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For $0<p<\infty$, an analytic Hardy space controls the integral means of a holomorphic function as it approaches a domain boundary. On the disk, $H^p$ requires ${\sup }_{0<r<1}\int |F(r{e}^{i\theta }){|}^{p}d\theta <\infty$; on the upper half-plane it requires ${\sup }_{y>0}{\int }_{\mathbb{R}}|f(x+iy){|}^{p}dx<\infty$. For $p=\infty$ use a supremum norm. These analytic spaces differ from the harmonic Hardy space of the upper half-space; the Hardy space of the circle is the boundary Hilbert-space realization of the disk $H^2$ space.

The Fredholm index measures the finite-dimensional obstruction to inverting an operator. For a bounded operator $T:H_1\to H_2$ between Hilbert spaces, the Fredholm conditions are that its range of a bounded linear operator is closed and its kernel and cokernel are finite-dimensional. The orthogonal complement of the range is $\ker T^{\ast }$, soAn invertible linear map has Fredholm index zero; the converse needs an additional finite-dimensional correction, since an operator can have equally large kernel and cokernel.

First construct an inverse up to finite-rank operators. On $(\ker T{)}^{\perp }$, the restriction of $T$ is a bounded bijection onto its closed range, with bounded inverse by the bounded inverse theorem. Extend that inverse by zero on $(\mathrm{ran}T{)}^{\perp }$, obtaining $S:H_2\to H_1$. ThenConversely, suppose $ST-I$ and $TS-I$ are compact operators. On $\ker T$, the first identity expresses the identity as a compact operator, so this kernel is finite-dimensional. If $T$ were not bounded below on $(\ker T{)}^{\perp }$, there would be unit vectors $x_j$ in that complement with $T{x}_j\to 0$. Write $ST=I+K$. Compactness of $K$ gives a subsequence for which $K{x}_j$ converges, hence $x_j=ST{x}_j-K{x}_j$ converges to a unit vector $x$ in $(\ker T{)}^{\perp }$ with $Tx=0$, a contradiction. This lower bound makes the range closed. Applying the analogous argument to $T^{\ast }$, using $S^{\ast }{T}^{\ast }=I+K^{\prime \ast }$, proves that its kernel is finite-dimensional. This proves the Atkinson theorem: a bounded operator is Fredholm exactly when it has an inverse modulo compact operators. For operators on one Hilbert space this says that its image in the Calkin algebra is invertible.

The Fredholm index is stable. Decompose the domain as $N\oplus N^{\perp }$, where $N=\ker T$, and the target as $C\oplus R$, where $C=\ker T^{\ast }$ and $R=\mathrm{ran}T$. A sufficiently small norm perturbation has block matrixMultiplying by invertible triangular block matrices reduces this matrix to the direct sum of $D$ and the Schur complement $a-b{D}^{-1}c:N\to C$. The latter's kernel dimension minus cokernel dimension is $\dim N-\dim C$, whatever its rank. Therefore the Fredholm set is open and the index is locally constant. If $K$ is compact, the same parametrix works modulo compact operators for every $T+tK$, $0\le t\le 1$. Local constancy along this path gives

For Fredholm operators $A:H_1\to H_2$ and $B:H_2\to H_3$, multiplication of parametrices shows that $BA$ is Fredholm. Its index is additive. Here is a direct proof, rather than an assertion of additivity. There is an exact sequence of finite-dimensional spacesThe map from $\ker B$ takes a vector to its class modulo $\mathrm{ran}A$; the next takes the class of $y$ to that of $By$; the last is the quotient map. For example, a class $y+\mathrm{ran}A$ lies in the kernel of the middle map exactly when $By=BAx$ for some $x$, so $y-Ax\in \ker B$. This verifies the only less immediate exactness assertion; the others follow directly from their definitions. Alternating the dimensions givesAlso $\mathrm{ind}{T}^{\ast }=-\mathrm{ind}T$, and indices add under direct sums. For $I+K$ with $K$ compact, the index is zero, so injectivity is equivalent to surjectivity. More generally $Tx=f$ is solvable exactly when $f$ is orthogonal to $\ker T^{\ast }$: these are the finitely many compatibility conditions in the Fredholm alternative.

Toeplitz operators exhibit how this analytic index detects a topological winding. Let $H^2\subset L^2(S^1)$ be the Hardy space of the circle, spanned by $1,z,z^2,\dots$, and let $P$ be its orthogonal projection. For a continuous symbol $g$, define $T_g=P{M}_g{|}_{H^2}$, where $M_g$ is multiplication by $g$. Then $\Vert T_g\Vert \le \Vert g{\Vert }_{\infty }$. The product defect isFor trigonometric polynomials the crossing of negative and nonnegative Fourier modes has finite rank. Uniform approximation of continuous symbols by trigonometric polynomials, together with the displayed norm estimate, therefore makes this defect a compact operator for all continuous $f,g$. If $g$ is nowhere zero, $1/g$ is continuous and $T_{1/g}$ is a two-sided parametrix modulo compact operators. Thus $T_g$ is Fredholm.

The converse is instructive. If $g(\zeta )=0$ at $\zeta \in S^1$, consider the normalized Hardy kernelsTheir norms are one, their individual Fourier coefficients tend to zero as $r\uparrow 1$, and hence they converge weakly to zero. Their squared moduli are the Poisson kernel, so continuity of $g$ givesA Fredholm operator is bounded below off its finite-dimensional kernel. The projection of these weakly vanishing vectors onto that kernel tends to zero, contradicting this lower bound. Thus a continuous-symbol Toeplitz operator is Fredholm exactly when its symbol never vanishes. Applying the same argument to $g-\lambda$ gives its essential spectrum as $g(S^1)$, with essential spectrum defined by failure of the Fredholm property.

For nonvanishing $g$, let $m=\mathrm{wind}(g,0)$. Lifting a continuous argument of $g(e^{it})$ on $0\le t\le 2\pi$ gives an argument change $2\pi m$. After division by $z^m$, its argument has matching endpoints, so there is a continuous function $h:S^1\to \mathbb{C}$ with $g(z)=z^m{e}^{h(z)}$. The nonvanishing-symbol homotopy $g_t=z^m{e}^{th}$ reduces the index to that of $T_{z^m}$. For $m\ge 0$ this is the $m$th power of the unilateral shift operator, whose kernel is zero and whose range has codimension $m$. For $m<0$ it is the $(-m)$th power of the adjoint shift, whose kernel has dimension $-m$ and whose range is all of $H^2$. ConsequentlyThe minus sign comes from counting missing nonnegative Fourier modes. This is a concrete bridge between the Fredholm index, stability under compact perturbations, and the winding number of a loop.

On the Hardy space of the circle $H^2(\mathbb{T})\subset L^2(\mathbb{T})$, let $P_{+}$ be the projection onto nonnegative Fourier modes and $T_f=P_{+}{M}_f{|}_{H^2}$. The Toeplitz index theorem for continuous symbols states: for continuous $f$, $T_f$ is Fredholm exactly when $f$ never vanishes, and thenwith counterclockwise winding positive.

First, $T_f{T}_g-T_{fg}=-P_{+}{M}_f(I-P_{+})M_g{P}_{+}$ is compact. For Laurent-polynomial symbols the negative-mode crossing involves only finitely many modes and is finite rank. Uniform approximation and $\Vert T_f\Vert \le \Vert f{\Vert }_{\infty }$ extend this to continuous symbols. If $f$ is nonvanishing, $T_{1/f}$ is therefore a two-sided parametrix, and part (b) gives the Fredholm property.

Writing $n=\mathrm{wind}(f,0)$, a continuous argument along the parametrized circle increases by $2\pi n$. Removing this increase gives a periodic continuous logarithm $h$ of $z^{-n}f(z)$, so $f=z^n{e}^h$. The symbols $z^n{e}^{th}$ give a continuous Fredholm path from $z^n$ to $f$. Part (c) makes their indices equal. For $n\ge 0$, $T_{z^n}$ is the $n$th unilateral shift, with zero kernel and $n$-dimensional cokernel. For negative $n$ it is the corresponding backward shift, with kernel dimension $-n$ and zero cokernel. In both cases the index is $-n$.

For necessity, if $f(\zeta )=0$ use the normalized Hardy kernelsThey have norm one and converge weakly to zero, as their inner products with each Fourier basis vector tend to zero. Their squared moduli are Poisson kernels concentrating at $\zeta$, so continuity gives $\Vert f{k}_{r,\zeta }{\Vert }_2\to 0$ and hence $\Vert T_f{k}_{r,\zeta }{\Vert }_2\to 0$. A Fredholm parametrix would imply $k_{r,\zeta }=B{T}_f{k}_{r,\zeta }+K{k}_{r,\zeta }\to 0$ in norm, because a compact operator takes this weakly null bounded family to zero in norm. This contradicts unit norm, proving the converse.

On the Hardy space of the circle, a continuous symbol gives a Fredholm Toeplitz operator exactly when it never vanishes. Nonvanishing gives a parametrix $T_{1/f}$ modulo compact semicommutators. Write $f(z)=z^n{e}^{h(z)}$, where $n$ is its counterclockwise winding number and $h$ is continuous. Homotopy through nonvanishing symbols and local constancy of the Fredholm index reduce the index to that of the shift $T_{z^n}$, namely $-n$. If $f$ vanishes at a point, normalized Hardy reproducing kernels concentrating there are weakly null unit vectors on which $T_f$ tends to zero, contradicting a Fredholm parametrix.