# Convergence rate for $L^2$ convergence

Let $f \in L^2(\mathbb R)$ then it is well-known that $$\widetilde{f}(x):=\sum_{n \in \mathbb Z} \frac{1}{\varepsilon}\int_{[n\varepsilon,(n+1)\varepsilon]} f(s) \ ds 1_{[n\varepsilon,(n+1)\varepsilon)}(x)$$

converges in the $L^2$ sense to $f.$

But even more is true, as we can write

$$(\widetilde{f}-f)(x):=\sum_{n \in \mathbb Z} \frac{1}{\varepsilon}\int_{[n\varepsilon,(n+1)\varepsilon]} f(s)-f(x) \ ds 1_{[n\varepsilon,(n+1)\varepsilon)}(x).$$

Then, it follows that if $f$ is absolutely continuous that $$(\widetilde{f}-f)(x):=\sum_{n \in \mathbb Z} \frac{1}{\varepsilon}\int_{[n\varepsilon,(n+1)\varepsilon]} \int_x^s f'(\tau) d \tau \ ds 1_{[n\varepsilon,(n+1)\varepsilon)}(x).$$

Hence, it is tempting to ask whether there are conditions on $f$ being in some Sobolev space such that

$$\left\lVert \widetilde{f}-f \right\rVert_{L^2}=\mathcal O(\varepsilon).$$

Moreover, perhaps this generalizes to higher dimensions, i.e. $\mathbb R^d$ rather than $\mathbb R.$

It seems that what would be sufficient is that $$\sum_{n \in \mathbb Z} \frac{1}{\varepsilon} \sup_{s \in [n \varepsilon, (n+1)\varepsilon)}\left\lvert f'(s)\right\rvert 1_{[n\varepsilon,(n+1)\varepsilon)}(x)$$ is square-integrable. However, I fail to see whether there is some natural Sobolev space in which functions have this property.


As you noted, \begin{equation*} \tf-f=\frac{1}{\ep}\sum_n a_n 1_{[n\ep,(n+1)\ep)}, \end{equation*} where \begin{equation*} a_n(x):=\int_{n\ep}^{(n+1)\ep} \int_x^s f'(\tau) d \tau \ ds. \end{equation*} For $x\in[n\ep,(n+1)\ep)$, \begin{equation*} |a_n(x)|\le\int_{n\ep}^{(n+1)\ep}\int_{n\ep}^{(n+1)\ep} |f'(\tau)| d \tau\ ds =\int_{n\ep}^{(n+1)\ep} |f'(\tau)| d \tau\ \ep\le b_n\ep\sqrt\ep, \end{equation*} where \begin{equation*} b_n:=\sqrt{\int_{n\ep}^{(n+1)\ep} |f'(\tau)|^2 d \tau}. \end{equation*} So, \begin{equation*} |\tf-f|\le\sum_n b_n \sqrt\ep \, 1_{[n\ep,(n+1)\ep)} \end{equation*} and \begin{equation*} \|\tf-f\|_2\le\sqrt{\sum_n b_n^2 \ep^2}=\ep\|f'\|_2\le\ep\|f\|_{W^{1,2}}, \end{equation*} as desired.

Consider now the general case when $f$ is a function on $\R^d$ with $d\ge2$ and \begin{equation*} \tf=\frac{1}{\ep^d}\sum_{n\in\Z^d} 1_{\de_n}\int_{\de_n} ds\,f(s), \end{equation*} where $\de_n:=\prod_1^d\,(n_i\ep,(n_i+1)\ep)$ for $n=(n_1,\dots,n_d)$. Take any real \begin{equation*} p>d \end{equation*} and let $q:=\frac p{p-1}$, so that $\frac1p+\frac1q=1$. We have
\begin{equation*} \tf-f=\frac{1}{\ep^d}\sum_n a_n 1_{\de_n}, \tag{!} \end{equation*} where \begin{equation*} a_n(x):=\int_{\de_n} ds \int_0^1 dt\, f'(x+t(s-x))\cdot(s-x) \end{equation*} $f':=\nabla f$ and $\cdot$ is the dot product. By the change of variables from $s$ to $\tau=x+t(s-x)$, for $x\in\de_n$ we have \begin{equation*} a_n(x)=\int_0^1 dt\int_{\de_n} \frac{d\tau}{t^d} \, f'(\tau)\cdot\frac{\tau-x}t\,1_{(t_*,\infty)}(t), \end{equation*} where \begin{align*} t_*:=t_*(n,x,\tau)&:=\max_1^d\Big(\frac{\tau_i-x_i}{x_i-n_i\ep}\vee\frac{x_i-\tau_i}{(n_i+1)\ep-x_i}\Big) \\ &\ge\max_1^d\frac{|\tau_i-x_i|}{\ep}\gg\frac{|\tau-x|}{\ep}, \end{align*} with the constants in $\gg$ and $\ll$ depending only on $d$ and $p$, and
$|\ \, |$ also denoting the Euclidean norm on $\R^d$. Note that \begin{equation*} \int_{t_*}^\infty\frac{dt}{t^{d+1}}\ll\frac1{t_*^d}\ll\frac{\ep^d}{|\tau-x|^d}. \end{equation*} It follows that for $x\in\de_n$ \begin{equation*} |a_n(x)|\ll\ep^d\,\int_{\de_n} \frac{d\tau}{|\tau-x|^{d-1}} \, |f'(\tau)| \le\ep^d\,b_n c^{1/q}, \tag{!!} \end{equation*} by H\"older's inequality, where \begin{equation*} b_n:=\Big(\int_{\de_n}d\tau\,|f'(\tau)|^p\Big)^{1/p}, \end{equation*} \begin{equation*} c:=\int_{|\tau-x|\le\ep\sqrt d}\frac{d\tau}{|\tau-x|^{(d-1)q}} \ll\int_0^{\ep\sqrt d}\frac{dr}{r^{(d-1)q}}\,r^{d-1} \ll\ep^{(p-d)/(p-1)}. \end{equation*} Thus, by (!) and (!!), \begin{equation*} \|\tf-f\|_p\ll\Big(\sum_n b_n^p \ep^{p-d}\ep^d\Big)^{1/p}=\ep\|f'\|_p\le\ep\|f\|_{W^{1,p}}, \end{equation*} as desired.

• thank you, but does this generalize to higher dimensions? – Selberg May 20 '18 at 10:30
• @Selberg : Yes, it does generalize. I have now provided this generalization as well. – Iosif Pinelis May 20 '18 at 17:40
• Nice answer. I cannot vote you up twice :-( – Piotr Hajlasz May 20 '18 at 18:12
• Thank you Piotr. Maybe I should have written down both parts of this answer at once, rather than in two steps. :-) – Iosif Pinelis May 20 '18 at 20:06
• By the way, you might be interested in the fact that the converse of this question is also, in a certain sense, true. That is, you can define Sobolev spaces and norms (even fractional and negative spaces, and for every exponent p) in terms of spatial averages of the function on every scale. For higher derivatives, one has to use a smoother function than a characteristic function. See for instance the appendix of math.ens.fr/~mourrat/lecturenotes.pdf – Scott Armstrong May 22 '18 at 19:37