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UPSC 2016 Maths Optional Paper 2 Q4b — Step-by-Step Solution

15 marks · Section A

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

Question

Let $f:\mathbb R\to\mathbb R$ be a continuous function such that $\displaystyle\lim_{x\to+\infty}f(x)$ and $\displaystyle\lim_{x\to-\infty}f(x)$ exist and are finite. Prove that $f$ is uniformly continuous on $\mathbb R$.

Technique

Three-region split — tails controlled by the limits (any two far-out points are close in value), compact middle by Heine–Cantor; combine with $\delta=\min(\delta_1,1)$, the "$1$" preventing the two opposite tails from interacting.

Solution

Strategy: the limits at $\pm\infty$ make $f$ “flat” far out, so uniform continuity is easy on the two tails; on a large compact middle interval, continuity gives uniform continuity (Heine–Cantor). Then we stitch the three uniform-continuity constants together.

Step 1 — Fix $\varepsilon$ and use the limits to control the tails

Let $\varepsilon>0$. Denote $L_+=\lim_{x\to+\infty}f(x)$, $L_-=\lim_{x\to-\infty}f(x)$ (finite).

By definition of the limit at $+\infty$, there is $M_+>0$ such that

Hence for any $x,y>M_+$,

Similarly there is $M_->0$ such that for any $x,y<-M_-$,

Set $M=\max(M_+,M_-)+1>0$. On each tail $(M,\infty)$ and $(-\infty,-M)$, any two points are mapped within $\varepsilon$ — no smallness of $|x-y|$ needed.

Step 2 — Uniform continuity on the compact middle (Heine–Cantor)

The interval $[-M-1,\,M+1]$ is closed and bounded, hence compact. A continuous function on a compact set is uniformly continuous (Heine–Cantor). So there is $\delta_1>0$ such that

(Heine–Cantor, proof.) Suppose not: then for some $\varepsilon>0$ and every $n$ there are $u_n,v_n\in[-M-1,M+1]$ with $|u_n-v_n|<1/n$ but $|f(u_n)-f(v_n)|\ge\varepsilon$. By Bolzano–Weierstrass $u_{n_k}\to c\in[-M-1,M+1]$; since $|u_n-v_n|\to0$, also $v_{n_k}\to c$. Continuity gives $f(u_{n_k}),f(v_{n_k})\to f(c)$, so $|f(u_{n_k})-f(v_{n_k})|\to0$, contradicting $\ge\varepsilon$. $\square$)

Step 3 — Stitch the three regions

Choose

Take any $x,y\in\mathbb R$ with $|x-y|<\delta\ (\le1)$. We show $|f(x)-f(y)|<\varepsilon$ in all cases. Because $|x-y|<1$, $x$ and $y$ cannot lie in opposite far tails (those are separated by the middle of width $\ge2$); so one of:

• Both $>M$: then both lie in the right tail; by (T+), $|f(x)-f(y)|<\varepsilon$.
• Both $<-M$: by (T−), $|f(x)-f(y)|<\varepsilon$.
• Otherwise: at least one of $x,y$ lies in $[-M,M]$; since $|x-y|<1$, both lie in $[-M-1,M+1]$. As $|x-y|<\delta\le\delta_1$, (C) gives $|f(x)-f(y)|<\varepsilon$.

In every case $|f(x)-f(y)|<\varepsilon$. Since $\delta$ depends only on $\varepsilon$ (not on $x,y$), $f$ is uniformly continuous on $\mathbb R$.

Answer