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Let $f:\mathbb{R}^2\to \mathbb{R}$ be $L$-Lipschitz. Let $f_\varepsilon:=f*\eta_\varepsilon$ be its smooth $\varepsilon$-mollification, where

  • $\eta_\varepsilon(x)=\frac{1}{C\varepsilon^2}\eta(|x|/\varepsilon)$ and
  • $\eta:\mathbb{R}\to\mathbb{R}$ is a standard smooth cut-off function taking values in $[0,1]$, integrating to $C$ and such that $\eta=1$ on $(-1,1)$, $\eta=0$ outside $(2,2)$, and $|\eta'|\leq 2$.

It is standard that $f_\varepsilon$ is also $L$-Lipschitz and that, as $\varepsilon\to 0$, $f_\varepsilon \to f$ pointwise and $Df_\varepsilon \to Df$ $\mathcal{L}^2$-a.e.

Clearly, $f$ restricted to $\mathbb{R}\times \{0\}$ is still Lipschitz, and hence (by Rademacher's theorem) for $\mathcal{H}^1$-a.e. $X\in P$ the derivative $\frac{\partial f}{\partial x}$ exists at $X$. I am trying to figure out if it is true that, for $f_\varepsilon$ defined as above (i.e. the mollification of $f$ on the whole $\mathbb{R}^2$), there exists a sequence $\varepsilon_j\searrow 0$ such that the following holds: $$ \frac{\partial f_{\varepsilon_j}}{\partial x}(X)\to \frac{\partial f}{\partial x}(X)\quad\quad\text{for $\mathcal{H}^{1}$-a.e. $X\in \mathbb{R}\times \{0\}$.} \label{1}\tag{$\star$} $$ Of course, if we parametrise the horizontal lines by their vertical component, i.e. $\mathbb{R}\times \{t\}$ for $t\in \mathbb{R}$, then Fubini implies that \eqref{1} holds for $\mathcal{L}^1$-a.e. $\mathbb{R}\times \{t\}$.

I am looking for either a proof that \eqref{1} holds in fact for all lines (with the sequence $(\varepsilon_j)$, and potentially also the mollifier $\eta$, depending on the line), or for an explicit counterexample.

Remark: I am aware of the fact that, for each given $t$, it is possibile (using Whitney's approximation theorem) to find a sequence of smooth functions (possibly different from the mollifications) such that the convergence holds on $\mathbb{R}\times\{t\}$, but here I am only interested in the sequence of the mollifications.

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  • $\begingroup$ It seems to me that this question should be a bit clarified. I start with your fonction $f$. Now I call $g$ the restriction of $f$ to $\mathbb R\times\{0\}$, which is indeed also Lipschitz-continuous. Your question seems to be: is it true that $\partial_1 f_\epsilon(x,0)$ converges to $g'(x)$, a.e. in $\mathbb R$. In particular, you cannot restrict the $L^\infty$ function $\partial_1f$ to a set of measure $0$. A hunch would be that it is true when the mollifier is a tensor product $\rho_1(x_1)\rho_2(x_2)$ and more difficult or untrue in the other cases. $\endgroup$ Commented May 14 at 14:49
  • $\begingroup$ @Bazin The mollified function $f_\varepsilon$ is smooth, so its restriction to $\mathbb{R}\times 0$ makes sense everywhere, and I am asking if it converges to $(f|_{\mathbb{R}\times 0})'$, which also makes sense $\mathcal{H}^1$-a.e. as $f$ is Lipschitz. I suspect the answer to be negative, by the way. $\endgroup$ Commented May 16 at 21:29
  • $\begingroup$ I agree with the formulation in your comment, but not completely with the way your question is written. I would also support your suspicion, except maybe in the tensor product case where the mollifier is tailored on the Line of restriction. $\endgroup$ Commented May 18 at 16:45
  • $\begingroup$ Please stop editing your question hence and forth - every edit bumps it to the front page, which is a nuisance for other people. $\endgroup$ Commented Oct 16 at 6:03

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What I want to write is too long for a comment, so let me write it as an answer. I want to point out a couple of basic facts related to your question. Let $f:\mathbb R^n\longrightarrow \mathbb R$ be a bounded measurable function such that its distribution-derivative $f'$ is also bounded measurable. I follow your notations (in $n$ dimensions) and set $$ f_\epsilon=f\ast\eta_\epsilon, \tag{1}$$ so that we have $ f_\epsilon(x) -f(x)=\int\bigl(f(x-\epsilon z)-f(x)\bigr) \eta(z)dz, $ yielding immediately $$ \sup_x\vert f_\epsilon(x) -f(x)\vert\le \epsilon L \int\vert z \eta(z)\vert dz. $$ The derivative estimate can be derived from the Lebesgue differentiation theorem: indeed we have $$ f'_\epsilon(x) -f'(x)= \int\bigl(f'(x-y)-f'(x)\bigr) \eta(y/\epsilon)dy\epsilon^{-n}, $$ so that, assuming that $\eta$ is supported in the unit ball, we get \begin{multline} \vert f'_\epsilon(x) -f'(x)\vert\le \epsilon^{-n}\Vert \eta\Vert_{L^\infty} \int_{\vert y-x\vert\le \epsilon}\vert f'(y)-f'(x)\vert dy \\\le \frac{C}{\vert B(x,\epsilon)\vert} \int_{B(x,\epsilon)}\vert f'(y)-f'(x)\vert dy. \end{multline} The Lebesgue differentiation theorem tells you that there is a set $\mathcal L$ (the Lebesgue points of $f'$) whose complement has zero Lebesgue measure such that for all $x\in \mathcal L$ $$ \lim_{\epsilon\rightarrow 0}\frac{1}{\vert B(x,\epsilon)\vert}\int_{B(x,\epsilon)}\vert f'(y)-f'(x)\vert dy=0. $$ Moreover you have obviously $ \Vert f'_\epsilon\Vert_{L^\infty}\le \Vert f'\Vert_{L^\infty}\Vert\eta\Vert_{L^1}. $

Note that the Lebesgue differentiation theorem is highly non-trivial and that the argument for the derivative works if you have only $f'$ locally integrable.

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Here is a counterexample. Let $\eta,\eta_\varepsilon,f_\varepsilon$ be as in the question.

First we consider scales close to $1$; consider the unit squares $S_{n}=[10000n,10000n+1]\times[1,2]$, for integer $n$. Consider copies $\phi_n(x)=\phi(x-10000n)$ of some $1$-Lipschitz non-negative bump-function $\phi$ with integral $\lambda\in[0.001,1]$, so that $\phi_n$ is supported at $S_n$. Let $f_0=\sum_n\phi_{n}$ and $(f_0)_\varepsilon=f_0*\eta_\varepsilon$

The function $f_0$ is just $0$ in the $\mathbb{R}\times\{0\}$. But for any $\varepsilon\in[3,300]$, the derivative of $(f_0)_\varepsilon=f_0*\eta_\varepsilon$ is not too close to $0$, as $(f_0)_\varepsilon(x,0)$ has value $>10^{-10}$ when $x\in10000\mathbb{Z}$ and $(f_0)_\varepsilon(x,0)=0$ for $x\in10000\mathbb{Z}+5000$. This implies, taking into account that $(f_0)_\varepsilon$ is $1$-Lipschitz, that the set $\left\{x\in\mathbb{R};\left|\frac{\partial}{\partial x}(f_0)_\varepsilon(x,0)\right|>10^{-100}\right\}$ has measure $>10^{-100}$ in each interval of length $10000$.

Now let's go down to smaller scales. For each $N\in\{0,1,2,\dots\}$, let $S_{N,n}=\frac{S_N}{100^N}$ and $\phi_{N,n}(x)=100^{-N}\phi_{n}(100^Nx)$ (so, the graph of $\phi_{N,n}$ is the graph of $\phi_n$ but scaled $\frac{1}{100^N}$ with respect to the origin). Let $f=\sum_{N,n}\phi_{N,n}$. From now I write $f(x)$ instead of $f(x,0)$ to abbreviate, similarly with $f_\varepsilon$.

In order to conclude, it will be enough to check that, for each $\varepsilon\in(0,1)$, the set $\left\{x\in[0,10000];|f_\varepsilon'(x)|>10^{-100}\right\}$ has measure $>10^{100}$; that would imply that for no sequence $(\varepsilon_j)\to0$ can we have $f_{\varepsilon_j}'(x)\to f'(x)=0$ for a.e. $x\in \mathbb{R}$.

Indeed, fix $\varepsilon>0$ and choose $N$ such that $\varepsilon\in\left[\frac{3}{100^{N}},\frac{300}{100^{N}}\right]$. Note that for all $x\in\frac{10^{6}\mathbb{Z}}{100^N}$, we have $f_\varepsilon(x)=f_\varepsilon(0)\geq\frac{10^{-10}}{100^N}$: we have $f_\varepsilon(x)=f_\varepsilon(0)$ because the squares $S_{k,n}$ with $k\leq N-2$ are too far from the real line to affect $f_\varepsilon$, and for $k\geq N-1$ the squares $(S_{k,n})_{n\in\mathbb{Z}}$ repeat in patterns with period dividing $\frac{10^{6}}{100^N}$. And we have $f_\varepsilon(0)\geq\frac{10^{-10}}{100^N}$ as the integral of $\phi_{N,n}$ is in $\left[\frac{0.001\lambda}{10^{6N}},\frac{\lambda}{10^{6N}}\right]$ and the maximum of $\eta_\varepsilon$ is $10^{4N}$.

But for all $x\in \frac{10000\mathbb{Z}+5000}{100^N}$, we have $f_\varepsilon(x)\leq \frac{1}{2}f_\varepsilon(0)$, because the ball $B(x,2\varepsilon)$ cannot intersect $S_{k,n}$ for any $n$ if $k\leq N$), and the integrals of the functions $\phi_{k,n}$ with $k>N$ have much smaller mass than $\phi_{N,n}$.

Thus, in each interval of length $\frac{10000}{100^N}$, the variation of $f_\varepsilon$ is at least $\frac{10^{-11}}{100^N}$. As $f_\varepsilon$ is $1$-Lipschitz, this implies that $\{x\in\mathbb{R};|\frac{\partial}{\partial x}f_\varepsilon(x,0)|>10^{-100}\}$ has measure $>\frac{10^{-100}}{100^N}$ in each interval of length $\frac{10000}{100^N}$. So $\{x\in\mathbb{R};|\frac{\partial}{\partial x}f_\varepsilon(x,0)|>10^{-100}\}$ has measure $>10^{-100}$ in each interval of length $10000$, so we are done.

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