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| $
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|$.