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Is there a differentiable function $f$ in which the differential function $f'$ is bounded but has no maximum on one closed interval? Thanks

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    $\begingroup$ Impossible, differentiable implies continuous. $\endgroup$ – Adam Hughes Jul 27 '14 at 22:53
  • $\begingroup$ @masoud: Are u looking for some differentiable function $f$ which $f'$ is bounded and has no max on any closed intervals? $\endgroup$ – mrs Jul 27 '14 at 23:15
  • $\begingroup$ I think something like $f(x) = -x|x|+x^2\sin(1/x)$ would also work. $\endgroup$ – Jonas Meyer Jul 28 '14 at 3:43
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Answer assuming looking for "at least one closed interval":

Define the sequence $a_{n}=\frac{1}{3^{n}}$. Then $s_{k}=\sum\limits_{i=k}^{\infty}a_{i}=\frac{3}{2\times 3^{k}}$. Hence $\lim\limits_{k\rightarrow\infty}\frac{s_{k}}{(\frac{1}{2^{k+1}})}=\lim\limits_{k\rightarrow\infty}3(\frac{2}{k})^{k}=0$.

Define a collection of function $g_{k}:[\frac{1}{2^{k}},\frac{1}{2^{k+1}}]\rightarrow\mathbb{R}$ to be a nonnegative continuous function that vanish at 2 endpoints and have the maximum value being $1-(\frac{1}{4})^{k}$ and also $\int\limits_{\frac{1}{2^{k}}}^{\frac{1}{2^{k+1}}}g(t)dt=a_{k}$. (it is easy to show that those $g_{k}$ exist, as we give it plenty of room to get the integral it needed, and obtaining a maximum value is just a matter of putting a small spike).

Define $g:[-1,1]\rightarrow\mathbb{R}$ to be $g(0)=0$ and $g(x)=g_{k}(|x|)$ if $|x|\in[\frac{1}{2^{k}},\frac{1}{2^{k+1}}]$.

Define $f:[-1,1]\rightarrow\mathbb{R}$ to be $f(x)=\int\limits_{0}^{x}g(t)dt$.

Now calculate $f^{\prime}(0)=\lim\limits_{h\rightarrow 0}\frac{\int\limits_{0}^{h}g(t)dt}{h}$. We have for $h\in[\frac{1}{2^{k}},\frac{1}{2^{k+1}}]$ then $0\leq\frac{\int\limits_{0}^{h}g(t)dt}{h}<\frac{s_{k}}{(\frac{1}{2^{k+1}})}$. Since $\lim\limits_{k\rightarrow\infty}\frac{s_{k}}{(\frac{1}{2^{k+1}})}=0$ by squeeze theorem we have $f^{\prime}(0)=\lim\limits_{h\rightarrow 0}\frac{\int\limits_{0}^{h}g(t)dt}{h}=0$.

As for $f^{\prime}(x)$ when $x\not=0$, notice that $g$ is continuous on $[-1,0)$ and $(0,1]$ so $f^{\prime}(x)=g(x)$.

Hence $f^{\prime}=g$.

Now notice that $g$ do not have any maximum on $[-1,1]$, since its values approach $1$ arbitrarily close, but it never take that values.

Hence $f$ is what we are looking for.

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  • $\begingroup$ @ Gina:Thanks a lot indeed $\endgroup$ – Finish Jul 28 '14 at 1:11
  • $\begingroup$ @masoud: sorry I got an error, that doesn't work on all closed interval. I added a proof on why no $f$ exist if we require that condition on all closed interval. $\endgroup$ – Gina Jul 28 '14 at 1:34
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    $\begingroup$ This function $f$ is only differentiable a.e. I believe the OP wanted a function that's differentiable everywhere. $\endgroup$ – user138530 Jul 28 '14 at 1:38
  • $\begingroup$ Good example, I didn't recognize this problem before. $\endgroup$ – Shine Jul 28 '14 at 1:47
  • $\begingroup$ @ Gina: I need only for one interval not for all nondegenerated intervals and also I need a function differentiable everywhere, not only almost every where. $\endgroup$ – Finish Jul 28 '14 at 1:48

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