An MCQ question on continuity.

Question

Let $f: \mathbb{R}\to\mathbb{R}$ be a continuous bounded function, then :

A. $f$ has to be uniform continuous.

B. There exists an $x\in\mathbb{R}$ such that $f(x)=x$

C. $f$ cannot be increasing.

D. $\lim_{x\to\infty} f(x)$ exists.

My attempt:

A counter-example for (D) can be $f(x)=\sin(x)$

Also (C) is not true. My example is

$f(x)= \mathbb{e}^x$ when $x\in (-\infty,0]\,\,\,\,\,$ ; and

$f(x)=2-\frac{1}{x+1}$ when $x\in (0,\infty)$

The function is bounded by $0$ and $2$.

Though I'd like you to give me a simpler example. This was the only thing that came to my mind.

Also i have no clue about $A$ and $B$

Though my intuition says, the correct choice is (B).

But you are not really allowed to "guess" in mathematics!!

Answer 1

A $f(x)=\sin x^2$

B Let $M$ be a bound for $f$, i.e. $|f(x)|<M$ for all $x$. Then for the continuous function $g(x)=f(x)-x$ we have $g(-M)>0$ and $g(M)<0$, hence by the IVT $g(x)=0$ (i.e. $f(x)=x$) for some $x\in(-M,M)$.

C $f(x)=\frac x{1+|x|}$

D $f(x)=\sin x$

Answer 2

• C. $x\mapsto \arctan x$ is increasing, continuous and bounded.
• B. By contradiction, assume (wlog) $f$ bounded and continuous such that $f(x) < x$ for all $x\in\mathbb{R}$ (indeed, $g\colon x \mapsto f(x)-x$ is continuous and never cancels, so it must be either always positive or always negative). Then $\lim_{x\to-\infty} f(x) = -\infty$, yielding a contradiction.
• A. $x\mapsto \sin x^2$ is not uniformly continuous.