DTFT and its convergence

Question

In the textbook "signals and systems", by prof. Simon Haykin, it says:　　

If $x[n]$ is not absolutely summable, but does satisfy square summable, then it can be shown that the following equation converges in a mean-square error sense, but not converges pointwise.
$$ X(e^{\,j\,Ω})=\sum_{n=-∞}^∞x[n]\,e^{\,-j\,Ω\,n} $$

What does "in a mean-square error sense, but not converges pointwise" exactly mean?

Answer 1

convergence in MMSE iS $$\lim_{n\to\infty }E[(X_n-\hat{X})^2]\to 0$$ and pointwise it is $$\lim_{n\to\infty }X_n(\omega)=X(w)\forall \omega\in\Omega$$

Answer 2

Or put another way,

• if $c=\{c[n]\}\in\ell^1(\Bbb Z)$, then the Fourier series converges in $C([-\pi,\pi])$,
• if $c\in\ell^2(\Bbb Z)$, then the Fourier series converges in $L^2([-\pi,\pi])$.

For instance, all piecewise linear periodic functions, jumps allowed, have coefficients in $\ell^2(\Bbb Z)$ and thus their Fourier series converges in the second sense, but not in the first. However, pointwise convergence holds at the points of continuity.