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  "body": "\n## Definition\n\nLet $\\mathbb{K}$ be $\\R$ or $\\C$, and let $V$ be a [[vector space|05EGHEQZAW]] over $\\mathbb{K}$.\nA *seminorm* on $V$ is a map $p \\colon V \\to \\R$ such that:\n\n1. $p(x + y) \\le p(x) + p(y)$ for all $x, y \\in V$.\n2. $p(\\lambda x) = \\abs{\\lambda} \\, p(x)$ for all $\\lambda \\in \\mathbb{K}$ and $x \\in V$, where $\\abs{\\lambda}$ is the [[absolute value|00S7C68D51]].\n\n## Remarks\n\nThe two conditions imply $p(0) = 0$ and $p(x) \\ge 0$ for all $x$.\nIndeed $p(0) = p(0 \\cdot 0) = 0 \\cdot p(0) = 0$, and then $0 = p(x - x) \\le p(x) + p(-x) = 2 p(x)$.\nA seminorm is a [[norm|08NYZ2FS6B]] exactly when $p(x) = 0$ implies $x = 0$.\nExample: on the space $C([0, 1])$ of continuous real functions, $p(f) = \\sup_{0 \\le t \\le 1/2} \\abs{f(t)}$ is a seminorm; both conditions follow from the same inequalities for the absolute value at each point.\nIt is not a norm, since it vanishes on a nonzero function supported in $(1/2, 1]$.\nNon-example: on $\\R$, $p(x) = x^2$ is not a seminorm, since $p(2 \\cdot 1) = 4 \\ne 2 = \\abs{2} \\, p(1)$.\nThe kernel $\\set{x : p(x) = 0}$ is a linear subspace, and $p$ induces a norm on the quotient of $V$ by it.\n",
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