Here are some inequalities involving the \(L_\infty\) or “maximum norm”, \(L_\infty(\x) = \max_j \abs{x_j}\):
\[\as{ \norm{\x}_\infty &\leq \norm{\x}_1 \leq d\norm{\x}_\infty \\ \norm{\x}_\infty &\leq \norm{\x}_2 \leq \sqrt{d}\norm{\x}_\infty }\]Proof: We will prove the two inequalities in the first line above. The proofs for the second line are essentially the same. Let \(m\) denote the index of the maximum element (in absolute value): i.e., \(\L_\infty(\x) = \abs{x_m}\).
\[\begin{alignat*}{2} \norm{\x}_\infty &= \abs{x_m} \\ &\le \abs{x_m} + \sum_{j \ne m} \abs{x_j} \\ &= \norm{\x}_1 \end{alignat*}\]Similarly,
\[\begin{alignat*}{2} \norm{\x}_1 &= \sum_j \abs{x_j} \\ &\le \sum_j \abs{x_m} && &\hspace{4em}& x_m \text{ is max} \\ &=d\norm{\x}_\infty \end{alignat*}\]See also: L1-L2 norm inequality