Conjugate function

Remarks

If $f$ is a trigonometric polynomial with Fourier series $\sum_k \hat f(k) e^{ik\theta}$, then $$\tilde f(\theta) = \sum_k (-i \sgn k) \, \hat f(k) e^{ik\theta},$$ with $\sgn 0 = 0$. For real $f$, let $u$ be the Poisson integral of $f$ in the unit disc, and let $v$ be its harmonic conjugate with $v(0) = 0$. Then $v(re^{i\theta}) \to \tilde f(\theta)$ as $r \to 1^-$ for almost every $\theta$.

Example: $f(\theta) = \cos\theta$ has $\tilde f(\theta) = \sin\theta$. Check: $\cos(\theta - t)\cot(t/2) = \cos\theta \cos t \cot(t/2) + \sin\theta \sin t \cot(t/2)$. The first term is odd in $t$, so its principal value integral is $0$. The second uses $\sin t \cot(t/2) = 2\cos^2(t/2) = 1 + \cos t$, whose integral over $[-\pi, \pi]$ is $2\pi$.

Non-example: the integral without the principal value need not exist. For $f = 1$ near $\theta$ the integrand behaves like $2/t$ near $t = 0$, which is not integrable, while the symmetric limit is $0$ since $\cot(t/2)$ is odd.

The function $\tilde f$ need not be integrable when $f$ is: the operator $f \mapsto \tilde f$ is bounded on $L^p(\mathbb{T})$ only for $1 < p < \infty$.