Why does DTFT start and end at the same magnitude?

1. Why is the DTFT continuous?
2. Is DTFT continuous in frequency domain?
3. What is the condition for the existence of DTFT?

Why is the DTFT continuous?

DTFT is continuous because the original time-domain signal that you sampled has well behaved transform which means it doesn't envolve impulses in the transformed domain (fourier domain). Because DTFT is just repeated version of the actual FT at every intervals.

Is DTFT continuous in frequency domain?

The DTFT itself is a continuous function of frequency, but discrete samples of it can be readily calculated via the discrete Fourier transform (DFT) (see § Sampling the DTFT), which is by far the most common method of modern Fourier analysis.

What is the condition for the existence of DTFT?

So existence means simply that the sum that defines a DTFT does not blow up. This is easy to prove for absolutely summable sequences. If you take the magnitude of the DTFT at any point omega, this is equal to the sum for n that goes from minus infinity to plus infinity of x[n] times e to the- j omega n in magnitude.