An MCQ question on continuity.

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Solution 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 $f(x)=\sin x$

Solution 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.
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Swapnil Tripathi
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Swapnil Tripathi

Updated on December 13, 2020

Comments

  • Swapnil Tripathi
    Swapnil Tripathi almost 3 years

    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!!

    • Clement C.
      Clement C. over 9 years
      for C, wouldn't $x\mapsto \arctan x$ do the job?
    • Swapnil Tripathi
      Swapnil Tripathi over 9 years
      @Clement C. : Haha! Yes! I knew there were simpler examples. Thanks. Any comments on A. and B.?
  • Swapnil Tripathi
    Swapnil Tripathi over 9 years
    I had a feeling about IVT. Thanks so much!!
  • Swapnil Tripathi
    Swapnil Tripathi over 9 years
    Thank you so much! That is why i love proofs by contradiction. Beautiful. :D