Let for $k = 1, \dots, K$, $A_k \in \mathbb{R}$ and $B_k \in \mathbb{R}$ be numbers that depend on $k$. I think the following holds true, but I don't know how to show it.

$$\left| \max_{k}|A_k| - \max_{k} |B_k| \right| \leq \max_{k}\left|A_k - B_k\right|.$$

Here $|\cdot|$ denotes absolute value. Using the triangle inequality doesn't help here, and I don't know what else can be used.


For $1\le k\le K$ you have $$\vert A_k \vert =\vert A_k-B_k +B_k \vert \le \vert A_k-B_k \vert +\vert B_k \vert $$ Hence $$\max\limits_k \vert A_k \vert \le \max\limits_k \vert A_k-B_k\vert + \max\limits_k \vert B_k \vert$$ or

$$\max\limits_k \vert A_k \vert - \max\limits_k \vert B_k \vert \le \max\limits_k \vert A_k - B_k\vert$$ by symmetry you also get the inequality

$$\max\limits_k \vert B_k \vert - \max\limits_k \vert A_k \vert \le \max\limits_k \vert A_k - B_k\vert$$ and therefore the desired inequality

$$\left| \max_{k}|A_k| - \max_{k} |B_k| \right| \leq \max_{k}\left|A_k - B_k\right|$$

  • 1
    $\begingroup$ I guess the triangle inequality does help here! This is exactly what I wanted. Thanks. $\endgroup$ – Greenparker Nov 29 '16 at 20:47

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