I posted this question on MSE a couple of days ago. Someone gave some hints, which, besides the fact that I struggle to understand them, go in a numerical analysis direction, which I am not interested in.

I was hoping someone here could help. If the question is not good for MO, just let me know and I'll delete it.

The goal I have in mind is to express the $L^2$ norm of the second derivative of a quite regular function as a sum of some coefficients. My idea was that such coefficients involved some system of wavelets. I'm not particularly attached to the equality (having the norm bounded below by such a sum is good enough for me), nor to the $L^2$ norm, as I am ok with being able some other quantity related to $f''$ (but not to $f$ or $f'$).

Below, I am going to try to explain why I'd hope for such a result to hold (as something analogous holds for the first derivative, with a geometric interpretation) and try to explain my failed attempt at getting such an inequality.

Let $h_I$ be the Haar wavelet on the interval $I$, normalized in $L^2$, i.e. $$ h_I(x)= \begin{cases} \dfrac{1}{\sqrt{2|I|}} & \text{if } x \in I_l, \\ - \dfrac{1}{\sqrt{2|I|}} & \text{if } x \in I_r, \\ 0 & \text{otherwise.} \end{cases} $$ where $|I|$ denotes the length of $I$, and $I_l$ and $I_r$ denote, respectively, the left and right half of $I$. The Haar wavelets forms a complete orthonormal system for $L^2(\mathbb R)$, which implies that, if $f$ is a function such that $f'$ exists almost everywhere and it is square integrable, we have that $$\|f'\|_2^2=\sum_{I \text{ dyadic}}\langle f', h_I\rangle^2.$$

If we look at the coefficients $a_I=\langle f, h_I\rangle$, we can get a feeling of what they represent. Given an interval $I$, let $x_l$, $x_r$ and $x_m$ denote, respectively, the left and right endpoints and the middle point of the interval. Computing explicitly the integral, one can check that they measure the change in slope between the line passing through the point $(x_l, f(x_l))$ and $(x_m, f(x_m))$ and the one passing through $(x_m, f(x_m))$ and $(x_r, f(x_r))$. More explicitly, we have $$a_I=\left(\frac{f(x_m)-f(x_l)}{\frac{|I|}{2}}\right)- \left(\frac{f(x_r)-f(x_m)}{\frac{|I|}{2}}\right).$$ It seems therefore natural that a quantity that measures change of slopes at any scale is related to the first derivative of the function.

Now, the idea is to get something related to the second derivative by considering changes of changes of slopes. This seemed to me a fairly natural idea. One of the things I tried was to subdivide an interval $I$ in four parts and consider the slope of the four lines passing through the five "subdivision" points to get something like $b_I=a_{I_l}-a_{I_r}$. The problem is that I couldn't relate that to a system of wavelets who gives me something about $f''$.

The only family of wavelets that came to my mind was an iteration of the previous $h_I$'s, so something like $\tilde{h}_I(x) = h_{I_l}(x)-h_{I_r}(x)$, opportunely renormalized to be an orthonormal system in $L^2$. The issues I have with that, is that, first I don't know completeness of the system, but more importantly, the only relation I can find is something like $$\sum_{I \text{ dyadic} } \langle f',\tilde{h}_I\rangle^2=\sum_{I \text{ dyadic} } b_I^2 \leq \|f'\|_2^2,$$ which still involves only the first derivative. I can't figure out a way to relate that to the second derivative of my function.

I was wondering: is there any way to relate the $b_I$ defined as above to the second derivative?

If not, that is they are not the right quantities to look at, can one find similar coefficients that relate to $f''$?

Edit: To rephrase the first question, the LHS in the last formula above should represent the changes of changes of slope, but what I can't figure out is how to relate the RHS to the second derivative of my function.

The problem I have with just using the $\tilde{h}_I$'s with $f''$ instead of $f'$ (or similarly, any other $L^2$ basis) is really about the lack of geometric interpretation of such coefficients. Indeed, while I can see what $\langle f', \tilde{h}_I \rangle$ represents, I can't for $\langle f'', \tilde{h}_I \rangle$.

I need to mention I am far from being an expert about wavelets, so I might be missing something quite easy to people familiar with them.

  • $\begingroup$ You seem to insist at expressing (or framing) some norm of $f''$ as a sum of wavelet coefficients, and this is certainly possible with any smooth enough wavelet (not just the Haar system, then). But why not just use the Fourier base on an interval ? Or do you mean $L^2(\mathbb R)$ ? $\endgroup$ Feb 18 '16 at 12:17
  • $\begingroup$ Yeah, I am in $\mathbb R$, and besides that, any coefficient that depends on something smoother (like Fourier ones), seems to me that the become much harder to compute (using the Haar system all you have to do is subtraction and division of some values of my function). But, maybe I am wrong, and my problem is exactly insisting in Haar wavelets. $\endgroup$ Feb 18 '16 at 14:03
  • 1
    $\begingroup$ Let me say it better: I am attached to the simplicity of computation, because I am attached to the nice geometric interpretation I can give it. @JeanDuchon $\endgroup$ Feb 18 '16 at 14:20
  • $\begingroup$ @SilviaGhinassi : the wavelet transform is not so simple to compute, that's why I told you to look at the discrete (Haar) wavelet transform (the normal Haar wavelet transform for a function being constant on any $[2^{-k}n,2^{-k}(n+1)[$ interval) which has the same complexity as the discrete Fourier transform, and indeed the algorithms are very very close, and for both there exists the fast transforms (FFT and fast wavelet transform) which are even closer $\endgroup$
    – reuns
    Mar 24 '16 at 22:57

In one of Yves Meyer's books on wavelets (Les ondelettes, algorithmes et applications, p.81), I read that the piecewise linear function $\psi$ satisfying $\psi(-1)=0$, $\psi(0)=-1$, $\psi(1/2)=3$, $\psi(1)=-1$ and $\psi(2)=0$ defines a Riesz (nonorthogonal) wavelet basis $\psi_{j,k}(x):=2^{j/2}\psi(2^jx-k)$, $j\in\mathbb Z$, $k\in\mathbb Z$. (Integrating $f''$ against $\psi_{j,k}$ gives a linear combination of values of $f$, and the sum of squares of these is equivalent to $||f''||_{L^2}^2$).

Meyer refers to Cohen-Daubechies-Feauveau for this result.


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