Boundedness of ratio of linear functions Consider the function 
\begin{eqnarray}
f(x_1,x_2,\cdots, x_n) = \frac{\sum_{i}^{n}a_ix_i}{\sum_{i}^{n}b_ix_i}, 
\end{eqnarray}
over the set $S = \{x := (x_1,x_2,\cdots, x_n):-1 \leq x_i \leq 1,\; \forall 1 \leq i \leq n,\; \sum_{i}^{n}b_ix_i \neq 0, \sum_{i=1}^{n}c_i^{k}x_i = 0, 1 \leq x_i \leq m,  \}$. It is given that the vector $b:= (b_1,b_2,\cdots,b_n)$ does not lie in the span of the vectors $c^{k}, \forall 1 \leq k \leq m$.  Assume that the constants $(a_i, b_i, c_i^{k} ),  1 \leq i \leq n$,, $1 \leq k \leq m$ are all strictly positive. Is the function $f(x)$ bounded over the set $S$. I am trying to guess sufficient conditions under on the constraint set under which this ratio of affine functions is bounded. 
Without the additional constraints $\sum_{i=1}^{n}c_i^{k}x_i = 0$ which are linearly independent of the vector $b$ , the function is does not seem to be bounded.   
 A: Your function $f$ is bounded on $S$ if and only if $a$ is in the span of $b$ and the $c^k$s.
First observation: You can drop the vectors $c^k$ (or take $m=0$). You only consider the function $f$ in the orthogonal complement $C$ of the span of $c^k$s. If $a^C$ and $b^C$ are orthogonal projections of $a$ and $b$ to $C$, then $f(x)=\langle a^C,x\rangle/\langle b^C,x\rangle$ for every $x\in C$.
The fact that $b\notin\text{span}\{c^1,\dots,c^k\}$ means that $b^C\neq0$. It can happen that $a^C=0$. I will thus assume that there are no vectors $c^k$, $b\neq0$ and $a$ is any vector.
Let $T$ denote the unit sphere in $\mathbb R^n$ (I'd usually use $S$ or $S^{n-1}$, but I want to avoid a notation conflict). It suffices to study whether $f$ is bounded on $S\cap T$ since $f(tx)=f(x)$ for all $t\neq0$. Now $S\cap T=T\setminus B$.
Claim:
The function $f$ is bounded on $S\cap T$ if and only if $a$ is a scalar multiple of $b$.
Proof:
If $a$ is a scalar multiple of $b$, the function $f$ is trivially bounded.
Suppose $a$ is not a scalar multiple of $b$. Let $D$ be the space of vectors orthogonal to both $a$ and $b$. Since $a$ and $b$ are linearly independent, $D$ has codimension two. Let $a^\perp$ be a vector such that $\langle a^\perp,a\rangle=0$ and $\langle a^\perp,b\rangle=1$. Let $b^\perp$ satisfy $\langle b^\perp,b\rangle=0$ and $\langle b^\perp,a\rangle=1$. Now any $x\in\mathbb R^n$ can be written uniquely as $x=ta^\perp+sb^\perp+d$ for $t,s\in\mathbb R$ and $d\in D$.
We get $f(ta^\perp+sb^\perp+d)=s/t$. Now $S\cap T=\{x=ta^\perp+sb^\perp+d;|x|=1,t\neq0\}$, so we can keep $s$ fixed at some nonzero value and take $t\to0$ without leaving $S\cap T$ (here $d$ has to depend on $s$ and $t$). Thus $f$ is unbounded.
A: One can write $a=(\lambda b + c) + \bar{a}$ where $\lambda \in \mathbb{R}$, $c\in span(c^1,\ldots,c^m)$, and $\bar{a} \in \{b,c^1,\ldots,c^m\}^\perp$.
Also, One can write $b=d+\bar{b}$ where $d\in span(c^1,\ldots,c^m)$, and $\bar{b} \in \{c^1,\ldots,c^m\}^\perp$ (Note that if $\bar{b}=0$ then $S=\emptyset$, thus we can suppose that $\bar{b} \neq 0$).
Now, for each $x\in S$ we have
$$f(x)=\lambda + \frac{\langle \bar{a},x \rangle }{\langle \bar{b},x \rangle}. $$
If $\bar{a}=0$ (i.e. $a \in span\{b,c^1,\ldots,c^m\}$), then $f$ is a constant function on $S$. therefore, $f$ is bounded.
Otherwise, If  $\bar{a}\neq 0$, then for all $t \in \mathbb{R}$ set $x^t=\bar{a} + t \bar{b}$. For sufficiently small values of $t \in \mathbb{R}$ We have $x^t\in S$, and
$$f(x^t)=\lambda + \frac{\langle \bar{a},\bar{a} \rangle }{t\langle \bar{b},\bar{b} \rangle}, $$
So, $|f(x^t)|\to \infty $ as $t\to 0$. Therefore f is not bounded.
