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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 $\log n$?

amj
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  • I see that you already saw this question, which I think goes a long way towards answering your question. Can you explain what remains unanswered? And are you considering only the maximum-norm on $\mathbb{R}$, or are you also interested in e.g. the Euclidean norm? – Daniel Fischer Dec 11 '19 at 15:46
  • Any norm would do. I know that the lipschitz continuity holds but I want to show it using a simpler proof technique. – amj Dec 11 '19 at 15:51
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    Simpler than what? I think that using the mean value theorem is the simplest solution. Using the fact that the sum of the partial derivatives (all of which are positive) is $1$ and Hölder's inequality gives $\lvert \operatorname{lse}(\mathbf{x}) - \operatorname{lse}(\mathbf{y})\rvert \leqslant \lVert \mathbf{x} - \mathbf{y}\rVert_p$ for all $p \in [1,\infty]$. (the inequality is strict for $\mathbf{x} \neq \mathbf{y}$ if $p < \infty$ and $n > 1$, but the Lipschitz constant is $1$ nevertheless). – Daniel Fischer Dec 11 '19 at 15:58
  • I meant something on the lines of what I was trying to do above. But nevermind. Can you please explain what you mean by 'strict for $p<\infty$'? – amj Dec 11 '19 at 16:05
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    For $p < \infty$, if $n > 1$ and $\mathbf{x} \neq \mathbf{y}$, then we have $\lvert\operatorname{lse}(\mathbf{x}) - \operatorname{lse}(\mathbf{y})\rvert < \lVert \mathbf{x} - \mathbf{y}\rVert_p$. ($a < b$ is a strict inequality, $a \leqslant b$ a non-strict inequality.) – Daniel Fischer Dec 11 '19 at 16:09

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