Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics II / 4

If $x\ne y$, one of $x<y$ and $y<x$ holds. Since $f$ is strictly increasing, the corresponding values are strictly ordered, so $f(x)\ne f(y)$. Thus $f$ is an injection. Every element of the image of a function $f([a,b])$ equals $f(x)$ for some $x\in [a,b]$, which is precisely surjectivity onto that image of a function. Hence $f:[a,b]\to f([a,b])$ is a bijection.

Let $u<v$ lie in $f([a,b])$, and put $x=g(u)$ and $y=g(v)$. Equality $x=y$ would imply $u=v$. If $x>y$, strict monotonicity of $f$ would give $u=f(x)>f(y)=v$. Both are impossible, so $x<y$. Therefore the inverse function $g$ is strictly increasing.

For $x_0>a$, defineThis nonempty set is bounded above by $f(x_0)$, so $l\le f(x_0)$. Given $\epsilon >0$, the supremum property supplies $s<x_0$ with $l-\epsilon <f(s)\le l$. Whenever $s<t<x_0$, monotonicity givesThis proves the left-hand one-sided limit $f^{-}(x_0)=l$.

For $x_0<b$, defineHere $r\ge f(x_0)$. Choose $u>x_0$ with $r\le f(u)<r+\epsilon$, using the infimum property. For $x_0<t<u$, monotonicity gives $r\le f(t)\le f(u)<r+\epsilon$, proving the right-hand one-sided limit $f^{+}(x_0)=r$. Thuswhere the relevant one-sided limits exist, including the indicated single one-sided limit at each endpoint.

At an interior point, continuity implies both one-sided limits equal $f(x_0)$. Conversely, equality $f^{-}(x_0)=f^{+}(x_0)$ and the displayed inequalities force both to equal $f(x_0)$, giving continuity. Therefore

If $f$ is a continuous function, the intermediate value theorem ensures that every value between $f(a)$ and $f(b)$ is attained. Monotonicity excludes values outside that range, so $f([a,b])=[f(a),f(b)]$.

Conversely, suppose this equality of sets holds. At an interior point, if $f^{-}(x_0)<f(x_0)$, any $y$ strictly between these two numbers lies in $[f(a),f(b)]$. Yet inputs below $x_0$ have values at most $f^{-}(x_0)$ and inputs at or above $x_0$ have values at least $f(x_0)$, so no input attains $y$. This contradicts surjectivity. The same argument excludes $f(x_0)<f^{+}(x_0)$, so part (b) gives continuity at every interior point.

At $a$, a strict gap $f(a)<f^{+}(a)$ likewise leaves every value between them unattained; hence $f^{+}(a)=f(a)$, giving right continuity. At $b$, the corresponding argument gives $f^{-}(b)=f(b)$ and left continuity. These are the appropriate endpoint notions on $[a,b]$. Hence

With the strict monotonicity assumed at the start of the question, parts (a) and (c) give a strictly increasing inverse functionIts image of a function is exactly the interval between its endpoint values. Apply part (c) to $g$, with domain $[f(a),f(b)]$, to conclude that $g$ is a continuous function, including at both endpoints.

If the word “increasing” in this subpart were interpreted as merely nondecreasing, an inverse function need not exist: a constant function is a counterexample. The preceding strict hypothesis supplies the required injectivity.