I'm stuck on the following problem:
Suppose that $\{V_j : j \in \mathbb{Z}\}$ is a multiresolution analysis with scaling function $\phi$, and that $\phi$ is continuous and compactly supported. Given:
$$u(x) = \left\{\begin{array} {1 1} 1, & \quad 0 \leq x \leq 1, \\ 0, & \quad x < 0 \text{ or } x > 1. \end{array} \right.$$
If $\int_{-\infty}^{\infty} \phi(x) dx = 0$, show that for all $j$ sufficiently large, $\| u - u_j \| \geq \frac{1}{2}$
Hint: $| \langle u, \phi_{j,k} \rangle| \leq 2^{-j/2} \int |\phi(y)| dy$
OK, so I really am quite clueless as to how to proceed here. I attempted to approach this problem as follows:
We have:
$$| \langle u, \phi \rangle - \langle u_j, \phi_j \rangle | = |(\langle u, \phi \rangle - \langle u, \phi_j \rangle) + (\langle u, \phi_j \rangle - \langle u_j, \phi_j \rangle)$$
$$\leq | \langle u, \phi - \phi_j \rangle | + | \langle u - u_j, \phi_j \rangle |$$
$$\leq \|u \| \|\phi - \phi_j\| + \|u - u_j\| \|\phi_j \|$$
However I don't see how to go from here, nor how the hint may help me. Also, since we are given that $\int_{-\infty}^{\infty} \phi(x) dx = 0$, won't that mean that the term $\|u - u_j\| \|\phi_j \|$ becomes $0$ in the above expression?
If anyone can help me along here, I would be very grateful!