TITLE: Showing that the best approximating linear map for a Lipschitz function is also Lipschitz
QUESTION [7 upvotes]: For $d,n \in \mathbb{N}$ with $1 \leq d<n$, let $f: \mathbb{R}^d \to \mathbb{R}^n$ be a Lipschitz map with some constant $L \geq 1$. Let $B=B(0,r)$ for some $r>0$ and define the quantity
$$ \Omega(B) = \inf_{A} \left( \frac{1}{|B|} \int_{B} \left(  \frac{|f(y)-A(y)|}{r} \right)^2 dy \right)^{\frac{1}{2}}, $$
where the infimum is over all linear maps $A: \mathbb{R}^{d} \to \mathbb{R}^{n}$ and $|B|$ is just the $d$-Lebesgue measure of $B$.
Suppose that $\Omega(B) < \epsilon$ where $\epsilon$ is as small as we wish. I am trying to show that the infimizing map $A$ for $\Omega(B)$ is also Lipschitz with constant, say, $2L$. This intuitively seems to be true as $\Omega(B)$ being small means that $A$ must approximate an $L$-Lipschitz function $f$ very well inside $B$ and so should not deviate much from $f$. I suspect even without $\Omega(B) < \epsilon$ assumption, the best approximating map should still be $2L$-Lipschitz.
For example if we had defined $``L^{\infty}"$ version of this quantity by
$$\Omega_{\infty}(B) = \inf_{A} \frac{|| f-A ||_{L^{\infty}(B)}}{r}$$
then $ \Omega_{\infty}(B) \leq \frac{|| f-f(0) ||_{L^{\infty}(B)}}{r}  \leq L$  so that for the map $A$ that realizes infimum in $\Omega_{\infty}(B)$,  $|f(x)-A(x)| \leq L r$ for any $x \in B$ and from here it is not difficult to show that $A$ is $2L$-Lipschitz. I am having troubles with the map $A$ when defined for integral version $\Omega(B)$ though. Some help would be appreciated.
References: These quantities originate from Dorronsoro's paper.

REPLY [2 votes]: Important: I prove only a rough estimate, and the result of this estimate is not as sharp as the $2L$ you wanted. However I post it as a weaker result.
As Cristophe Leuridan, we observe that the minimizing function $A$ is the orthogonal projection of $f$ on the subspace of linear functions.
Let's define $E_{ij}$ as the linear function such that
$$E_{ij}[{e_k}]=\delta_{ik}e_j$$
We note that if $(i,j)\neq (i',j')$ then
$$\langle E_{ij},E_{i'j'}\rangle=\int_{B} E_{ij}(x)\cdot E_{i'j'}(x) dx=0$$
As a consequence we can express $A$ as
$$ \sum_{i,j}\frac{\langle f,E_{ij}\rangle}{\langle E_{ij},E_{ij}\rangle}E_{ij}$$
Then
$$\vert \vert {A}\vert \vert_2\leq \sum_{i,j}\frac{\vert \langle f,E_{ij}\rangle\vert}{\langle E_{ij},E_{ij}\rangle}\vert \vert E_{ij}\vert \vert_2 $$
We now note that $\langle E_{ij}, E_{ij}\rangle$ does not depend on $(i,j)$. Let's call $K=\langle E_{ij},E_{ij}\rangle$: The previous inequality becomes
$$\vert \vert A\vert \vert_2\leq \frac{1}{K}\sum_{i,j}\vert\langle f,E_{ij}\rangle\vert$$
We notice that
$$ \sum_{i,j}\vert\langle f,E_{ij}\rangle\vert=\sum_{i,j}\vert\int_{B} x_i f(x)\cdot e_j dx\vert=\sum_{i,j}\vert\int_{B} x_i f_j(x)dx\vert\leq \sum_{i,j}\int_{B}\vert x_i\vert \cdot \vert f_j(x)\vert dx $$
We now apply Cauchy Schwarz inequality, obtaining
$$\sum_{i,j}\int_{B}\vert x_i\vert \cdot \vert f_j(x)\vert dx \leq \sqrt{nd}\int_{B} \vert \vert x\vert \vert_2 \cdot \vert \vert f(x) \vert \vert_2 dx \leq \sqrt{nd} (\vert \vert f(0)\vert \vert_2+Lr)\int_{B}\vert \vert x\vert \vert_2 dx$$
Then
$$\vert \vert A \vert \vert_2\leq \frac{ \sqrt{nd}(\vert f(0)\vert+Lr)}{K}\int_{B}\vert \vert x\vert \vert_2 dx$$
Now it's only a matter of computing the RHS. We notice that
$$K=\int_{B} x_1^2 dx$$
and so
$$ dK=\int_{B} \vert \vert x \vert \vert _2 ^2 dx $$
which gives
$$ \vert \vert A\vert \vert_2\leq  n^{\frac{1}{2}}d^{\frac{3}{2}} (\vert f(0)\vert +Lr) \frac{\int_{B} \vert \vert x\vert \vert_2 dx}{\int_{B}\vert \vert x\vert \vert_2^2 dx}$$
If I'm not mistaken, the result of the fraction in RHS is
$$ \frac{\int_{0}^{r} z^d dz}{\int_{0}^{r} z^{d+1} dz}=\frac{d+2}{(d+1)r}$$
As a result, we have
$$ \vert \vert A\vert \vert_2\leq \frac{d+2}{d+1}n^{\frac{1}{2}}d^{\frac{3}{2}} \frac{\vert f(0)\vert +Lr}{r}  $$
Edit: as Cristophe Leuridan noticed, $A$ is not only the orthogonal projection of $f$, but more in general of $f+m$ where $m$ is an arbitrary constant. As a consequence, we can assume $f(0)=0$, obtaining
$$ \vert \vert A \vert \vert _2 \leq \frac{d+2}{d+1}n^{\frac{1}{2}}d^{\frac{3}{2}}L$$
Edit: Now I show an estimate starting from the last part of Cristophe's answer. Let's say that we are searching for a constant $\alpha$ and a constant $m$ such that
$$ \langle f+m, f+m \rangle \leq \alpha^2 L^2 K$$
The role of constant $m$ is to fix the value that $f$ takes in $0$ (for example). So let's assume that $f(0)=0$, that is equivalent to taking $m=-f(0)$. We are then searching $\alpha$ such that
$$ \alpha^2 \geq \frac{\langle f ,f\rangle}{KL^2}$$
We notice that
$$ \langle f,f\rangle =\int_{B} f(x)^2 dx$$
and that $g(x)=f(x)\cdot f(x)$ is $2rL^2$ Lipschitz on $B$, since it is the composition of $\vert \vert \cdot \vert \vert_2^2$ (which is $2rL$- Lipschitz on $B(0,\sup_{x\in B}\vert \vert{f(x)}\vert \vert_2)$) and $f$ (which is $L$-Lipschitz). As a consequence
$$ \alpha^2 \geq \frac{2rL^2 d}{L^2}\frac{\int_{B}\vert \vert x \vert \vert _2 dx}{\int_{B} \vert \vert x \vert \vert_2 ^2 dx}$$
and so
$$ \alpha^2 \geq \frac{2(d+2)}{d+1}\cdot d$$
So we can take
$$ \alpha =\sqrt{\frac{2(d+2)}{d+1}}d^{\frac{1}{2}}$$