# Showing the Lipschitz continuity of a coordinates function

Notations.

Let $$E$$ be a subspace of dimension $$k$$ of $$\mathbb R^n$$, let $$(e_1,\ldots,e_k)$$ be an orthonormal basis of $$E$$.

Let $$(X_1,\ldots,X_{k-1})$$ be an orthonormal family of $$E$$. We can write the $$X_i$$ in terms of the $$e_j$$:

$$\forall i\in\{1,\ldots,k-1\},\quad X_i=\sum_{j=1}^k x_{i,j}e_j.$$

Since $$\dim E=k$$, there is a unique vector $$Y\in E$$ such that $$(X_1,\ldots,X_{k-1},Y)$$ is orthonormal and $$\det(X_1,\ldots,X_k)>0$$.

We can then define

$$f_1,\ldots,f_k\colon K\to \mathbb R$$

where $$K$$ is the compact of $$\mathbb R^{k(k-1)}$$ where $$(x_{1,1},\ldots,x_{k-1,k})$$ correspond to an orthonormal family $$(X_1,\ldots,X_{k-1})$$, such that $$f_i$$ is the $$i$$-th coordinates of $$Y$$ in the basis $$(e_1,\ldots,e_k)$$, i.e.

$$Y=\sum_{i=1}^k f_i(x_{1,1},\ldots,x_{k-1,k})e_i.$$

The question.

Let $$i\in\{1,\ldots,k\}$$.

Is $$f_i$$ Lipschitz continuous on $$K$$?

This means: does there exist $$L$$ such that

$$\forall x,y\in K,\quad \vert f_i(x)-f_i(y)\vert\leqslant L\Vert x-y\Vert?$$

What we know.

We know that the $$f_i$$ are continuous, and since $$K$$ is compact, Heine's theorem tells us that the $$f_i$$ are uniformly continuous. This is why I have great hopes that they will be Lipschtiz continuous, but I don't know how to prove it, and it seems really complicated to do a direct computation of the $$f_i$$.

• $B(0,M) \setminus B(0,\epsilon)$ is not compact, if $B$ denotes either an open ball, or a closed ball. Also it seems $f_i$ are not defined on all of $\mathbb R^{k(k-1)}$, but only when those coordinates correspond to those of an orthonormal family of $k-1$ vectors Commented Sep 22, 2019 at 8:55
• @CalvinKhor You are right, it was poorly formulated. I edited thanks to your comment. Commented Sep 22, 2019 at 10:37
• It seems for me that the situation is similar to that when $k=3$ and $Y=X_k$ is collinear to the vector product $X_1\times X_2$, so we can express it as a linear combination of minors of the matrix $(x_{i,j})$. Unfortunately, the condition $X_k\cdot e_k>0$ can break the continuity of $f_i$. For instance, when $k=n=2$, and $X_1=(\cos\varphi,\sin\varphi)$, we have that $X_2$ flips over the $y$-axis when $\varphi$ is at $\frac {\pi}2$. Commented Sep 23, 2019 at 14:31
• @AlexRavsky You are right, the issue here is the condition $X_k\cdot e_k$. The good thing is that I don't really care about this arbitrary condition, it is just here to give a unique condition so we can define $X_k$. Do you think it would work if we replace this condition by $\det(X_1,\ldots,X_k)>0$? Commented Sep 23, 2019 at 15:08

Consider $$k\times k$$ matrix $$X=\|x_{ij}\|$$. By othonormality of the family $$(X_i)$$, we have $$XX^t=I$$, where $$X^t$$ is the transpose of the matrix $$X$$. Thus $$(\det X)^2=\det X\cdot \det X^t=\det I=1$$. Since $$\det X$$ is a positive real number, it equals $$1$$. Thus $$X=(X^t)^t=(X^{-1})^t=C$$ (see, for instance, Wikipedia), where $$C=\|C_{ij}\|$$ is the cofactor matrix of $$X$$. It follows that for each $$l$$, $$f_l=C_{k,l}$$, so $$f_l$$ is a polynomial function depending of variables $$x_{i,j}$$, $$1\le i\le k-1$$, $$1\le j\le k$$ defined on a compact subset $$K$$ of $$\Bbb R^{k(k-1)}$$, thus $$f_l$$ is Lipschitz.