Showing that a function is surjective onto a set

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.

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fresh_42
Mentor
Just an idea: maybe it helps that ##|f(z)-z|\leq 0.04##.

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
Homework Helper
Gold Member
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.