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Given a multiresolution analysis $(V_j)_{j\in \mathbb{Z}}$ in $L²(ℝ)$ with scaling function $\varphi \in L²(ℝ)$. As $(2^{1/2}\varphi(2t-k))_{k\in \mathbb{Z}}$ is an orthonormal basis of $V_1$ and $\varphi \in V_0\subseteq V_1$ we have $$\varphi(t)= \sum\limits_{k\in \mathbb{Z}}^{}\langle\varphi, 2^{1/2}\varphi(2t-k)\rangle 2^{1/2}\varphi(2t-k).$$ This equation is called scaling equation. Does this equation hold for almost every $t\in ℝ$? And if yes, why?

cmk
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anjo1659
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  • If $(e_k)$ is an orthonromal basis of a Hilbert space $V$, then for each $f\in V$ we have $f = \sum_k\langle f,e_k\rangle e_k$. – amsmath Oct 09 '19 at 20:42
  • So does it hold almost everywhere or not? Because sometimes I can read this! – anjo1659 Oct 09 '19 at 20:43
  • Hmm, I doubt it because a celebrated result by Carleson says that this is true if $(e_k)$ is the Fourier basis. Here, the basis is rather arbitrary. – amsmath Oct 09 '19 at 20:48
  • Then I don't understand the pages 153 and 154 here (https://books.google.de/books?id=EUH4DQAAQBAJ&printsec=frontcover&hl=de&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false) – anjo1659 Oct 09 '19 at 20:53
  • What exactly don't you understand? – amsmath Oct 09 '19 at 20:55
  • At the beginning of the page 154 it says that this scaling equation holds for almost every $t\in ℝ$. Thats exact the opposite what you have said. – anjo1659 Oct 09 '19 at 20:58
  • Equality is as elements of $L^2$. Which means that the points of difference form a set of measure zero. So the equality is almost everywhere. – Lutz Lehmann Oct 09 '19 at 20:59
  • What they mean is not that the series converges for a.e. $t$. You have to understand the series as an $L^2$-limit $g$, so $\phi = g$ in $L^2$ and that means equality almost everywhere. – amsmath Oct 09 '19 at 21:01
  • ahhh thank you, but it is wrong to say, if $g$ is the series, that $\varphi(t)=g(t)$ for almost every t holds? – anjo1659 Oct 09 '19 at 21:03
  • No, that's correct. But it's wrong that $\phi(t) = \sum_ka_k\phi_k(t)$ for a.e. $t$ (where the series is a different one for each $t$). – amsmath Oct 09 '19 at 21:03
  • But where is there the difference? It is just somehow renaming the series with a different letter. I don't get it. But thank you so far! – anjo1659 Oct 09 '19 at 21:09
  • $L^p$ convergence implies pointwise almost everywhere convergence of a subsequence, not the original sequence per se. – Cameron Williams Oct 09 '19 at 21:46
  • See here : https://math.stackexchange.com/questions/138043/does-convergence-in-lp-implies-convergence-almost-everywhere – Cameron Williams Oct 09 '19 at 21:47
  • The scaling equation on the Fourier side but holde in L² sense and almost everywhere because of Carlsons theorem or? – anjo1659 Oct 10 '19 at 06:33

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