Let \(f:\mathbb R\to\mathbb R\) be continuous and suppose \[ |f(x)-f(y)|\ge |x-y| \] for all \(x,y\in\mathbb R\). Prove that the range of \(f\) is all of \(\mathbb R\).
Solution
1. The function \(f\) is injective.
If \(x\ne y\), then \[ \abs{x-y}>0. \] The hypothesis gives \[ \abs{f(x)-f(y)}\ge\abs{x-y}>0, \] so \(f(x)\ne f(y)\).
2. The function \(f\) is either strictly increasing or strictly decreasing.
A continuous injective real-valued function on an interval is strictly monotone. Apply this standard theorem to the interval \(\RR\) using step 1.
3. If \(f\) is strictly increasing, then\[ \lim_{x\to\infty}f(x)=\infty, \qquad \lim_{x\to-\infty}f(x)=-\infty. \]
For \(x>0\), monotonicity gives \(f(x)>f(0)\), so the hypothesis with \(y=0\) yields \[ f(x)-f(0)=\abs{f(x)-f(0)}\ge x. \] Thus \(f(x)\ge f(0)+x\to\infty\).
For \(x<0\), monotonicity gives \(f(x)<f(0)\), so \[ f(0)-f(x)=\abs{f(x)-f(0)}\ge -x. \] Hence \(f(x)\le f(0)+x\to-\infty\) as \(x\to-\infty\).
4. If \(f\) is strictly decreasing, then\[ \lim_{x\to\infty}f(x)=-\infty, \qquad \lim_{x\to-\infty}f(x)=\infty. \]
For \(x>0\), one has \(f(x)<f(0)\), and therefore \[ f(0)-f(x)=\abs{f(x)-f(0)}\ge x, \] so \(f(x)\le f(0)-x\to-\infty\).
For \(x<0\), one has \(f(x)>f(0)\), and hence \[ f(x)-f(0)=\abs{f(x)-f(0)}\ge -x, \] so \(f(x)\ge f(0)-x\to\infty\) as \(x\to-\infty\).
5. The range of \(f\) is all of \(\RR\).
Let \(y\in\RR\). In either monotonicity case, steps 3 and 4 show that there exist \(a<b\) such that \(y\) lies between \(f(a)\) and \(f(b)\). Since \(f\) is continuous, the intermediate value theorem gives \(c\in[a,b]\) with \[ f(c)=y. \] Thus every real number lies in the range of \(f\).
6. Q.E.D.
Step 5 proves that \(f(\RR)=\RR\).