Uniform Continuity of Power Functions on (0,1)
An exam problem asks: for which α∈R is f_α(x)=x^α uniformly continuous on (0,1)? A student writes: 'For α>0, f_α is continuous and 0<f_α(x)<1, so it is bounded. Every continuous bounded function on (0,1) is uniformly continuous; hence all α>0 work. For α=0 the function is constant, and for α<0 it blows up near 0, so those fail.' The TA awards most of the credit but notes the justification is not acceptable. Another student objects that sin(1/x) is continuous and bounded on (0,1) but not uniformly continuous, so the boundedness sentence cannot be a theorem. The instructor asks you to adjudicate the classification for all α, decide whether the student's final answer is correct, and explain what valid argument replaces the false boundedness lemma. Include explicit sequences for the failing cases and for the sin(1/x) objection.
1. Classify all α∈R for which x^α is uniformly continuous on (0,1), with proof.
Read the worked answer
The classification is α≥0. For α>0, x^α extends continuously to [0,1] by setting f(0)=0. A continuous function on the compact interval [0,1] is uniformly continuous there, so its restriction to (0,1) is uniformly continuous. For α=0, f is the constant 1. Thus every α≥0 works.
For α<0, write β=−α>0, so f(x)=x^{−β}. Take x_n=1/n and y_n=1/(2n). Then |x_n−y_n|=1/(2n)→0, but |f(x_n)−f(y_n)|=|n^β−(2n)^β|=n^β(2^β−1)→∞. Hence f is not uniformly continuous.
The student's final set is correct, but the reason is not. The lemma 'continuous and bounded on (0,1) implies uniformly continuous' is false. What actually saves α>0 is the existence of a continuous extension to the compact endpoint 0; boundedness alone gives no control over oscillations or over the behavior near a missing endpoint. For α>0 the extension closes the domain, so compactness can be used.
2. Use sin(1/x) to show why the boundedness lemma fails, and explain the connection to the missing endpoint.
Read the worked answer
Let g(x)=sin(1/x) on (0,1). It is continuous and bounded by 1. Choose x_n=1/(2πn) and y_n=1/(2πn+π/2). Then x_n−y_n is about 1/(8π n^2), so |x_n−y_n|→0. But g(x_n)=sin(2πn)=0 and g(y_n)=sin(2πn+π/2)=1, so |g(x_n)−g(y_n)|=1 for every n. Therefore g is not uniformly continuous.
The failure is tied to the missing endpoint 0: g has no continuous extension to [0,1], because sin(1/x) oscillates infinitely often as x→0+. Boundedness does not prevent those oscillations from producing pairs of points arbitrarily close together with outputs separated by 1. By contrast, x^α for α>0 has a finite limit 0 at the missing endpoint, so it extends continuously and inherits uniform continuity from the compact interval. The sin(1/x) example is therefore not a counterexample to the correct classification; it is a counterexample to the student's general lemma.