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 $f''$?

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. It's also worth mentioning that the main reason which led me to ask this question is curiosity (about what behavior of a function can wavelets capture).

  • $\begingroup$ go to the discrete wavelet transform instead and define precisely what you want. $\delta(n-1) - \delta(n)$ is a finite difference filter and is also an approximation of a derivative filter, $(\delta(n-1) - \delta(n)) \ast (\delta(n-1) - \delta(n)) \ast \delta(n+1)$ is an approximation to the second derivative, but there is also the ideal (in the Shannon sense) derivative filter : $\sum_{k=1}^\infty \frac{\delta(n-k-1)- \delta(n+k)}{k}$ $\endgroup$ – reuns Feb 15 '16 at 23:25
  • $\begingroup$ @user1952009 sorry I have no idea what the discrete wavelet transform is, nor what $\delta(n)$ or a finite difference filter are. Could you explain? $\endgroup$ – Silvia Ghinassi Feb 15 '16 at 23:27
  • $\begingroup$ @Silver : no you'll understand yourself, and $\delta(n) = 1$ only if $n= 0$ otherwise $\delta(n) = 0$ : it is the discrete Dirac delta $\endgroup$ – reuns Feb 15 '16 at 23:29
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    $\begingroup$ What I don't know, as I mentioned in the question, is the theory of wavelets, so without explanations this means nothing to me. Even less discrete wavelets transform. Why are you so rude to me"? (e.g. "no you'll understand yourself") And my name is not a metal. $\endgroup$ – Silvia Ghinassi Feb 15 '16 at 23:33
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    $\begingroup$ It is very rude. I'm glad you have an answer for me, but if you don't want to spend the time to give it to me, just say it. I'm ok with that. Throwing terms I don't know at my face while insulting me won't help me, nor you. $\endgroup$ – Silvia Ghinassi Feb 15 '16 at 23:38

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