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I saw on a book to following claim:
Given an Absolutely Summable Series $ \sum_{n = -\infty }^{\infty}\left | x\left [ n \right ] \right | \leqslant \infty $, Namely, $ l_1 $ series it is possible to show its DTFT (Discrete Time Fourier Transform) is continuous.
Where the DTFT is given by: $$ DTFT\left \{ x\left [ n \right ] \right \} = X\left ( {e}^{j \omega} \right ) = \sum_{n = -\infty}^{\infty} x\left [ n \right ] {e}^{-j \omega n} $$

My question is, how could that be proven?
Is it also differentiable?

Thank You.

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1 Answer 1

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Defining $ f(n) = x \left [ n \right ] {e}^{-j \omega n} $ and rewriting the DTFT:

$$ DTFT \left \{ x \left [ n \right ] \right \} = X \left ( {e}^{j \omega} \right ) = \sum_{n = -\infty}^{\infty} x \left [ n \right ] {e}^{-j \omega n} = \sum_{n = -\infty}^{\infty} f(n) $$

Since $ f \left ( n \right ) $ is continuous for every $ n $ by the Uniform Limit Theorem the limit function is also continuous if the sequence converges uniformly.

$$ \sum_{n = -\infty}^{\infty} x \left [ n \right ] {e}^{-j \omega n} \leq \left | \sum_{n = -\infty}^{\infty} x \left [ n \right ] {e}^{-j \omega n} \right | \leq \sum_{n = -\infty}^{\infty} \left | x \left [ n \right ] {e}^{-j \omega n} \right | = \sum_{n = -\infty}^{\infty} \left | x \left [ n \right ] \right | \left | {e}^{-j \omega n} \right | = \sum_{n = -\infty}^{\infty} \left | x \left [ n \right ] \right | \leq \infty $$

The last equality $ \sum_{n = -\infty}^{\infty} \left | x \left [ n \right ] \right | \leq \infty $ come from the definition of the sequence. Since the convergence doesn't depend on $ \omega $ this is a Uniform Convergence and hence the limit function is continuous.

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