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  "body": "\n## Definition\n\nLet $\\mathbb{T}$ be the unit circle with normalised Lebesgue measure, and let $L^2 = L^2(\\mathbb{T})$ ([[Lp space|SASP14MA1Z]]).\nLet $H^2 \\subseteq L^2$ be the [[Hardy space|PNNNFJVF6W]], the closed subspace of functions whose Fourier coefficients of negative index vanish.\nLet $P \\colon L^2 \\to H^2$ be the [[orthogonal projection|0Q8HZTJC7G]].\nFor $\\varphi \\in L^\\infty(\\mathbb{T})$, the *Toeplitz operator* with symbol $\\varphi$ is the operator $T_\\varphi \\colon H^2 \\to H^2$ given by\n$$T_\\varphi f = P(\\varphi f).$$\n\n## Remarks\n\n$T_\\varphi$ is a bounded linear operator, and in fact $\\norm{T_\\varphi} = \\norm{\\varphi}_\\infty$.\nWith respect to the orthonormal basis $e_n(t) = e^{int}$, $n \\ge 0$, its matrix entries are $\\ip{T_\\varphi e_k}{e_j} = \\hat\\varphi(j - k)$, so the matrix is a [[Toeplitz matrix|GKY2WSHE26]].\nIf $\\varphi$ is continuous and has no zeros on $\\mathbb{T}$, then $T_\\varphi$ is a [[Fredholm operator|8HWGKWD6ZM]] whose index is minus the winding number of $\\varphi$ around $0$.\n\nExample: $\\varphi(e^{it}) = e^{it}$.\nFor $f \\in H^2$ the product $\\varphi f$ has no negative Fourier coefficients, so $P(\\varphi f) = \\varphi f$ and $T_\\varphi$ is the shift operator $f \\mapsto z f$.\n\nNon-example: for $\\varphi(e^{it}) = e^{-it}$, plain multiplication by $\\varphi$ does not map $H^2$ into itself, since it sends $1$ to $e^{-it} \\notin H^2$.\nSo the projection $P$ in the definition cannot be omitted; the Toeplitz operator $T_\\varphi$ is the backward shift.\n",
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