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Let $g: S^n \to R^n$ be a continuous odd function (i.e. $g(-x)=-g(x)$ for all $x$). The Borsuk-Ulam theorem implies that $g$ has a zero, i.e. there is an $x$ such that $g(x)=(0,0,...,0)$.

Suppose $g$ is (1,1,...,1) on the positive orthant (i.e. when all its $n$ arguments are non-negative) and (-1,-1,...,-1) on the negative orthant. Is this true that for every constant $r\in [-1,1]$, there is an $x$ such that $g(x)=(r,r,...,r)$?

For $n=1$, this is obviously true by the intermediate value theorem. Is this true for $n>1$?

Let $g: S^n \to R^n$ be a continuous odd function (i.e. $g(-x)=-g(x)$ for all $x$). The Borsuk-Ulam theorem implies that $g$ has a zero, i.e. there is an $x$ such that $g(x)=(0,0,...,0)$.

Suppose $g$ is (1,1,...,1) on the positive orthant (i.e. when all its $n$ arguments are non-negative) and (-1,-1,...,-1) on the negative orthant. Is this true that for every constant $r\in [-1,1]$, there is an $x$ such that $g(x)=(r,r,...,r)$?

For $n=1$, this is obviously true by the intermediate value theorem. Under what conditions is it true for $n>1$?

The question seems closely related to the Poincare-Miranda theorem, which is a generalization of the IVT to multi-dimensional cubes, but so far I haven't found the connection.

Let $f$ be a continuous function from $S^n$ to $R^n$. The Borsuk-Ulam theorem states that there is an $x$ such that $f(x)=f(-x)$.

Under what conditions is this true that for every constant ''r>0'' there is an $x$ such that $f(x)=r f(-x)$?

1. In general this is of course not true, for example when $f$ is a constant function.

2. What if $f$ is a non-negative function with at least one zero in $S^n$? As commented by @FrancescoPolizzi, this is insufficient, for example when $f(x)=\text{Re}(x)^2$, $f$ has two zeros, but for every $x$ on the unit sphere: $f(x)=f(-x)$.

3. What if there is at least one point $x_0$ for which $f(x_0)=0$ and $f(-x_0) > 0$ componentwise? I think for $n=1$ this is true. Choose a small $\epsilon>0$. Define $g(x)=f(x)/[\epsilon+f(-x)]$. Then $g(x_0)=0$ but $g(-x_0)=f(-x_0)/\epsilon$. This can be made arbitrarily large by a proper selection of $\epsilon$. The function $g$ is continuous, so by the Intermediate Value Theorem it must accept every value between 0 and $\infty$ somewhere between $x_0$ and $-x_0$.

Is this true in general? If not, what conditions should be added to make it true?

Let $g: S^n \to R^n$ be a continuous odd function (i.e. $g(-x)=-g(x)$ for all $x$). The Borsuk-Ulam theorem implies that $g$ has a zero, i.e. there is an $x$ such that $g(x)=(0,0,...,0)$.

Suppose $g$ is (1,1,...,1) on the positive orthant (i.e. when all its $n$ arguments are non-negative) and (-1,-1,...,-1) on the negative orthant. Is this true that for every constant $r\in [-1,1]$, there is an $x$ such that $g(x)=(r,r,...,r)$?

For $n=1$, this is obviously true by the intermediate value theorem. Is this true for $n>1$?

Let $f$ be a continuous function from $S^n$ to $R^n$. The Borsuk-Ulam theorem states that there is an $x$ such that $f(x)=f(-x)$.

Under what conditions is this true that for every constant ''r>0'' there is an $x$ such that $f(x)=r f(-x)$?

1. In general this is of course not true, for example when $f$ is a constant function.

2. What if $f$ is a non-negative function with at least one zero in $S^n$? As commented by @FrancescoPolizzi, this is insufficient, for example when $f(x)=\text{Re}(x)^2$, $f$ has two zeros, but for every $x$ on the unit sphere: $f(x)=f(-x)$.

3. What if there is at least one point $x_0$ for which $f(x_0)=0$ and $f(-x_0) > 0$ componentwise? I think for $n=1$ this is true. Choose a small $\epsilon>0$. Define $g(x)=f(x)/[\epsilon+f(-x)]$. Then $g(x_0)=0$ but $g(-x_0)=f(-x)/\epsilon$. This can be made arbitrarily large by a proper selection of $\epsilon$. The function $g$ is continuous, so by the Intermediate Value Theorem it must accept every value between 0 and $\infty$ somewhere between $x_0$ and $-x_0$.

Is this true in general? If not, what conditions should be added to make it true?

Let $f$ be a continuous function from $S^n$ to $R^n$. The Borsuk-Ulam theorem states that there is an $x$ such that $f(x)=f(-x)$.

Under what conditions is this true that for every constant ''r>0'' there is an $x$ such that $f(x)=r f(-x)$?

1. In general this is of course not true, for example when $f$ is a constant function.

2. What if $f$ is a non-negative function with at least one zero in $S^n$? As commented by @FrancescoPolizzi, this is insufficient, for example when $f(x)=\text{Re}(x)^2$, $f$ has two zeros, but for every $x$ on the unit sphere: $f(x)=f(-x)$.

3. What if there is at least one point $x_0$ for which $f(x_0)=0$ and $f(-x_0) > 0$ componentwise? I think for $n=1$ this is true. Choose a small $\epsilon>0$. Define $g(x)=f(x)/[\epsilon+f(-x)]$. Then $g(x_0)=0$ but $g(-x_0)=f(-x_0)/\epsilon$. This can be made arbitrarily large by a proper selection of $\epsilon$. The function $g$ is continuous, so by the Intermediate Value Theorem it must accept every value between 0 and $\infty$ somewhere between $x_0$ and $-x_0$.

Is this true in general? If not, what conditions should be added to make it true?