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. 



Boundedness of ratio of linear functions
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.


