We say that a norm $\|\cdot\|$ on $\mathbb{R}^n$ is symmetric if it is invariant under sign changes and permutations of the components. Suppose that $$\|(1, 0, ..., 0)\| = 1$$ for a symmetric norm.
Does it follow that $$ \|\cdot\|_\infty \leq \|\cdot\| \leq \|\cdot\|_1?$$
Here $\|\cdot\|_\infty$ and $\|\cdot\|_1$ stand for the usual max and $\ell^1$-norm on $\mathbb{R}^n.$
Yes, the inequalities hold. Let $B$ be the closed unit ball for the norm $\|\cdot \|$. The assumptions imply that $\pm e_j$, the standard basis vectors, are in $B$. Hence, their convex hull is contained in $B$. This convex hull is the unit ball for the $\ell^1$-norm, which implies $\|\cdot \|\le \|\cdot \|_1$.
Similarly, we need to prove that $B$ is contained in the unit ball for $\ell^\infty$ norm. To do this, observe that $B$ admits a supporting plane at every $\pm e_j$. The symmetry with respect to coordinate planes implies that such a plane (when not unique) can be chosen to be $x_j = \pm 1$. The conclusion follows.