UPSC 2020 Maths Optional Paper 2 Q2b — Step-by-Step Solution

15 marks · Section A

Uniform continuity · Real Analysis · asked 2× in 13 yrs · Read the full method →

Question

Prove that the function $f(x)=\sin x^2$ is not uniformly continuous on the interval $[0,\infty)$ .

Technique

Negation of uniform continuity via two sequences $x_n,y_n$ with $x_n-y_n\to0$ but $|f(x_n)-f(y_n)|=1$ ; rationalize the surd difference to show the gap vanishes.

Solution

Here $f(x)=\sin(x^2)$ . We disprove uniform continuity using the negation of the definition:

$f$ is not uniformly continuous on $I$ iff there exists $\varepsilon_0>0$ such that for every $\delta>0$ there exist points $x,y\in I$ with $|x-y|<\delta$ but $|f(x)-f(y)|\ge\varepsilon_0$ .

A clean way to exhibit such points is via two sequences whose arguments get arbitrarily close while the function values stay apart.

Step 1 — Choose two sequences

For $n\ge1$ let

x_n=\sqrt{\,2n\pi+\tfrac{\pi}{2}\,},\qquad y_n=\sqrt{\,2n\pi\,}.

Both lie in $[0,\infty)$ . Then

x_n^2=2n\pi+\tfrac{\pi}{2},\qquad y_n^2=2n\pi,

f(x_n)=\sin\!\big(2n\pi+\tfrac{\pi}{2}\big)=1,\qquad f(y_n)=\sin(2n\pi)=0.

Hence for every $n$ ,

|f(x_n)-f(y_n)|=|1-0|=1.

Step 2 — The arguments get arbitrarily close

Compute the gap $x_n-y_n$ and show it $\to 0$ . Rationalize:

x_n-y_n=\sqrt{2n\pi+\tfrac{\pi}{2}}-\sqrt{2n\pi} =\frac{\big(2n\pi+\tfrac{\pi}{2}\big)-2n\pi}{\sqrt{2n\pi+\tfrac{\pi}{2}}+\sqrt{2n\pi}} =\frac{\pi/2}{\sqrt{2n\pi+\tfrac{\pi}{2}}+\sqrt{2n\pi}}.

As $n\to\infty$ the denominator $\to\infty$ , so

0< x_n-y_n=\frac{\pi/2}{\sqrt{2n\pi+\tfrac{\pi}{2}}+\sqrt{2n\pi}}\ \longrightarrow\ 0.

Step 3 — Conclude non-uniform continuity

Take $\varepsilon_0=1$ . Let $\delta>0$ be arbitrary. Since $x_n-y_n\to0$ , choose $n$ large enough that $x_n-y_n<\delta$ ; then

|x_n-y_n|<\delta\quad\text{yet}\quad |f(x_n)-f(y_n)|=1\ge\varepsilon_0.

This contradicts the definition of uniform continuity for that $\delta$ . Since $\delta>0$ was arbitrary, no single $\delta$ works for $\varepsilon_0=1$ . Therefore $f(x)=\sin(x^2)$ is not uniformly continuous on $[0,\infty)$ .

Answer

\boxed{\;\text{With }x_n=\sqrt{2n\pi+\tfrac\pi2},\ y_n=\sqrt{2n\pi}:\ \ |x_n-y_n|\to0\text{ but }|f(x_n)-f(y_n)|=1.\;}