partial R_s}{\partial x_i}C_{ai}V^a+\sum_{a,i=1}^nR_s\frac{\partial C_{ai}}{\partial x_i}V^a+\sum_{a,i=1}^nR_sC_{ai}\frac{\partial V^a}{\partial x_i}. \end{aligned}$

The middle term vanishes by $\sum_i\partial_{x_i}C_{ai}=0$ . For the first term, the chain rule gives

$\displaystyle \frac{\partial R_s}{\partial x_i}=\sum_{b=1}^n\frac{\partial\rho_\varepsilon}{\partial y_b}(F_s)A_{bi}.$

Therefore the cofactor identity $\sum_iA_{bi}C_{ai}=J_s\delta_{ab}$ turns that first term into

$\displaystyle J_s\sum_{a=1}^n\frac{\partial\rho_\varepsilon}{\partial y_a}(F_s)V^a.$

The last term is exactly $R_s\partial_sJ_s$ . Hence the derivative of the signed density is a divergence:

$\displaystyle \frac{\partial}{\partial s}\bigl(\rho_\varepsilon(F_s(x))\det DF_s(x)\bigr)=\sum_{i=1}^n\frac{\partial}{\partial x_i}\left(R_s(x)\sum_{a=1}^nC_{ai}(x,s)V^a(x,s)\right).$

We name the expression inside the divergence:

$\displaystyle G_s^i(x):=R_s(x)\sum_{a=1}^nC_{ai}(x,s)V^a(x,s).$

Thus $\displaystyle \partial_s(R_sJ_s)={\text{div}}G_s$ . The important point is that $G_s$ is forced by the attempt to rewrite the $s$ -derivative of the signed zero-density as a flux through the boundary.

This is the key conservation law: the change of the signed zero-density is a divergence. Integrating over the ball and applying the divergence theorem gives

$\displaystyle \frac{d}{ds}Z_s(\varepsilon)=\int_{\partial B^n}G_s\cdot\nu dS.$

But $|F_s(x)|\ge a>\varepsilon$ on the boundary, so $\rho_\varepsilon(F_s(x))=0$ there. Hence $R_s=0$ , therefore $G_s=0$ , and the boundary integral vanishes. We conclude that

$\displaystyle Z_s(\varepsilon)=Z_0(\varepsilon)\quad\text{for every }0\le s\le1.$

The initial value is immediate: since $F_0(x)=x$ and $DF_0=I$ ,

$\displaystyle Z_0(\varepsilon)=\int_{B^n}\rho_\varepsilon(x) dx=1.$

Therefore

$\displaystyle \int_{B^n}\rho_\varepsilon(x-f(x))\det(I-Df(x))dx=1.$

In particular, the integrand cannot vanish everywhere. Hence there is some $x_\varepsilon\in B^n$ with $\rho_\varepsilon(x_\varepsilon-f(x_\varepsilon))\ne0$ , and the support condition on the bump gives

$\displaystyle |x_\varepsilon-f(x_\varepsilon)|<\varepsilon.$

Taking a sequence $\varepsilon_k\to0$ , compactness of $B^n$ gives a subsequence with $x_{\varepsilon_k}\to x_{*}$ . Continuity of $f$ then implies $x_{\varepsilon_k}-f(x_{\varepsilon_k})\to x_{*}-f(x_{*})$ , while the left-hand side tends to zero. Thus $x_{*}=f(x_{*})$ , proving Brouwer’s fixed-point theorem.

When all fixed points are nondegenerate, so that $\det(I-Df(x_*))\ne0$ at each fixed point, the same argument has a sharper interpretation. For sufficiently small $\varepsilon$ , every fixed point contributes precisely its local orientation sign, and one obtains

$\displaystyle \sum_{f(x)=x}{\text{sgn}}\det(I-Df(x))=1.$

Individual fixed points may be created or annihilated during a deformation, but only in oppositely signed pairs. Their ordinary number can change; their total signed count cannot. It begins at $+1$ for the identity map, and it remains $+1$ because no zero is allowed to cross the boundary.

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