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The discrete analogue of the $L_p$ norm for the mesh function $V$ is $$\|V\|_{l^p(\bar{\Omega}^N)}=\left(\sum_{i=0}^NV_i^p\bar{h}_i\right)^{1/p}$$ where $\bar{\Omega}^N$ is an arbitrary mesh, $\bar{h}_i=(h_{i+1}+h_i)/2$ and $h_i=x_i-x_{i-1}$. The error $e$ of a certain approximation is such that $e_i=-e^{-x_i/\epsilon}$ for $1\leq i\leq N-1$ and 0 for $i=0$ or $i=N$

For any $p$, $1\leq p\leq \infty$ the $l^p(\bar{\Omega}^N)$ norm of the error on a general non-uniform mesh is $$\|e\|_{l^p(\bar{\Omega}^N)}=\left(\sum_{i=1}^{N-1}e^{-px_i/\epsilon}\bar{h}_i\right)^{1/p}$$ My first question is why is this expression $O(N^{-1/P})$.

On a uniform mesh, why is $$\left\vert {e_i}^p\right\vert \leq N\sum_{j=1}^{N-1}\left\vert e_j\right\vert^ph$$ How does this imply $$\|e\|_{\bar{\Omega}^N}\leq N^{1/P}\|e\|_{l^p(\bar{\Omega}^N)}$$

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