(January 2012 #1b) Let \(y\in\mathbb R\) and \(f:\mathbb R\to\mathbb R\) be given. Suppose that for every sequence \(\{x_n\}\) we have \[ \liminf_{n\to\infty}|f(x_n)-f(y)|\le \liminf_{n\to\infty}|x_n-y|. \] Prove that \(f\) is continuous at \(y\).
Solution
1. Prove the contrapositive: if \(f\) is not continuous at \(y\), the hypothesis fails.
2. Discontinuity gives a sequence violating the liminf condition.Proof: suppose \(f\) is discontinuous at \(y\).Then there is \(\epsilon > 0\) and a sequence \(x_n \to y\) with \(|f(x_n) - f(y)| \ge \epsilon\) for all \(n\) (take \(x_n \in (y - 1/n, y + 1/n)\) with \(|f(x_n) - f(y)| \ge \epsilon\), which exists since \(f\) is not continuous at \(y\)).
3. Compare the two liminfs.Proof: for this sequence, \(\liminf_{n\to\infty}|x_n - y| = 0\) (as \(x_n \to y\)), while \(\liminf_{n\to\infty}|f(x_n) - f(y)| \ge \epsilon > 0\).Hence \[\liminf |f(x_n) - f(y)| \ge \epsilon > 0 = \liminf |x_n - y|,\] violating the hypothesis for this particular sequence.
4. Q.E.D. Proof: contrapositive established: hypothesis ⟹ \(f\) continuous at \(y\).