Definition 1 Let us fix n\in \mathbb{N}. We say that a function z:int{\mathrm{\Delta}}^{n}\to {\mathbb{R}}^{n}, z(p)=({z}_{1}(p),\dots ,{z}_{n}(p)), is an excess demand function, if it satisfies the following conditions:

1.
z is continuous on int{\mathrm{\Delta}}^{n},

2.
Walras’ law holds, that is, pz(p)=0 for p\in int{\mathrm{\Delta}}^{n},

3.
the boundary condition holds: if {p}^{j}\in int{\mathrm{\Delta}}^{n}, j\in \mathbb{N}, {lim}_{j\to +\mathrm{\infty}}{p}^{j}=p\in \partial ({\mathrm{\Delta}}^{n}) and {p}_{i}=0, i\in [n], then {lim}_{j\to +\mathrm{\infty}}{z}_{i}({p}^{j})=+\mathrm{\infty},

4.
z is bounded from below: {inf}_{p\in int{\mathrm{\Delta}}^{n}}{z}_{i}(p)>\mathrm{\infty}, i\in [n].
Definition 2 Let z:int{\mathrm{\Delta}}^{n}\to {\mathbb{R}}^{n} be an excess demand function, n\in \mathbb{N}. A point p\in int{\mathrm{\Delta}}^{n} is called an equilibrium point for z, if z(p)=0.
The main goal of the paper is to give a new proof of the fact that for each excess demand function there exists an equilibrium point. First, we are going to characterize the behavior of z near the (relative) boundary of its domain, which is crucial for the theorem to follow. The intuition for the lemma below is as follows: if the price {p}_{i} of a good i is low (in comparison to some other price  prices are standardized; they sum up to 1) then the demand significantly exceeds the supply of that good; if the price {p}_{i} is (relatively) high  so all the other prices are low  then the demand for the i th good is considerably less than its supply.
Lemma 2 Let z:int{\mathrm{\Delta}}^{n}\to {\mathbb{R}}^{n} be an excess demand function. Then there exists {\epsilon}_{1}>0 such that for i\in [n] and p\in int{\mathrm{\Delta}}^{n} we have
({p}_{i}\le {\epsilon}_{1}\Rightarrow {z}_{i}(p)>0)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}({p}_{i}\ge 1{\epsilon}_{1}\Rightarrow {z}_{i}(p)<0).
Proof Suppose that the former implication is not true. Then there exist i\in [n] and a sequence {p}^{j}\in int{\mathrm{\Delta}}^{n}, j\in \mathbb{N}:{lim}_{j\to +\mathrm{\infty}}{p}^{j}=p, {p}_{i}=0, and {lim}_{j\to +\mathrm{\infty}}{z}_{i}({p}^{j})\le 0 , which contradicts the boundary condition. This implies that there exists {\epsilon}_{1}>0 for which the just considered implication is true and without loss of generality we can assume that {\epsilon}_{1}<1{\epsilon}_{1}. To prove the latter implication, observe that {p}_{i}\ge 1{\epsilon}_{1} implies {p}_{{i}^{\prime}}\le {\epsilon}_{1}, i\ne {i}^{\prime}, so the first implication guarantees that {z}_{{i}^{\prime}}(p)>0, {i}^{\prime}\ne i. Now, from Walras’ law, we get 0<{\sum}_{{i}^{\prime}\ne i}{p}_{{i}^{\prime}}{z}_{{i}^{\prime}}(p)={p}_{i}{z}_{i}(p), and {z}_{i}(p)<0 is satisfied. □
Lemma 3 Let z and {\epsilon}_{1} be as in Lemma 2. Let {S}_{1}:=\{p\in int{\mathrm{\Delta}}^{n}:{p}_{n}\in (0,1{\epsilon}_{1}/2]\} and define the function \tilde{z}:int{\mathrm{\Delta}}^{n}\to {\mathbb{R}}^{n1} as follows:
\mathrm{\forall}p\in int{\mathrm{\Delta}}^{n}\phantom{\rule{1em}{0ex}}\tilde{z}(p):={((1{p}_{n}){z}_{i}(p)+{p}_{n}{z}_{n}(p))}_{i=1}^{n1}.
(1)
Then

1.
\tilde{z} is continuous,

2.
\tilde{z} is bounded from below: {inf}_{p\in int{\mathrm{\Delta}}^{n}}{\tilde{z}}_{i}(p)>\mathrm{\infty}, i\in [n1],

3.
{p}_{1}{\tilde{z}}_{1}(p)+\cdots +{p}_{n1}{\tilde{z}}_{n1}(p)=0 for p\in int{\mathrm{\Delta}}^{n},

4.
if {p}^{j}\in {S}_{1}, j\in \mathbb{N}, {lim}_{j\to +\mathrm{\infty}}{p}^{j}=p\in \partial ({\mathrm{\Delta}}^{n}) and {p}_{i}=0, i\in [n1], then {lim}_{j\to +\mathrm{\infty}}{\tilde{z}}_{i}({p}^{j})=+\mathrm{\infty},

5.
\mathrm{\exists}{\epsilon}_{2}\in (0,{\epsilon}_{1}/2] \mathrm{\forall}p\in {S}_{1} \mathrm{\forall}i\in [n1]:
({p}_{i}\le {\epsilon}_{2}\Rightarrow {\tilde{z}}_{i}(p)>0)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}({p}_{i}\ge 1{\epsilon}_{2}\Rightarrow {\tilde{z}}_{i}(p)<0).
Proof The continuity of \tilde{z} is obvious. The boundedness from below of \tilde{z} stems from the fact that z is bounded from below and the weights {p}_{n}, 1{p}_{n}, are positive and less than 1 for all {p}_{n}\in (0,1). The following equalities show that property (3) is met:
If {p}^{j}\in {S}_{1}, j\in \mathbb{N}, converges to a point p with {p}_{i}=0 for some i\in [n1] then (1{p}_{n}^{j}){z}_{i}({p}^{j}) diverges to +∞ and since the product {p}_{n}^{j}{z}_{n}({p}^{j}) is bounded from below it holds: {lim}_{j\to +\mathrm{\infty}}{\tilde{z}}_{i}({p}^{j})=+\mathrm{\infty}. To prove that (5) is true it suffices to observe that for p\in {S}_{1} we have 1{p}_{n}\ge {\epsilon}_{1}/2 and to proceed as in the proof of Lemma 2 with \tilde{z} in place of z. □
The formula used to define the function \tilde{z} resembles the linear homotopy between functions
{((1{\epsilon}_{1}/2){z}_{1}(\cdot ,{\epsilon}_{1}/2)+({\epsilon}_{1}/2){z}_{n}(\cdot ,{\epsilon}_{1}/2))}_{i=1}^{n1},
and
{(({\epsilon}_{1}/2){z}_{1}(\cdot ,1{\epsilon}_{1}/2)+(1{\epsilon}_{1}/2){z}_{n}(\cdot ,1{\epsilon}_{1}/2))}_{i=1}^{n1};
just put t in place of {p}_{n}, assume that t changes from {\epsilon}_{1}/2 through 1{\epsilon}_{1}/2 and the ‘homotopy’ is
H({p}_{1},\dots ,{p}_{n1},t):={((1t){z}_{i}(\cdot ,t)+t{z}_{n}(\cdot ,t))}_{i=1}^{n1}.
But H is not a homotopy since the domain of H(\cdot ,t) changes as t changes.
The important thing which Lemma 3 reveals is that at each fixed {p}_{n}\in (0,1) the function \tilde{z}(\cdot ,{p}_{n}) is an excess demand function defined on a simplex of dimension n2 instead of n1.^{i}
Now suppose that {\epsilon}_{1} and {\epsilon}_{2} satisfy the statement of Lemma 3 and let for i\in [n]:
{\overline{e}}^{i}:=(\frac{{\epsilon}_{2}}{n1},\dots ,\frac{{\epsilon}_{2}}{n1},\underset{i\text{th coordinate}}{\underset{\u23df}{1{\epsilon}_{2}}},\frac{{\epsilon}_{2}}{n1},\dots ,\frac{{\epsilon}_{2}}{n1})\in int{\mathrm{\Delta}}^{n}.
(2)
We can assume that the vectors {\overline{e}}^{i}, i\in [n], are linearly independent; it suffices to take sufficiently small {\epsilon}_{2}>0. The set {S}_{2}:=\u3008{\overline{e}}^{i}:i\in [n]\u3009\subset int{\mathrm{\Delta}}^{n} is an (n1)simplex with the vertices {\overline{e}}^{i}, i\in [n]. If p\in {S}_{2}\cap {S}_{1}, then {p}_{i}\in [{\epsilon}_{2}/(n1),1{\epsilon}_{2}], i\in [n1] and if {\alpha}_{i}^{p}=0 (i.e., {p}_{i}={\epsilon}_{2}/(n1)<{\epsilon}_{1}/2) then {\tilde{z}}_{i}(p)>0; similarly, if {\alpha}_{i}^{p}=1 (i.e. {p}_{i}=1{\epsilon}_{2}>1{\epsilon}_{1}/2) then {\tilde{z}}_{i}(p)<0. Moreover, if p\in {S}_{2} and {p}_{n}\ge 1{\epsilon}_{1} then {z}_{n}(p)<0 and if {p}_{n}\le {\epsilon}_{1} then {z}_{n}(p)>0 (see Lemma 2). We are now in a position to prove the main result of the paper.
Theorem Let z be as in Lemma 3. For each \epsilon >0 there exists p\in int{\mathrm{\Delta}}^{n}:{z}_{i}(p)\le \epsilon ,i\in [n].
Proof If n=1, then there is nothing to prove: int{\mathrm{\Delta}}^{1}=\{1\}\subset \mathbb{R}, and by Walras’ law, z(p)=0 at p=1. Suppose that n\ge 2. Let us fix \epsilon >0 and define {\epsilon}^{\prime}:=\epsilon {\epsilon}_{1}, where {\epsilon}_{1} comes from Lemma 2. Let also {S}_{1} be as in the hypothesis of Lemma 3 and let {S}_{2} be the (n1)simplex with vertices given by (2). By the continuity of the restriction of \tilde{z} to the compact set {S}_{2}, there exists \delta >0 such that if p,{p}^{\prime}\in {S}_{2} and p{p}^{\prime}<\delta, then \tilde{z}(p)\tilde{z}({p}^{\prime})<{\epsilon}^{\prime}. Choose an integer m\ge 2 for which all simplices in K({S}_{2},m) have diameter less than min\{\delta ,{\epsilon}_{1}/4\}. Let {k}_{1} denote the smallest integer in [m] for which (1\frac{{k}_{1}}{m})\frac{{\epsilon}_{2}}{n1}+(1{\epsilon}_{2})\frac{{k}_{1}}{m}\ge 1\frac{{\epsilon}_{1}}{2}  this ensures that a point p\in {S}_{2} whose last barycentric coordinate in {S}_{2} is greater than or equal to {k}_{1}/m satisfies {p}_{n}\ge 1{\epsilon}_{1}/2. To justify this statement, observe that 1{\epsilon}_{2}\frac{{\epsilon}_{2}}{n1}\ge 12{\epsilon}_{2}\ge 1{\epsilon}_{1}>0 and {\alpha}_{n}^{p}\ge {k}_{1}/m entail
\begin{array}{rcl}{p}_{n}& =& (1{\alpha}_{n}^{p})\frac{{\epsilon}_{2}}{n1}+(1{\epsilon}_{2}){\alpha}_{n}^{p}=\frac{{\epsilon}_{2}}{n1}+(1{\epsilon}_{2}\frac{{\epsilon}_{2}}{n1}){\alpha}_{n}^{p}\\ \ge & \frac{{\epsilon}_{2}}{n1}+(1{\epsilon}_{2}\frac{{\epsilon}_{2}}{n1})\frac{{k}_{1}}{m}=(1\frac{{k}_{1}}{m})\frac{{\epsilon}_{2}}{n1}+(1{\epsilon}_{2})\frac{{k}_{1}}{m}\\ \ge & 1{\epsilon}_{1}/2.\end{array}
The minimality of {k}_{1} assures that for any nonnegative integer k<{k}_{1} if p\in {S}_{2} and {\alpha}_{n}^{p}\le k/m, then {p}_{n}<1{\epsilon}_{1}/2 and p\in {S}_{1}; the latter implies that the claim of Lemma 3(5) applies to p. Notice that if p\in {S}_{2} and {p}_{n}\ge 1{\epsilon}_{1}/2 then {z}_{n}(p)<0 and if {p}_{n}<{\epsilon}_{1}/2 then {z}_{n}(p)>0 (see Lemma 2). Let us define a function l from the set of vertices V({S}_{2},m) to [n]\cup \{0\} as follows:^{j}
l(p)=\{\begin{array}{cc}n,\hfill & \text{if}{\alpha}_{n}^{p}=1,\hfill \\ 0,\hfill & \text{if}{\alpha}_{n}^{p}=0,\hfill \\ min\{i\in [n1]:{\alpha}_{i}^{p}0\},\hfill & \text{if}1{\alpha}_{n}^{p}\ge {k}_{1}/m,\hfill \\ min\{i\in [n1]:{\tilde{z}}_{i}(p)\le 0\},\hfill & \text{if}{k}_{1}/m{\alpha}_{n}^{p}0,\hfill \end{array}
(3)
where \tilde{z} is defined in (1). For i\in [n1], if p\in V({S}_{2},m), 1>{\alpha}_{n}^{p}\ge {k}_{1}/m, and {\alpha}_{i}^{p}=0 then it is clear that l(p)\ne i, since if l(p)=i, then we would obtain {\alpha}_{i}^{p}>0. Assume that p\in V({S}_{2},m) and 0<{\alpha}_{n}^{p}<{k}_{1}/m. Since p\in int{\mathrm{\Delta}}^{n}, Lemma 3(3) ensures that {\tilde{z}}_{i}(p)\le 0 for some i\in [n1]  so, l(p) is well defined. Moreover, {\alpha}_{n}^{p}<{k}_{1}/m implies {\alpha}_{n}^{p}=k/m for some nonnegative integer k such that k<{k}_{1} and, therefore, due to Lemma 3(5), it holds that {\tilde{z}}_{i}(p)>0 for {\alpha}_{i}^{p}=0 from which we obtain l(p)\ne i whenever {\alpha}_{i}^{p}=0. Therefore, the assumptions of the combinatorial Lemma 1 are satisfied. Hence, there exists a sequence of simplices {\sigma}_{1},\dots ,{\sigma}_{J} in K({S}_{2},m) such that {\sigma}_{j} and {\sigma}_{j+1} are adjacent and n\in l({\sigma}_{1}), 0\in l({\sigma}_{J}), [n1]\subset l({\sigma}_{j}), j\in [J]. There exists the first simplex in that sequence, call it {\sigma}_{{j}_{1}}, such that for all j>{j}_{1} the last barycentric coordinate of all vertices of {\sigma}_{j} in {S}_{2} are less than {k}_{1}/m. Simplices {\sigma}_{{j}_{1}}\cap {\sigma}_{{j}_{1}+1} are adjacent, i.e. they share an (n2)face, and in other words, they differ by one vertex only. By the choice of {j}_{1} all vertices p\in V({\sigma}_{{j}_{1}+1}) satisfy {\alpha}_{n}^{p}<{k}_{1}/m, and there is a vertex \overline{p}\in V({\sigma}_{{j}_{1}})\mathrm{\setminus}V({\sigma}_{{j}_{1}+1}) such that {\alpha}_{n}^{\overline{p}}\ge {k}_{1}/m. Now, the adjacency of {\sigma}_{{j}_{1}} and {\sigma}_{{j}_{1}+1}, the fact that all simplices in K(S,m) have diameters less than \epsilon /4 and the inequality {\overline{p}}_{n}\ge 1{\epsilon}_{1}/2 entail that {p}_{n}\ge 1{\epsilon}_{1} for p\in V({\sigma}_{{j}_{1}+1}), which implies {z}_{n}(p)<0 for p\in V({\sigma}_{{j}_{1}+1}). Reasoning analogously, we get for the last simplex, {\sigma}_{J}, that it holds: {z}_{n}(p)>0, p\in V({\sigma}_{J}). By the choice of {j}_{1}, all simplices {\sigma}_{j}, j\ge {j}_{1}+1, are contained in {S}_{1}\cap {S}_{2}. Moreover, their diameters are less than δ so p,{p}^{\prime}\in {\sigma}_{j}, j\ge {j}_{1}+1, implies {\tilde{z}}_{i}(p){\tilde{z}}_{i}({p}^{\prime})\le {\epsilon}^{\prime}, i\in [n1]. Since {\bigcup}_{j\ge {j}_{1}}{\sigma}_{j} is (arcwise) connected and V({\sigma}_{{j}_{1}})\cap {z}_{n}^{1}((\mathrm{\infty},0))\ne \mathrm{\varnothing} and V({\sigma}_{J})\cap {z}_{n}^{1}((0,+\mathrm{\infty}))\ne \mathrm{\varnothing} then by the continuity of \tilde{z} there exists a simplex {\sigma}_{{j}_{2}},{j}_{2}\ge {j}_{1}+1:0\in {z}_{n}({\sigma}_{{j}_{2}}). Let p\in {\sigma}_{{j}_{2}}:{z}_{n}(p)=0. So p{p}^{\prime}<\delta, {p}^{\prime}\in V({\sigma}_{{j}_{2}}).Since for each i, there exists a vertex {p}^{i} of {\sigma}_{{j}_{2}} such that {\tilde{z}}_{i}({p}^{i})\le 0 (by the inclusion [n1]\subset l({\sigma}_{{j}_{2}})), (1{p}_{n}){z}_{i}(p)={\tilde{z}}_{i}(p)\le {\tilde{z}}_{i}({p}^{i})+{\epsilon}^{\prime}\le {\epsilon}^{\prime}, i\in [n1]. Further, {z}_{i}(p)\le \frac{{\epsilon}^{\prime}}{(1{p}_{n})}\le \frac{{\epsilon}^{\prime}}{{\epsilon}_{1}}=\epsilon, i\in [n1], since {p}_{n}\in [{\epsilon}_{1},1{\epsilon}_{1}], if {z}_{n}(p)=0, due to Lemma 2. We have found a point p\in int{\mathrm{\Delta}}^{n}:{z}_{i}(p)\le \epsilon, i\in [n], which ends the proof. □
Figure 2 illustrates the proof.
Corollary Let z be as in the above theorem. There exists an equilibrium point for z.
Proof Let {\epsilon}^{q}>0, q\in \mathbb{N}, be a sequence converging to 0. In view of the proof of the theorem, for each q\in \mathbb{N} there exists a point {p}^{q}\in {S}_{2} such that {z}_{i}({p}^{q})\le {\epsilon}^{q}, i\in [n]. The BolzanoWeierstrass theorem and compactness of {S}_{2} imply that there exists a convergent subsequence {p}^{{q}^{\prime}} of {p}^{q}, such that {lim}_{{q}^{\prime}\to +\mathrm{\infty}}{p}^{{q}^{\prime}}=p\in {S}_{2}. From the continuity of z, it follows that {z}_{i}(p)\le 0, for i\in [n]. Since p\in {S}_{2}\subset int{\mathrm{\Delta}}^{n}, {p}_{i}>0, i\in [n]. Walras’ law ensures that z(p)=0. □