# The Continuity of the Discrete Time Fourier Transform of Absolutely Summable Series

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

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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