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 $f_i$.

  • $\begingroup$ $B(0,M) \setminus B(0,\epsilon)$ 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 $\endgroup$ Commented Sep 22, 2019 at 8:55
  • $\begingroup$ @CalvinKhor You are right, it was poorly formulated. I edited thanks to your comment. $\endgroup$
    – E. Joseph
    Commented Sep 22, 2019 at 10:37
  • $\begingroup$ 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 $\frac {\pi}2$. $\endgroup$ Commented Sep 23, 2019 at 14:31
  • 1
    $\begingroup$ @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 $\det(X_1,\ldots,X_k)>0$? $\endgroup$
    – E. Joseph
    Commented Sep 23, 2019 at 15:08

1 Answer 1


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.

  • 1
    $\begingroup$ Great work, thanks a lot! $\endgroup$
    – E. Joseph
    Commented Sep 23, 2019 at 19:32

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