Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 1 / 10D (Analysis I)

We use the Bolzano-Weierstrass theorem: every bounded real sequence has a convergent subsequence. If $f$ were unbounded on $[a,b]$, there would be $x_n\in [a,b]$ with $|f(x_n)|\ge n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_k}\to c$, and the closed interval contains $c$. Since $f$ is a continuous function, $f(x_{n_k})\to f(c)$, contradicting $|f(x_{n_k})|\ge n_k\to \infty$. Thus $f$ is bounded.

Let $s={\sup }_{x\in [a,b]}f(x)$. By the definition of the supremum, choose $y_n$ with $s-1/n<f(y_n)\le s$. Again the Bolzano-Weierstrass theorem gives $y_{n_k}\to d\in [a,b]$. Continuity yields $f(d)=s$. Applying the same argument to $-f$ gives a point where $f$ attains its infimum. Therefore both the supremum and the infimum are attained. This proves the extreme value theorem for a closed bounded interval.

Away from integer multiples of $\pi$, the function is a composition of continuous functions. At zero, the bound $|g(x)|\le \sqrt{|x|}$ gives $g(x)\to 0=g(0)$ by the squeeze theorem. This is oscillation multiplied by a vanishing amplitude.

Fix $n\ne 0$ and set $t_k=\pi /2+2\pi k$ and $x_k=n\pi +(-1{)}^n\arcsin (1/t_k)$. Then $x_k\to n\pi$, $\sin x_k=1/t_k$, andSo $g$ is discontinuous at every nonzero integer multiple of $\pi$. Its set of continuity points is

On $[0,\pi ]$, $g(x)\le \sqrt{x}<\sqrt{\pi }$ for $x<\pi$, while $g(\pi )=0$. With $x_k=\pi -\arcsin (1/t_k)$ as above, $g(x_k)=\sqrt{x_k}\to \sqrt{\pi }$. Consequently the supremum is $\sqrt{\pi }$ and is not attained. This illustrates that supremum can fail to be attained at an oscillatory endpoint.

On $[\pi ,3\pi /2]$, set $x_0=\pi +\arcsin (2/(3\pi ))$. It lies strictly inside the interval, and $1/\sin x_0=-3\pi /2$, so $g(x_0)=\sqrt{x_0}>\sqrt{\pi }$. Choose $0<\delta <x_0-\pi$. On $[\pi ,\pi +\delta ]$,On $[\pi +\delta ,3\pi /2]$, $g$ is a continuous function, so the extreme value theorem gives a maximum there, at least $g(x_0)$. It exceeds every value on the discarded interval. Therefore $g$ does attain its supremum on $[\pi ,3\pi /2]$. This is the interior-value criterion for an attained maximum.