Merge pull request #26 from Nikoleta-v3/edits_ch

CH: Fixing a minor detail in the Proof of Theorem 1

Author: Nikoleta-v3

si.pdf

@@ -1040,7 +1040,7 @@ \subsection{Proof of Theorem~\ref{theorem:nash_against_pure_self_reactive}: Suff
 \begin{equation}
 f'_{\mathbf{x},j}(t) = \frac{\partial}{\partial t}  f(x_1,\ldots,x_j+t,\ldots,x_k) = \frac{a_j\left( \sum_{i\neq j} b_i x_i\right) - b_j \left(\sum_{i\neq j} a_i x_i\right)}{\big(b_0+b_1x_1+\ldots+b_j(x_j+t)+\ldots+b_kx_k\big)^2}.
 \end{equation}
-Because $f$ is bounded on the entire domain, this expression for $f'_{\mathbf{x},j}(t)$ is finite. It hence follows fthat the denominator of $f'_{\mathbf{x},j}(t)$ is always positive, and that the numerator is independent of $t$. Thus, depending on the sign of the numerator, $f'_{\mathbf{x},j}(t)$ is either monotonically increasing, monotonically decreasing, or constant. 
+Because $f$ is bounded on the entire domain, the denominator in this expression for $f'_{\mathbf{x},j}(t)$ is strictly positive. Moreover, we note that the numerator is independent of $t$. Thus, depending on the sign of the numerator, $f'_{\mathbf{x},j}(t)$ is either monotonically increasing, monotonically decreasing, or constant. 
 \end{proof}
 
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