= One-sided Fourier spectrum excludes jumps
{title2=$\widehat f(n)=0\ (n<0)\Longrightarrow f(t^-)=f(t^+)$}
Let an integrable function have finite one-sided limits $A,B$ at zero. Oddness gives $(Q_r*f)(0)=(2\pi)^{-1}\int_0^\pi Q_r(t)[f(-t)-f(t)]\,dt$. Also $\int_0^\delta Q_r(t)\,dt=\log[(1-2r\cos\delta+r^2)/(1-r)^2]=2\log(1/(1-r))+O_\delta(1)$. Thus $(Q_r*f)(0)/\log(1/(1-r))\to(A-B)/\pi$: make the near-zero error in $f(-t)-f(t)$ arbitrarily small and bound the remaining integral uniformly. When negative <Fourier coefficients> vanish, the conjugate-kernel identity makes this convolution bounded because $P_r*f$ tends to $(A+B)/2$. Therefore $A=B$. A <removable discontinuity> caused solely by altering a point value is not excluded.
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