Distance from nonnegativity of some orthonormal vectors Suppose that $1 < k < n$. Does there exist a constant $\beta > 0$, such that for every $k$ orthonormal vectors $f_1,\ldots,f_k \in \mathbb R^n$,
there exist $k$ orthonormal vectors with nonnegative elements, $x_1,\ldots,x_k\in \mathbb R_+^n$, such that 
$$\sum_{i=1}^k \|x_i - f_i\|^2_2 \leq \beta \sum_{i=1}^k\|f_i^-\|_2^2$$
where $f_i^- := \max\{-f_i,0\}$ is the negative part of the vectors $f_i$?

In another way, I am interested in the estimation of the distance of a $n\times k$ dimensional matrix $F$ whose columns are orthonormal from the set of $n\times k$ matrices whose columns are nonnegative and orthonormal, i.e., 
$$
\mathrm{dist}(F;St_+(n,k))
$$
Where $St_+(n,k)$  is the set of $n\times k$ matrices whose columns are nonnegative and orthonormal. If we drop orthonormal condition and compute $\mathrm{dist}(F;\mathbb{R}^{n\times k }_+)$, we obtain $\|F^-\|$ as a lower bound for the above distance. In this term, my question is  as follows: Is a multiple of $\|F^-\|$ an upper bound for the above distance, i.e., 

Is there  a constant $c>0$, such that
  $$
\|F^-\| \leq \mathrm{dist}(F;St_+(n,k)) \leq c \|F^-\|
$$
  for every $n\times k$ dimentional matrix $F$ whose columns are orthonormal?

In the special case $k=1$, the above statement is true, with $c = 2$. 
I'm interested in the special case of small values of $k$, such as $k=2$. Experimentally,
for $k>1$ and random matrices $F$ and by using  Frobenius norm, I get an upper bound for $\mathrm{dist}(F; St_+(n,k))$ by alternating projection to nonnegative matrices and orthonormal matrices. I guess that the above statement is true for $c \approx 2$, $(\beta \approx 4)$.
 A: $\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\varepsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\thh}{\theta}
\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\PP}{\operatorname{\mathsf P}}
\newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$
The following two simple lemmas are crucial. 

Lemma 1. For any nonnegative numbers $a_1,\dots,a_k$,
  \begin{equation*}
 \sum_1^k a_i^2-\max_1^ka_i^2\le\sum_{1\le i<i'\le k}a_i a_{i'}. 
\end{equation*}

Proof. Without loss of generality, $a_1=\max_1^ka_i$. Then $a_i^2\le a_1a_i$ for all $i=2,\dots,k$. So, Lemma 1 follows.
For any $u\in\R^n$, let $u^+ :=\max\{u,0\}$ and $u^- :=\max\{-u,0\}$, so that $u=u^+-u^-$.  

Lemma 2. For any orthonormal vectors $u$ and $v$, 
  \begin{equation*}
 u^+\cdot v^+\le \|u^-\|+\|v^-\|,  
\end{equation*}
  where $\cdot$ denotes the dot product. 

Proof. We have $0=u\cdot v=u^+\cdot v^+ - u^+\cdot v^- -u^-\cdot v^+ + u^-\cdot v^-
\ge u^+\cdot v^+ - \|u^+\|\,\|v^-\| -\|u^-\|\,\| v^+\|$, whence
Lemma 2 follows. 
As in the question, let now $f_1,\ldots,f_k$ be any orthonormal vectors in $\R^n$. Write $f_i=(f_{ij})_{j=1}^n$ and $f^+_i=(f^+_{ij})_{j=1}^n$. Let $(J_1,\dots,J_k)$ be any partition of the set $[n]:=\{1,\dots,n\}$ such that for all $i\in[k]$ and $j\in[n]$ we have the implication 
\begin{equation*}
 j\in J_i\implies f^+_{ij}=\max_{q\in[k]}f^+_{qj}. 
\end{equation*}
Define $y_i=(y_{ij})_{j=1}^n$ by 
\begin{equation*}
 y_{ij}:=f^+_{ij}\,\ii{j\in J_i}, 
\end{equation*}
where $\ii{}$ denotes the indicator; so, $y_{ij}=\max_{q\in[k]}f^+_{qj}$ for $j\in J_i$ and $y_{ij}=0$ for $j\in[n]\setminus J_i$. 
Hence, in view of Lemmas 1 and 2, 
\begin{multline*}
 \sum_1^k\|y_i-f^+_i\|^2
 =\sum_{j\in[n]}\Big(\sum_{i\in[k]}(f^+_{ij})^2-\max_{i\in[k]}(f^+_{ij})^2\Big)
 \le\sum_j\sum_{i<i'}f^+_{ij}f^+_{i'j} \\ 
 =\sum_{i<i'}f^+_i\cdot f^+_{i'}  
 \le\sum_{i<i'}(\|f^-_i\|+\|f^-_{i'}\|)
 =2(k-1)\sum_{i\in[k]}\|f^-_i\|.  \tag{1}
\end{multline*}
Also, $\sum_1^k\|f_i-f^+_i\|^2=\sum_1^k\|f^-_i\|^2\le\sum_1^k\|f^-_i\|$. So, by (1) and Minkowski's inequality,
\begin{equation*}
 \sum_{i\in[k]}\|y_i-f_i\|^2
 \le(\sqrt{2(k-1)}+1)^2\sum_{i\in[k]}\|f^-_i\|\le3k\sum_{i\in[k]}\|f^-_i\|=:\ep. \tag{2}
\end{equation*}
Next, 
\begin{equation}
 0\le1-\|y_i\|=\|f_i\|-\|y_i\|\le\|y_i-f_i\|, \tag{3}
\end{equation} 
by the triangle inequality. 
Consider now two possible cases: 
Case 1: $\ep<1$. (This is hopefully the main case.) Then, by (2), $\|y_i-f_i\|<1$ for all $i$, whence, by (3), $y_i\ne0$ for all $i$, so that we can let
\begin{equation*}
 x_i:=y_i/\|y_i\|. 
\end{equation*}
Then $x_1,\dots,x_k$ are orthonormal vectors in $\R_+^n$, and
\begin{equation*}
 \sum_{i\in[k]}\|x_i-y_i\|^2=\sum_{i\in[k]}(1-\|y_i\|)^2\le\sum_{i\in[k]}\|y_i-f_i\|^2\le\ep  
\end{equation*}
by (3) and (2), 
which yields 
\begin{equation*}
 \sum_{i\in[k]}\|x_i-f_i\|^2
\le4\ep=12k\sum_{i\in[k]}\|f^-_i\|. 
\end{equation*}
Case 2: $\ep\ge1$. Here for any orthonormal $x_1,\dots,x_k$ we have 
\begin{equation*}
 \sum_{i\in[k]}\|x_i-f_i\|^2\le 2\sum_{i\in[k]}(\|x_i\|^2+\|f_i\|^2)=4k\le4k\ep
 =12k^2\sum_{i\in[k]}\|f^-_i\|.  
\end{equation*}
Thus, 
\begin{equation*}
 \sum_{i\in[k]}\|x_i-f_i\|^2\le
 \left\{
 \begin{aligned}
 12k\sum_{i\in[k]}\|f^-_i\|&\text{ if }\ep<1,\\
 12k^2\sum_{i\in[k]}\|f^-_i\|&\text{ if }\ep\ge1. 
 \end{aligned}
 \right.
\end{equation*}
(As follows from the comment by user fedja, here $\|f^-_i\|$ cannot be replaced by $\|f^-_i\|^{1+\ep}$, for any real $\ep>0$.) 
