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  "body": "\n## Proof\n\nWe compare each norm with the sup norm of the coordinates in a basis, and use compactness of the unit sphere of that norm.\n\n1. Let $\\mathbb{K}$ be $\\mathbb{R}$ or $\\mathbb{C}$. If $V = \\{0\\}$, each norm is the zero function, and any two norms are [[equivalent|0J3578562S]] with $c = C = 1$. Let $V \\neq \\{0\\}$.\n2. $V$ is [[finite-dimensional|018W3V3TCT]], so a finite spanning set [[contains a basis|06MRNP4BNT]] $e_1, \\ldots, e_n$, with $n \\ge 1$. Write $x = \\sum_{j} a_j(x) e_j$. The coordinates $a_j \\colon V \\to \\mathbb{K}$ are linear, by the [[isomorphism with the coordinate space|059THXBRBB]].\n3. Let $\\|x\\|_\\infty = \\max_j |a_j(x)|$. This is a norm. It vanishes only when all coordinates vanish, that is, at $x = 0$. Also $\\|\\lambda x\\|_\\infty = |\\lambda| \\|x\\|_\\infty$, and $|a_j(x + y)| \\le |a_j(x)| + |a_j(y)|$ gives the triangle inequality.\n4. Equivalence of norms is symmetric and transitive. If $c_a \\|x\\|_\\infty \\le \\|x\\|_a \\le C_a \\|x\\|_\\infty$ and $c_b \\|x\\|_\\infty \\le \\|x\\|_b \\le C_b \\|x\\|_\\infty$, then $(c_b / C_a) \\|x\\|_a \\le \\|x\\|_b \\le (C_b / c_a) \\|x\\|_a$.\n5. So it suffices to fix a norm $\\|\\cdot\\|$ on $V$ and find $c, C \u003e 0$ with $c \\|x\\|_\\infty \\le \\|x\\| \\le C \\|x\\|_\\infty$ for all $x \\in V$.\n6. Upper bound. Let $C = \\sum_j \\|e_j\\| \u003e 0$. By the [[triangle inequality|0FJZMBX0ZZ]], $\\|x\\| \\le \\sum_j |a_j(x)| \\, \\|e_j\\| \\le C \\|x\\|_\\infty$.\n7. Let $m = n$ if $\\mathbb{K} = \\mathbb{R}$ and $m = 2n$ if $\\mathbb{K} = \\mathbb{C}$. Define $\\Lambda \\colon \\mathbb{R}^m \\to V$ by $\\Lambda(t) = \\sum_j t_j e_j$ in the real case, and $\\Lambda(t) = \\sum_j (t_{2j-1} + \\mathrm{i} \\, t_{2j}) e_j$ in the complex case.\n8. $\\Lambda$ is a bijection and is additive. Let $d_2$ be the [[Euclidean metric|0E1MMCAY5B]] on $\\mathbb{R}^m$. Each coordinate of $\\Lambda(t)$ has modulus at most $d_2(t, 0)$, so $\\|\\Lambda(t)\\|_\\infty \\le d_2(t, 0)$.\n9. Let $g(t) = \\|\\Lambda(t)\\|$ and $h(t) = \\|\\Lambda(t)\\|_\\infty$. By the reverse triangle inequality in [[this result|0FJZMBX0ZZ]] and steps 6 and 8, $|g(t) - g(s)| \\le \\|\\Lambda(t - s)\\| \\le C d_2(t, s)$.\n10. In the same way, $|h(t) - h(s)| \\le \\|\\Lambda(t - s)\\|_\\infty \\le d_2(t, s)$. So $g$ and $h$ are [[Lipschitz|0TMZ21WKCN]].\n11. A Lipschitz function satisfies the epsilon-delta condition with $\\delta = \\varepsilon / (C + 1)$. By [[continuity via open sets|0YF82GBTND]], $g$ and $h$ are continuous on $\\mathbb{R}^m$.\n12. Let $\\Sigma = h^{-1}(\\{1\\})$. The set $\\{1\\}$ is closed in $\\mathbb{R}$, so $\\Sigma$ is closed, by [[continuity via closed sets|0B8GPXFSP7]].\n13. $\\Sigma$ is [[bounded|0P953MG05E]]. For $t \\in \\Sigma$, each $|t_k|$ is at most the modulus of a coordinate of $\\Lambda(t)$, so $|t_k| \\le 1$. Then $d_2(t, 0) \\le \\sqrt{m}$ and $d_2(t, s) \\le 2\\sqrt{m}$ for $t, s \\in \\Sigma$.\n14. By the [[Heine–Borel theorem|07PFYV65MB]], $\\Sigma$ is compact. It is nonempty, since $\\Lambda^{-1}(e_1) \\in \\Sigma$.\n15. By [[this result|0V4DET59NW]], $g$ attains a minimum $c$ on $\\Sigma$, at some $t^\\ast \\in \\Sigma$. As $h(t^\\ast) = 1$, $\\Lambda(t^\\ast) \\neq 0$, so $c = \\|\\Lambda(t^\\ast)\\| \u003e 0$.\n16. Lower bound. Let $x \\in V$ with $x \\neq 0$, and let $y = x / \\|x\\|_\\infty$. Then $\\|y\\|_\\infty = 1$, so $y = \\Lambda(t)$ with $t \\in \\Sigma$, and $\\|y\\| = g(t) \\ge c$.\n17. Multiplying by $\\|x\\|_\\infty$ gives $\\|x\\| \\ge c \\|x\\|_\\infty$. This also holds for $x = 0$.\n18. With step 6, $\\|\\cdot\\|$ is equivalent to $\\|\\cdot\\|_\\infty$. By steps 4 and 5, any two norms on $V$ are equivalent.\n\n## Idea\n\nEvery norm is bounded above by a multiple of the sup norm of the coordinates, by the triangle inequality. The lower bound needs compactness: the unit sphere of the sup norm is closed and bounded in $\\mathbb{R}^m$, so the continuous positive function $\\|\\cdot\\|$ has a positive minimum on it. Scaling then gives the bound everywhere.\n",
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