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  "body": "\n## Definition\n\nLet $\\mathbb{K}$ be $\\R$ or $\\C$, and let $X$ be a [[Banach space|0BDTTM24DP]] over $\\mathbb{K}$ with norm $\\norm{\\cdot}$.\nA sequence $(e_n)_{n \\ge 1}$ in $X$ is a *Schauder basis* of $X$ if the following holds.\nFor every $x \\in X$ there is a unique sequence $(\\alpha_n)_{n \\ge 1}$ in $\\mathbb{K}$ such that\n$$\\lim_{N \\to \\infty} \\norm{x - \\sum_{n=1}^{N} \\alpha_n e_n} = 0,$$\nthat is, the [[series|WN6M8B3ZKB]] $\\sum_n \\alpha_n e_n$ converges to $x$ in $X$.\n\n## Remarks\n\nExample: in $\\ell^p$ with $1 \\le p \u003c \\infty$, see [[spaces ℓ^p|08P6T2WH6R]], the unit vectors $e_n$ (with $1$ in place $n$ and $0$ elsewhere) form a Schauder basis.\nFor $x \\in \\ell^p$ and $\\alpha_n = x_n$, one has $\\norm{x - \\sum_{n \\le N} x_n e_n}_p = \\left( \\sum_{n \u003e N} \\abs{x_n}^p \\right)^{1/p} \\to 0$.\nThe coefficients are unique, because a norm-convergent series converges coordinatewise, so the $n$-th coordinate of $\\sum_k \\alpha_k e_k$ is $\\alpha_n$.\n\nNon-example: in $\\ell^\\infty$ the unit vectors are not a Schauder basis.\nFor $x = (1, 1, 1, \\dots)$ the only possible coefficients are $\\alpha_n = 1$, and $\\norm{x - \\sum_{n \\le N} e_n}_\\infty = 1$ for every $N$.\nIn fact $\\ell^\\infty$ has no Schauder basis at all, because it is not [[separable|0Z2P41SVDP]].\n\nThe order of the basis matters, since the series need not converge after a rearrangement.\nA Schauder basis of an infinite-dimensional Banach space is never a [[Hamel basis|0VTQTPQGAV]].\nThe coefficient functionals $x \\mapsto \\alpha_n$ are always continuous.\n",
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