Showing that a function is surjective onto a set

[mathstudent34]

Homework Statement

Let $B(0,1)\subseteq\mathbb{R^n}$ be the ball of radius $1$ in $\mathbb{R^n}$. Suppose $f:B(0,1)\to\mathbb{R^n}$ satisfies $f(0)=0$ and
$$\forall x\neq y\in B(0,1),~~~|f(x)-f(y)-(x-y)|\leq 0.1|x-y|.$$
Show that $f$ is onto $B(0,0.4)$.

Relevant Equations

$f:X\to Y$ is surjective if $\forall y\in Y,\ \exists x\in X$ such that $f(x)=y$.

I have to show that $\forall z\in B(0,0.4)$, there exists an $x\in B(0,1)$ such that $f(x)=z$ but I am not sure how to show this. From the reverse triangle inequality
$$-|f(x)-f(y)|+|x-y|\leq 0.1|x-y|\implies |f(x)-f(y)|\geq 0.9|x-y|$$
im not sure if this helps.

 

[fresh_42]

Just an idea: maybe it helps that $|f(z)-z|\leq 0.04$.

 

[William Crawford]

I assume (since you are not precisely specifying it) that $B(0,r)\subseteq\mathbb{R}^n$ is the closed ball of radius $r>0$ centered at $0$.

You are given that $0\in B(0,1)$ is a fixed point of $f$. This immediately tells us that there exist a point in $x\in B(0,1)$ for which $f(x) = 0$, namely the point $x=0$. Thus, the problems is now to show that there for every point $z\in B^\ast(0,\tfrac{4}{10})$ exist at least one point $x\in B(0,1)$ s.t. $f(x)=z$, where $B^\ast(0,r) = B(0,r)\setminus\{0\}$ denotes the punctured closed ball.

Now, note that
$$\forall x\in B^\ast(0,\tfrac{4}{10})\ :\ \vert f(x) - x\vert \leq \frac{1}{10}\vert x\vert.$$
is a special case of your original inequality satisfied by $f$ (Hint, tak $y=0$ and use that $B^\ast(0,\tfrac{4}{10})\subset B^\ast(0,1)$).

Try to proceed from here (Hint, what it the farthest two points in $B^\ast(0,\tfrac{4}{10})$ can be from each other?).

 

[epenguin]

Did you read online what you wrote? As you got part right not too difficult to get the rest. (Though I don't know why a bit came out red.)

Show that $f$ is onto $B(0,0.4)$.
Relevant Equations:: $f:X\to Y$ is surjective if $\forall y\in Y,\ \exists x\X$ such that $f(x)=y$.

I have to show that $\forall z\in B(0,0.4)$, there exists an $x\in B(0,1)$ such that $f(x)=z$ but I am not sure how to show this. From the reverse triangle inequality
$$-|f(x)-f(y)|+|x-y|\leq 0.1|x-y|\implies |f(x)-f(y)|\geq 0.9|x-y|$$
im not sure if this helps.