0
$\begingroup$

Prove that the family of all Haar functions $h_{j,k}$,$(j \in \mathbb{N}, 0\leq k < 2^j) $, is an orthonormal family in $L^2([0,1])$.

I`d like to ask will this link math.stackexchange.com/questions/2101403/ help me in the solution ? or anyone could have an advice for me?

The Haar function associated with $I$, a dyadic interval, is defined by:

enter image description here

$\endgroup$
4
  • 1
    $\begingroup$ Could you offer a little on what you have tried and which part (orthogonality or normality or both) is stopping you? If you are completely lost I would suggest writing out the dyadic intervals for $j = 1$ and $j = 2$, and then drawing a graph of, say, $h_{1,0}$. $\endgroup$ Jun 9, 2017 at 17:35
  • $\begingroup$ ok I `ll do, but I do not know the shape of the Haar function.@JasonKnapp $\endgroup$
    – Emptymind
    Jun 9, 2017 at 19:12
  • 1
    $\begingroup$ A little more on a starting point then: $I_L$ and $I_R$ are the left and right halves of a given dyadic interval, and are disjoint. $\chi_{I_R} - \chi_{I_L}$ is therefore $0$ outside $I$, so all you have to think about is what it looks like on $I_L$ and $I_R$. $\endgroup$ Jun 9, 2017 at 19:34
  • $\begingroup$ may be this link will help math.stackexchange.com/questions/2101403/… $\endgroup$
    – user426277
    Jun 19, 2017 at 14:54

1 Answer 1

1
+25
$\begingroup$

Consider a Haar wavelet $h_{n,m}$. The way to visualise these wavelets is as follows:

Subdivide the interval $[0,1]$ into $2^n$ different subintervals $I_{n,m}$ of length $2^{-n}$. Each of these subintervals $I_{n,m}$ is split into two halves, the function $h_{n,m}$ is $+C_{n}$ on the left half and $-C_{n}$ on the right half, here $C_{n}$ is a constant. Outside of $I_{n,m}$ the function $h_{n,m}$ is zero. Compare this with the images in the post you have linked.

It is important to note that if you now look at $h_{n+1,m'}$ that the intervals $I_{n+1,m'}$ all lie in an $I_{n,m}$. Indeed each of the $I_{n +1,m'}$ is either a left or a right half of an $I_{n,m}$.

If you look at $\langle h_{n,m},h_{n+1,m'}\rangle$ there are then three cases to consider, the first is that the supports of the two functions are disjoint, the second is that the support of $h_{n+1,m'}$ is the left half of $I_{n,m}$ and the third is that the support of $h_{n+1,m'}$ is the right half of $I_{n,m}$.

The scalar product is obviously zero in the first case, in the second case you have that $\langle h_{n,m},h_{n+1,m'}\rangle=\int \overline{h_{n,m}} h_{n+1,m'} = C_{n}\int h_{n+1,m'}$. Since $\int h_{i,j}$ is always $0$ you get the scalar product is zero in this case too. The third case can be seen in exactly the same way.

I hope these examples make it clear what is going on. In general you want to consider $\langle h_{n,m}, h_{k,l}\rangle$, assume $n≥k$. If $n=k$ and $m\neq l$ you must see that the supports are disjoint. If $n>k$ you need to generalise the above example to see that the support of $h_{k,l}$ is either disjoint from that of $h_{n,m}$ or contained entirely in its positive or negative part.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .