Here are some inequalities involving the $L_\infty$ or “maximum norm”, $L_\infty(\x) = \max_j \abs{x_j}$:
\[\begin{align*} \norm{\x}_\infty &\leq \norm{\x}_1 \leq d\norm{\x}_\infty \\ \norm{\x}_\infty &\leq \norm{\x}_2 \leq \sqrt{d}\norm{\x}_\infty \end{align*}\]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., $\L_\infty(\x) = \abs{x_m}$.
\[\begin{align*} \norm{\x}_\infty &= \abs{x_m} \\ &\le \abs{x_m} + \sum_{j \ne m} \abs{x_j} \\ &= \norm{\x}_1 \end{align*}\]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