A Log-Sum-Exponent function of $\mathbf{x}$ is given by: \begin{equation*} \text{lse}(\mathbf{x})=\log\sum_{i=1}^n\exp(x_i) \end{equation*} I have read in literature that it is a contraction but I am not sure how to prove it. My progress so far is:
Let $\mathbf{x}'$ and $\mathbf{x}''$ be two vectors, then \begin{eqnarray*} \left|\text{lse}(\mathbf{x}')-\text{lse}(\mathbf{x}'')\right|&\leq&\left|\max(\mathbf{x'})+\log n-\max(\mathbf{x''})\right|\\ &{\leq}&\|\mathbf{x}'-\mathbf{x''}\|_\infty+\log n \end{eqnarray*} Is there a cleaner approach where I dont get this extra factor of $\log n$?