Do the closed unit disk $D$ and $f(D)$ intersect, if $||f(x)-x||\le2$ for all $x\in D$? In $\Bbb R^n$ let $D=\{x:||x||\le1\}$, and let $f:D\to\Bbb R^n$ be continuous with the property that $||f(x)-x||\le2$ for all $x\in D$.  
Is it true that $D\cap f(D)\neq\varnothing$ (where $f(D)=\{f(x):x\in D\}$) ? In other words, is there $x\in D$ such that $f(x)\in D$ ? 
Clearly $f:D\to3D:=\{3x:x\in D\}$, and if $g(x)=\frac{f(x)}3$ then $g:D\to D$ and $g$ has a fixed point but I do not see if I could employ this to answer my question. 
If $f$ is contracting, or if $f(D)\subseteq D$ then we are lucky (since $f$ would have a fixed point). If $f^{-1}$ exists (and is continuous) then we are lucky again, since $f^{-1}:3D\to D\hookrightarrow3D$ has a fixed point $p$, which necessarily would be a fixed point of $f$ too. (Then $p=f(p)\in D\cap f(D)\neq\varnothing$.) 
If $f$ is a translation distance $2$ (any direction) then $x$ and $f(x)$ would be two diametrically opposite points of $D$, for some $x$, so $f(x)\in D\cap f(D)\neq\varnothing$. 
I came up with this question since it relates (I would think) to a possible "topological" proof of another question I posted earlier (which, in turn, relates to yet another question someone else posted). 
Given a positive semi-definite matrix $B$ does there exist a non-zero vector $z$ with all components non-negative such that all components of $Bz$ are non-negative?
 A: Because $\Vert f(x)-x\Vert\leq2$, the function $x\mapsto \frac12(x-f(x))$ maps $D$ continuously into itself and therefore has a fixed point $a\in D$. Then $f(a)=-a$ is in $f(D)\cap D$.
A: This is true. Let's fix $n$ and let $S=\{x \in \mathbb R^n : \|x\| = 1\}$ and $D=\{x\in\mathbb R^n : \|x\| \leq 1\}$.  Let's suppose that we had some map $f:D \rightarrow \mathbb R^n\setminus D$ such that $\|f(x)-x\| \leq 2$ everywhere. We make the following observation: if $x$ is on $S$, then $f(x)$ cannot be antipodal to $x$.
More precisely, define $C:\mathbb R^n\setminus D\rightarrow S$ by mapping $x$ to $\frac{x}{\|x\|}$ and let $f'$ be the restriction of $f$ to $S$. We observe that
$$C(f'(x)) \neq -x.$$
This implies that $C\circ f'$ is homotopic to the identity because we can, for each $x$, draw a great circle segment connecting $C(f'(x))$ to $x$ and then linearly interpolate upon that segment for each $x$ to create a homotopy from $C\circ f'$ to $\operatorname{id}_S$. However, this contradicts that $f'$ must be homotopic to a constant function because it is the boundary of a function on $D$, so no such $f$ exists.
