Monthly Archives: May 2017

Counterexamples in differentiation

Exercise. Prove or disprove. If it is not true, find a counterexample.

  1. L\in \mathbb{R} and \lim_{n\rightarrow\infty} n \left( f\left(1+\frac{1}{n}\right) -f(1)\right) =L, then f^\prime(1)=L.
  2. If f+g is differentiable at a, f or g are differentaible at a.
  3. If f and g are infinitely differentiable on \mathbb{R} and fg=0 on \mathbb{R}, then f=0 or g=0.
  4. If f is differentiable at x=a and g\circ f is differentiable at x=a, then g is differentiable at y=f(a).
  5. If f is differentiable at x=a, then f^\prime is continuous at x=a
  6. Let f:D\rightarrow \mathbb{R}. If f^\prime (x)\geq 0 for all x\in D, then f is monotone increasing on D.
  7. If a function f:D\rightarrow \mathbb{R} satisfies f^\prime (x)=0 for all x\in D, then f is constant on D.
  8. If f is differentiable on \mathbb{R} and monotone increasing, then f^\prime is monotone on \mathbb{R}.
  9. If D is bounded and f is differentiable on D and uniformly continuous on D, then f^\prime is bounded on D.
  10. If f is differentiable on (1,\infty) and \lim_{x\rightarrow\infty} f(x)=0, then \lim_{x\rightarrow \infty} f^\prime (x)=0.
  11. If f:[a,b]\rightarrow \mathbb{R} is differentiable on (a,b), then there exists c\in (a,b) such that

        \[ \frac{f(b)-f(a)}{b-a} =f^\prime(c). \]

  12. If f and g are continuous on [a,\infty) and differentiable on (a,\infty) and \lim_{x\rightarrow \infty} f(x)=\infty=\lim_{x\rightarrow\infty} g(x), then

        \[ \lim_{x\rightarrow\infty}\frac{f(x)}{g(x)} =\lim_{x\rightarrow\infty}\frac{f^\prime(x)}{g^\prime(x)}. \]

  13. Suppose that f^\prime(0)=f''(0)=f'''(0)=0, f^{(4)}(0)<0 and f^{(5)} is bounded near 0. Then f(0) is local maximum.

Solution. (13)번만 제외하고 모두 거짓입니다. (13)만 제외하고 반례를 적어두겠습니다. 학생분들은 반례가 왜 명제에 어긋나는지 정확하게 기술해야 합니다.

(1) f=\chi_\mathbb{Q}.


    \[ f(x) = \begin{cases} 1 & x\in \mathbb{Q}\\ -1 & x\notin \mathbb{Q} \end{cases}\quad g(x) = \begin{cases} -1 & x\in \mathbb{Q}\\ 1 & x\notin \mathbb{Q} \end{cases}\quad \]


    \[ f(x) = \begin{cases} 0 & x\leq 0\\ e^{-1/x^2} & x>0 \end{cases}\quad g(x) = \begin{cases} e^{-1/x^2} & x<0\\ 0 & x \geq 0. \end{cases}\quad \]

(4) f(x)=x^2, g(x)=|x|.

(5) f(x)=x^2 \sin (1/x) if x\neq 0 f(0)=0.

(6) D=\mathbb{R}\setminus\{0\}, f(x)=-\frac{1}{x^2}

(7) D=(0,1)\cup (2,3). Define f:D\rightarrow \mathbb{R} by f(x)=1 if x\in (0,1) and f(x)=2 if x\in (2,3).

(8) f(x)=x+\sin x

(9) D=(0,1), f(x)=\sqrt{x}.

(10) f(x)=\frac{1}{x} \sin (x^2)

(11) Define f:[-1,1]\rightarrow \mathbb{R} by

    \[ f(x) = \begin{cases} 0 & x=-1\\ x & -1<x<1 \\ \frac{1}{2} & x=1 \end{cases}.  \]

(12) Define f(x)=3x+\sin x, g(x)=2x-\sin x.

(13) Assume f^{(5)} is bounded on (-\delta,\delta) for some \delta>0, say |f^{(5)}(x)|\leq M for all x\in (-\delta,\delta). For x\in (0,\delta), by Taylor theorem, we have f(x)=f(0)+\frac{f^{(4)}(0)}{4!} x^4 +\frac{f^{(5)}(c)}{5!} x^5 for some c between 0 and x.


    \[   \left({f^{(4)}(0)} +\frac{f^{(5)}(c)}{5} x \right)\frac{x^4}{4!}\leq 0 \]


    \[      f^{(4)}(0) -\frac{M}{5} \leq f^{(4)} (0) + \frac{f^{5}(c)}{5}x \leq +f^{(4)}(0) + \frac{M}{5}x \]

Choose x>0 so small that f^{(4)}(0) + \frac{M}{5}x<0. This is possible since f^\prime(4)(0)<0. So there is \delta_1>0 with \delta_1<\delta such that f(x)\leq f(0) for all x\in (0,\delta_1).

Similarly, for the case x<0, there is \delta_2>0 with \delta_2<\delta such that f(x)\leq f(0) for all x\in (-\delta_2,0). Choose \delta^\prime =\min\{\delta_1,\delta_2\}>0. Then for x\in (-\delta,\delta), we have f(x)\leq f(0). This shows f(0) is a local maximum.