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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.