The zero(s) of a vector-valued multivariable function

Question

Suppose that $h_{1},cdots,h_{n}in mathbb{R}^{p}$ are $n$ constant vectors, $g(lambda)$ is a
$p-$demensional function of a vector $lambda in mathbb{R}^{p}$ defined as below:
begin{equation}
g(lambda) = sum_{i=1}^{n}frac{h_{i}}{1+lambda^{prime}h_{i}},
end{equation}
which has negative-definite derivative matrix
$frac{partial g}{partial lambda}(lambda)=-sum_{i=1}^{n}frac{h_{i}h_{i}^{prime}}{(1+lambda^{prime}h_{i})^{2}}<0$ ($lambda^{prime}$ denotes the transpose of the vector $lambda$).
My question is whether $g(lambda)=0$ has roots within the convex set ${lambda| 1+lambda^{prime}hi>0 text{  for  } i=1,cdots,n}$. And if there is a unique root.
Thank you a lot for your help!

Vishnu Mangalath · Answer

Here is an answer for a specific case. Suppose that $h_1,ldots h_n$ are linearly independent. Set $c_i(lambda) = 1/(1+lambda'h_i)$ for each $i$. Now if $g(lambda_0)= 0$ in the region $1+lambda'h_i>0$, then $c_i>0$ for each $i$ and so
$$
  g(lambda_0) = sum_{i=1}^nc_i(lambda_0)h_i = 0
$$
which implies that $h_i$ are not linearly independent. Thus in this case there are no roots in the region ${lambda in mathbb{R}^pmid 1+lambda'h_i>0,,forall i}$.

The zero(s) of a vector-valued multivariable function

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