Let $\phi$ be a real compactly supported smooth function on $\mathbb R$ with total integral zero. Define $\phi_t=\frac{1}{t} \phi(\frac{x}{t})$. I also suspect that they must be even, but the notes I am working from are imprecise.

I want to show that $$f(x) \left(\int_0^\infty |\hat\phi(t)|^2 \frac{dt}{t}\right)=\int_0^\infty (\phi_t\ast\phi_t\ast f)(x) \frac{dt}{t}.$$

This can be thought of as a reproducing formula for $f$ because the integral on the left side is a constant (with respect to $x$).

I am attempting to prove this but coming up short. I began by taking the Fourier transform of the right hand side and recalling that it takes convolutions into pointwise multiplication. We get

$$f(\xi) \int_0^\infty \hat\phi_t(\xi)^2 \frac{dt}{t}.$$

We can rewrite the term being integrated because the Fourier transform of ${\frac{1}{t}\phi(\frac{x}{t})}$ is $\hat\phi(t\xi)$.

$$f(\xi) \int_0^\infty \hat\phi(t\xi)^2 \frac{dt}{t}.$$

I do not see how to continue. In particular, I do not see how to make use of the fact that the total integral of $\phi$ is 0.

EDIT: A proof has been given in the comments, but I still don't know why the integral of $\phi$ must be zero! This condition appears in other statements of this formula I have found, not just this simplified one, but it does not seem necessary. I still welcome a definitive answer on this matter.

  • 1
    $\begingroup$ Aren't you done since $$\int_0^{\infty} \hat{\phi}(t\xi)^2 \dfrac{dt}t = \int_0^{\infty} \hat{\phi}(t)^2 \dfrac{dt}t?$$ $\endgroup$
    – user17762
    Jan 9 '13 at 6:04
  • $\begingroup$ @Marvis What about the absolute values? And what about the fact that the total integral of the kernels is zero? We don't know that he Fourier transform is real, right? $\endgroup$
    – Potato
    Jan 9 '13 at 6:06
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    $\begingroup$ I suspect $\phi$ must be even since then $\hat{\phi}(t)$ is real and hence $\hat{\phi}(t)^2 = \vert \hat{\phi}(t) \vert^2$. I do not know how the total integral being zero matters. I think if you are studying harmonic analysis in the context of imaging, typically mean zero functions are considered to eliminate the background and since we are interested only in differences between points in space. $\endgroup$
    – user17762
    Jan 9 '13 at 6:11
  • $\begingroup$ @Marvis If you copy your comments and give them as an answer below, I'd be happy to accept it. $\endgroup$
    – Potato
    Jan 9 '13 at 8:37

The first thing you should determine is in what sense do you want this integral to hold. Do you want it pointwise:

For all $x\in\mathbb R$,


Or do you want it in $L^p$:


Or do you want it in the topology of $\mathcal S(\mathbb R)$, the Schwartz class:

For all multi-indices $\alpha,\beta\in\mathbb N_0^n$,


Each of these things are true for appropriate functions $f$. It is not hard to show that the formula holds in $L^2$ when $f\in L^2$. Plus, if it holds in $L^2$, then it holds pointwise a.e. First fix an $L^2$ function $f$. Then you must show


Since $f\in L^2$ and $\phi_t$ is "nice," it follows that


is in $L^2$ for all $R>0$ (a priori with $L^2$ norm depending on $R$). Then by Plancherel's theorem, it is enough to show $\hat f_R\rightarrow \hat f$ in $L^2$ as $R\rightarrow\infty$, i.e. it is enough to show the following goes to zero as $R\rightarrow \infty$

\begin{align*} \int_{\mathbb R}\left|\hat f_R(\xi)-\hat f(\xi)\right|^2d\xi&=\int_{\mathbb R}|\hat f(\xi)|^2\left|\int_{1/R}^R\hat\phi(t\xi)^2\frac{dt}{t}-1\right|^2d\xi. \end{align*}

One way to do this is to apply dominated convergence in the following way: First uniformly bound

\begin{align*} \left|\int_{1/R}^R\hat\phi(t\xi)^2\frac{dt}{t}-1\right|&\leq1+\int_0^{|\xi|^{-1}}\hat\phi(t\xi)^2\frac{dt}{t}+\int_{|\xi|^{-1}}^\infty\hat\phi(t\xi)^2\frac{dt}{t}\\ &\leq1+||\nabla\hat\phi||_{L^\infty}^2\int_0^{|\xi|^{-1}}(t\xi)^2\frac{dt}{t}+||\nabla\phi||_{L^\infty}^2\int_{|\xi|^{-1}}^\infty(t\xi)^{-2}\frac{dt}{t}\\ &=1+\frac{1}{2}||\nabla\hat\phi||_{L^\infty}^2+\frac{1}{2}||\nabla\phi||_{L^\infty}^2. \end{align*} I'm not completely sure that the $||\nabla\phi||_{L^\infty}$ term is correct here, but it should be some constant depending on the smoothness of $\phi$. Then

$$|\hat f_R(\xi)-\hat f(\xi)|^2\leq|\hat f(\xi)|^2\left(1+\frac{1}{2}||\nabla\hat\phi||_{L^\infty}^2+\frac{1}{2}||\nabla\phi||_{L^\infty}^2\right)^2,$$

which is an $L^1$ function independent of $R$ since $f\in L^2$. So by dominated convergence,

$$\lim_{R\rightarrow\infty}||\hat f_R-\hat f||_{L^2}^2=\int_\mathbb R|\hat f(\xi)|^2\lim_{R\rightarrow\infty}\left|\int_{1/R}^R\hat\phi(t\xi)^2\frac{dt}{t}-1\right|^2d\xi=0.$$

This proves that the formula holds in $L^2$, and hence it follows that it holds pointwise. To prove the $L^p$ convergence is a little more difficult since we cannot make use of Plancherel's theorem, and to prove convergence in the topology of Schartz functions is even more difficult since you must work with Schwartz semi-norms.

A word of caution: In order for this argument to work, you must know that $f\in L^2 $ to start. If $f$ is not in $L^2$ and you prove that $f_R$ (as defined above) is Cauchy in $L^2$, then indeed it does converge to some function $\tilde f\in L^2$, but it is possible that $f\neq\tilde f$. In particular, these two function may differ by a polynomial. This is essentially because $\hat\phi(0)=0$ and the distributional Fourier transform of polynomials are supported at the origin.


Albeit this question is already quite old, I'll give another answer that focuses more on the origin of this task rather then answering the actually posed question in the hope that it will be useful to readers coming from Google or using the search function.

Let's look at classical Littlewood-Palye theory. Let $\varphi$ be a bump function which equals $1$ on $B(0,1)$ and is supported in $B(0,2)$. Put $\varphi_j(\xi) := \varphi(\xi/2^j)$, $\psi_j := \varphi_j - \varphi_{j-1}$ and define Fourier multiplier $P_j := \psi_j(D)$. By Plancherel and dominated convergence, $f=\sum_{j\in \mathbb{Z}} P_j f$ in $L^2$. The $P_j$ operators localize the frequencies of $f$ at $|\xi|\approx 2^j$, which has an effect on how fast $P_j f$ varies: In fact, $f$ is almost constant around $x$ where around means $|x-y| \ll 2^{-j}$ and almost constant can be made precise using the Hardy-Littlewood maximal operator.

So we want to localize functions into frequency patches and if we have two functions localized in frequency we can compare them and use techniques as indicated above to get estimates, and for this comparison it is essential to localize with annuli and not merely balls which would already suffice to bound the varying.

Now we can rewrite $P_j$ into a convolution operator with some kernel $k_j$ given by inverse Fourier transform of the bump on the annulus. What kind if function is $k_j$? Clearly, it is a Schwartz function because the bump was, but it is moreover mean value free because $\int k_j \,dx = \mathcal{F}(f_j)(0) = \psi_j(0) = 0$. So in the discrete case, the mean value property of the convolution kernels doesn't come for technical reasons but is the essential feature why one wants to do such a construction.

And from this point of view, I would always encourage to construct the bumps and therefore convolution kernels in such a way that its easy to show things (which includes showing such representation formulae). So you could for example start with a smooth radial bump on a ball and take a derivative to directly get a bump for an annulus. In such a way it will be easy to calculate the integral using the fundamental theorem. In the case from the original post where the Littlewood-Paley operators should be squared, you have to take roots somewhere, but you can just start with a bump on a ball whose derivatives decay sufficiently rapid to avoid any issues. Just look what you want to achieve and then build your stuff as it's need and most convenient.


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