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I have the following question. Let $ g_1,\ldots,g_k: \mathbb{R}^n\rightarrow \mathbb{R} $ be Lipschitz continuous (with respective constants $ L_1,\ldots,L_k>0 $). How can I proove the Lipschitz-continuity of the function $ h(x):=\left(\sum_{i=1}^k |g_i(x)|^d\right)^{\frac{1}{d}} $ for $ d>1$?

I don't know how to proceed here in order to use the Lipschitz condition for the functions $ g_i $: $ |h(x)-h(y)|=\left|\left(\sum_{i=1}^k |g_i(x)|^d\right)^{\frac{1}{d}}-\left(\sum_{i=1}^k |g_i(y)|^d\right)^{\frac{1}{d}} \right| $

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    $\begingroup$ Are you aware of $p$-norms? Of the Minkowski inequality? $\endgroup$ – Omnomnomnom Jun 18 '16 at 11:00
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    $\begingroup$ Hello, of course I am aware of the Minkowski inequality. Unfortunately, I did not manage to apply it here (since I tried to merge the two sums somehow which was not successfull). But now, I saw the trick $g_i(x)=g_i(x)-g_i(y)+g_i(y) $. Afterwards, the Minkowski inequality can be applied and the new Lipschitz constant should result to $ \left( \sum_{i=1}^k L_i^d\right)^{\frac{1}{d}} $, right? $\endgroup$ – Phil Jun 18 '16 at 11:38
  • $\begingroup$ That's right. In particular, we have $$ \left|\left(\sum_{i=1}^k |g_i(x)|^d\right)^{\frac{1}{d}}-\left(\sum_{i=1}^k |g_i(y)|^d\right)^{\frac{1}{d}} \right| \leq \left(\sum_{i=1}^k |g_i(x) - g_i(y)|^d\right)^{\frac{1}{d}} $$ By the triangle inequality $\endgroup$ – Omnomnomnom Jun 18 '16 at 12:16
  • $\begingroup$ Ok, thank you very much. Then this problem is solved! $\endgroup$ – Phil Jun 18 '16 at 12:36
  • $\begingroup$ great! If you want, you can write up an answer and accept it to resolve this question post $\endgroup$ – Omnomnomnom Jun 18 '16 at 14:07
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On the basis of the previous comments I was able to find the answer. As mentioned there, the trick $ g(x)=g(x)-g(y)+g(y) $ and the triangle inequality lead to the following: $ |h(x)-h(y)|=\left|\left(\sum_{i=1}^k |g_i(x)|^d\right)^{\frac{1}{d}}-\left(\sum_{i=1}^k |g_i(y)|^d\right)^{\frac{1}{d}} \right| \leq \left(\sum_{i=1}^k |g_i(x) - g_i(y)|^d\right)^{\frac{1}{d}} \leq \left(\sum_{i=1}^k L_i|x-y|^d\right)^{\frac{1}{d}} = \left( \sum_{i=1}^k L_i^d\right)^{\frac{1}{d}}\cdot |x-y|$.

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