Haar wavelets are defined as:
$$\phi_{0,0}(t) = \begin{cases} 1, & \text{ for } 0<t< 1/2\\ -1, & \text{ for } 1/2<t<1 \\ 0, & \text{ otherwise } \end{cases} $$
Where the mother wavelet is $$\phi_{k,n}(t) = 2^{k/2} \phi_{0,0}(2^k t -n)$$
I know that this wavelet set is orthonormal and as such the coefficients $c_{k,n}$ in:
$$x(t) = \sum_{k=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} c_{k,n} \phi_{k,n} (t)$$
can be found as $\langle x(t),\phi_{k,n(t)}\rangle$.
However, I have a question which asks:
Let $x_k(t) = \sum_{n=-\infty}^{\infty} c_{k,n} \phi_{k,n}(t)$ for a given $k$.
Show that the $c_{k,n}'s$ can be found as:
Find $h_k(t)$.
Now clearly $h_k(t)$ is some combination of the basis function.
My attempt:
\begin{align*} y(t) &= \int_{-\infty}^{\infty} x(\tau) h(t-\tau) d\tau \quad \text{Desired Form}\\ c_{k,n} &= \int_{-\infty}^{\infty} x(t) \phi_{k,n}(t) dt\\ &= 2^{k/2} \int_{-\infty}^{\infty} x(t)\phi_{0,0}(2^k t -n) dt \end{align*}
But now if I sample $t=\frac{n}{2^k}$ I will be left with:
$$ 2^{k/2} \int_{-\infty}^{\infty} x(\frac{n}{2^k})\phi_{0,0}(n-n) dt$$
and I don't know how to change the integral to a summation in this case. Plus $\phi_{0,0}(n-n)$ makes no sense.
How can I proceed? Any help is appreciated!