Krdytkyu

Past year paper question.

Let $(X, d)$ be a metric space and let $f: X to mathbb{R}$ be a continuous function, where $mathbb{R}$ is given the standard metric. Assume that for any $epsilon > 0$, the set ${ x in X : |f(x)| geq epsilon }$ is a compact metric subspace of $X$. Show that $f$ is uniformly continuous on $X$.

Attempt:

Let $epsilon >0$ be given. Let $K := { x in X : |f(x)| geq frac{epsilon}{2} }$. $f$ is continuous on $K$ $implies f$ is uniform continuous on $K$. Then there exists $delta$ such that $d(x,y)<delta implies |f(x)-f(y)| < frac{epsilon}{2}$.

Take any $2$ points $x,y in X$, such that $d(x,y)<delta$. If $x,y in K$, then $|f(x)-f(y)| < frac{epsilon}{2} < epsilon$. If $x,y notin K$, then $|f(x)-f(y)| leq |f(x)|+|f(y)| leq frac{epsilon}{2} + frac{epsilon}{2} leq epsilon$.

But I am stuck here, because what if $x in K$ and $y notin K$.

Suppose $f$ is not uniformly continuous. Then there exists $delta >0$ and sequences ${x_n},{y_n}$ such that $d(x_n,y_n) to 0$ but $|f(x_n)-f(y_n)| geq delta$ for each $n$. For each $n$ either $|f(x_n)| geq delta /2$ or $|f(y_n)| geq delta /2$. One of these holds for infinitely many $n$. Suppose $|f(x_n)| geq delta/2 $ along a subsequence. The hypothesis tells you that the subsequence lies in a compact set, so it has a convergent subsequence. Along this subsequence both sequences ${x_n},{y_n}$ converge to the same limit $x$ because $d(x_n,y_n) to 0$. This contradicts continuity of $f$ at $x$.