Showing that the best approximating linear map for a Lipschitz function is also Lipschitz 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.
 A: Partial answer.
The quantity $\Omega(B)$ is $r^{-1}$ times the $L^2$ norm of $f-A$ associated to the uniform probability measure on the ball $B(0,r)$ (denoted by $\mu$).
I use the notation $$||f||_{\mathrm{Lip}} = \sup_{x \ne y }\frac{||f(y)-f(x)||}{||y-x||}$$ for the best Lipschitz constant.
On needs only to consider the case where $n=1$. Indeed, restriction, since minimizing
$$||f-A||_2^2 = \sum_{i=1}^n ||f_i-A_i||_2^2$$
is equivalent to minimizing separately each $||f_i-A_i||_2^2$ for $1 \le i \le n$, and one has
$$||f||_{\mathrm{Lip}}^2 = \sum_{i=1}^n ||f_i||_{\mathrm{Lip}}^2$$
and the same relation for $A$.
Call $(b_1,\ldots,b_d)$ the canonical basis in $\mathbf{R}^d$ and $(e_1,\ldots,e_d)$ its dual basis. Then $(e_1,\ldots,e_d)$ is an orthogonal family in $L^2(\mu)$ and a basis of the subspace of all linear functions. All vectors of this basis have the same norm, and
$$||e_1||_2^2+\ldots+||e_d||_2^2 = |B|^{-1} \int_B (x_1^2+\ldots+x_d^2) \mathrm{d x} = (v_d r^d)^{-1} \int_0^r r^2 dv_dr^{d-1} \mathrm{d}r = \frac{r^2}{d+2}.$$
The linear map $A$ which minimizes $||f-A||_2^2$ is the orthogonal projection of $f$ on the subspace of all linear functions. Hence
$$A = \sum_{j=1}^d  \frac{\langle f,e_j \rangle}{\langle e_j,e_j \rangle} e_j.$$
Given $x$ and $y$ in $B$,
$$(A(y)-A(x))^2 = \sum_{j=1}^d \Big(\frac{\langle f,e_j \rangle}{\langle e_j,e_j \rangle} (y_j-x_j) \Big)^2.$$
By Cauchy-Schwarz inequality,
$$(A(y)-A(x))^2 \le \sum_{j=1}^d \Big(\frac{\langle f,e_j \rangle}{\langle e_j,e_j \rangle}\Big)^2 ||y-x||^2 = \frac{||A||_2^2}{||e_1||_2^2} \times ||y-x||^2.$$
Since the linear maps have null average on $B$ (by imparity), $A$ is also the orthogonal projection of $f-E(f)$, where $E(f)$ denotes the mean value (expectation) of $f$ with regard to $\mu$. Hence
$$|A(y)-A(x)| \le \frac{||f-E(f)||_2}{||e_1||_2} ||y-x||.$$
Now, what we remains to be proven is that $||f-E(f)||_2 \le L||e_1||_2$ when $f$ is $L$-Lipschitz. In other words, the functions $L\langle u,\cdot\rangle$ where $u$ is a unit vector minimize the norm in $L^2(\mu)$ among all $L$-Lispchitz functions with null average on $B$. Here is an intuitive reason: using Fubini's theorem, one sees that
$$2||f-E(f)||_2^2 = \int_B \int_B \big(f(y)-f(x)\big)^2 \mathrm{d}\mu(x)\mathrm{d}\mu(y).$$
One may assume that $f$ is $\mathcal{C}^1$. Then, the way to maximize this quantity under the constraint that $f$
is $L$-Lispchitz is that the gradient of $f$ is constant.
A: 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
$$ \alpha =\sqrt{\frac{2(d+2)}{d+1}}d^{\frac{1}{2}}$$
