In an electromagnetic waveguide, there is generally a "cutoff frequency." Electromagnetic waves with a frequency that is lower than this cutoff frequency will not propagate at all -- i.e., they will be exponentially attenuated.
Suppose that there is a source of electromagnetic radiation at the center of a waveguide (e.g., an oscillating electric dipole inside a rectangular waveguide), and that the emitted waves have a frequency below the cutoff frequency of the waveguide. All of the waves will be exponentially attenuated and thus will not propagate within/outside of the waveguide.
But, before the waves are sufficiently exponentially attenuated, they will be carrying away some energy from the radiating source in the form of Poynting flux. What happens to that radiation energy immediately before and after attenuation -- where does it go? Maybe the radiation is very quickly absorbed by the conducting boundary of the waveguide and thus heats up the waveguide's surface?
Most people would say that the Poynting flux is zero if the frequency of the waves is below the cutoff frequency of the waveguide, but the language of "exponential attenuation" suggests to me that the waves did travel for some short amount of time before being exponentially attenuated (and thus the Poynting flux was not zero before exponential attenuation).