# Showing that the best approximating linear map for a Lipschitz function is also Lipschitz

For $$d,n \in \mathbb{N}$$ with $$1 \leq d, let $$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.

• Ok, just a preliminary question, how can you be sure that some "infimizing" map exists? I am asking because that should give useful information. (Also, the scaling factor $r$ seems to be irrelevant). May 3, 2022 at 18:11
• Deleted my last comment, was too late to edit. Should exists as we can take a subsequence that would converge to a function realizing the quantity $\Omega(B)$. The map must not be unique though I think. May 3, 2022 at 18:40
• It seems that this question did not get much traction, unfortunately. It is rather technical but interesting. (Your edit with the $L^\infty$ norm is a nice toy model, though.) To simplify the exposition, I suggest that you drop that denominator $r$ in the inf. It is irrelevant to the present problem and it makes formulas messier. May 9, 2022 at 17:39
• @GiuseppeNegro Yeah sadly. For the purpose of this problem $r$ is not relevant indeed but I still prefer keeping it in its original form - stemming from the work of Dorronsoro in "A Characterization of Potential Spaces" originally I believe. May 10, 2022 at 8:34
• Then by all means include that citation in the main text, possibly with a link. May 10, 2022 at 10:26

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.

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 $$\alpha =\sqrt{\frac{2(d+2)}{d+1}}d^{\frac{1}{2}}$$

• Thanks for the extra estimates - I wish I could share the bounty. May 16, 2022 at 20:51