filters - For complex values, why use complex conjugate in convolution?

Taken from Adaptive Filter Theory (2014) written by Haykin page 110 :

$$y(n) = \sum_{k=0}^{\infty} w_k^*u(n-k), \quad n=0,1,2,...$$

where $u$ and $w$ are complex values. My question is why use complex conjugate of $w_k$? The answer found in the book says "..., in complex terminology, the term $w_k^*u(n-k)$ represents the scalar version of an inner product of the filter coefficient $w_k$ and the filter input $u(n-k)$ ". I still don't understand, can you elaborate more on this answer?

Answer

Turns out that convolution and correlation are closely related. For real signals (and finite energy signals):

Convolution: $\qquad y[n] \triangleq h[n]*x[n] = \sum\limits_{m=-\infty}^{\infty} h[n-m] \, x[m]$

Correlation: $\qquad R_{yx}[n] \triangleq \sum\limits_{m=-\infty}^{\infty} y[n+m] \, x[m] = y[-n]*x[m]$

Now, in metric spaces, we like to use this notation:

$$ R_{xy}[n] \triangleq \Big\langle x[m], y[n+m] \Big\rangle = \sum\limits_{m=-\infty}^{\infty}x[m] y[n+m] $$

The $\langle \mathbf{x},\mathbf{y} \rangle$ is the Inner product of the vectors $\mathbf{x}$ and $\mathbf{y}$ where $\mathbf{x} =\{x[n]\}$ and $\mathbf{y} =\{y[n]\}$. Then we also like to define the norm of a vector as

$$\begin{align} \| \mathbf{x} \| &\triangleq \sqrt{\big\langle \mathbf{x},\mathbf{x} \big\rangle} \\ &= \sqrt{\sum\limits_{m=-\infty}^{\infty}x[m] x[m]} \\ &= \sqrt{\sum\limits_{m=-\infty}^{\infty}x^2[m]} \\ \end{align}$$

and that looks a lot like the Euclidian length of a vector with an infinite number of dimensions. All this works very well for the case where the elements $x[n]$ of the vector $\mathbf{x}$ are all real. The norm $\| \mathbf{x} \|$ is always real and non-negative.

So, if we generalize and allow the elements of $\mathbf{x}$ to be complex-valued, then if the same definition of norm is to be used,

$$\| \mathbf{x} \| \triangleq \sqrt{\big\langle \mathbf{x},\mathbf{x} \big\rangle}$$

then the definition of the inner product needs to be modified a little:

$$\Big\langle \mathbf{x},\mathbf{y} \Big\rangle = \sum\limits_{m=-\infty}^{\infty}x[m] y^*[m]$$

Then if $\mathbf{x}$ has complex-valued elements, the norm comes out as:

$$\begin{align} \| \mathbf{x} \| &\triangleq \sqrt{\big\langle \mathbf{x},\mathbf{x} \big\rangle} \\ &= \sqrt{\sum\limits_{m=-\infty}^{\infty}x[m] x^*[m]} \\ &= \sqrt{\sum\limits_{m=-\infty}^{\infty}\Big|x[m]\Big|^2} \\ \end{align}$$

So, evidently, Haykin is just walking that definition of inner product back to the definition of convolution.