# Why downsample exhibit wider bandwidth when doing discrete time fourier transform?

Based on Nyquist sampling theorem, sampling would have a convolution in frequency domain, and reasonably the bandwidth of each convolution would be as that of the original signal. Nevertheless, the discrete time Fourier transform of a signal ,which has been downsampled , has the $$M$$ multiples bandwidth of each convolution.($$M$$ is the downsample factor). I could not have got the idea, what is the true relation between time/frequency domain with underlying discrete time Fourier transform?

Sampling gets convolution in frequency domain:

Downsampling gets more wider bandwidth:

You don't get "more" bandwidth. That's simply a misunderstanding of your x-axis. Your signal after downsampling occupies a $M$ times larger part of the Nyquist rate, but that's because the Nyquist rate got reduced by a factor of $M$, not because the signal got wider!
To understand this, you need to know the relation between the continuous and discrete frequency axes. $$f = \frac{F}{F_s}$$ where $f$ is the discrete frequency, $F$ is the continuous frequency and $F_s$ is the sample rate. Once you downsample a signal, the actual signal bandwidth does not change; however, the discrete frequency axis on which you drew the first spectrum no longer holds. You have a new discrete frequency axis that is scaled by a factor of $M$; so $$f' = \frac{F}{F_s/M} = M\cdot \frac{F}{F_s}$$
Taking your figures as example, assume that a signal with a maximum frequency $F = 200$ kHz was sampled at a rate of $F_s = 800$ kHz to yield $f = 1/4$ cycles/sample (which is the same as $\omega = \pi/2$ approx). Downsampling it by $M=2$ yields $$f' = 2\cdot \frac{1}{4} = \frac{1}{2}$$ cycles/sample, which is the same as $\omega' = \pi$ approx in your second figure.