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Let us consider in one dimension. Let $a>0$ be a given constant, $$ C[0,a]:=\{f:[0,a]\to\mathbb{R} \mid \text{$f$ is continuous in $[0,a]$}\}, $$ $$ C^{1}(0,a]:=\{f:(0,a]\to\mathbb{R} \mid \text{$f$ is once differentiable in $(0,a]$ & $f'$ is conti. in $(0,a]$}\} $$ and $$ L^{1}(0,a):=\{f:(0,a)\to\mathbb{R} \mid \text{$f$ is Lebesgue integrable in $(0,a)$}\}. $$ Here $f'$ denotes the first derivative of $f$.

My question : Is there function $f$ satisfying $f\in C^{1}(0,a]\cap C[0,a]$ but $f'\notin L^{1}(0,a)$?

For example, functions like square root $f(x)=x^{b}$, where $b\in(0,1)$, are failed since these belong to not only $C^{1}(0,a]\cap C[0,a]$ but also $L^{1}(0,a)$. The function

・$f(x)=x\sin(1/x)$ if $x\neq0$, $0$ if $x=0$

is also failed.

To begin with, I think there is not since $$ \left|\int_{0}^{a}f'(x)dx\right|=|f(a)-f(0)|<\infty. $$

I'm glad if you give examples as many as possible.

user
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  • Are you sure the last function doesn't work? – zhw. Sep 22 '15 at 03:49
  • Thank you for comment, zhw. Yes, it doesn't work. Please see the url : http://www.wolframalpha.com/input/?i=improper+integral+calculator&f1=sin(1%2Fx)+-+cos(1%2Fx)+%2Fx&f=Integral.integrand_sin(1%2Fx)+-+cos(1%2Fx)+%2Fx&f2=0&f=Integral.rangestart%5Cu005f0&f3=1&f=Integral.rangeend%5Cu005f1&a=FVarOpt.1-_-.Integral.variable---.Integral.rangestart-.*Integral.rangeend--- – user Sep 22 '15 at 03:54
  • You should know Wolfram often spits out nonsense. More to the point, I can't see why you think that link shows what you think it does. – zhw. Sep 22 '15 at 04:01
  • I'm sorry. I pasted the Wolfram's url because I visit to here for the first time and am not used, but it can be shown in theoretical in fact. – user Sep 22 '15 at 04:08
  • I think you're confused about something. See my answer below. – zhw. Sep 22 '15 at 04:10

1 Answers1

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Your last function, $f(x) = x\sin (1/x),$ actually provides the example you seek. To prove this you need to show $[\cos (1/x)]/x \not \in L^1(0,1).$ Think about the integral of this function over $[1/(2n\pi+\pi/4), 1/2n\pi].$

zhw.
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  • Thank you for giving your answer. I thought based on your advise but $f(x)=x\sin(1/x)$ is failed on second thought, that is, $[\cos(1/x)]/x\in L^{1}(0,1)$. Indeed, $$ \int_{0}^{1}\frac{\cos(\frac{1}{x})}{x}dx=\int_{1}^{\infty}\frac{\cos x}{x}dx=-Ci(1), $$ where $Ci(x)$ denotes the cosine integral (see https://en.wikipedia.org/wiki/Trigonometric_integral), and it is known that $Ci(1)$ converges. – user Sep 22 '15 at 12:53
  • But I got $[\cos(1/x)]/x^{2}\notin L^{1}(0,1)$ and so I knew $f(x)=\sin(1/x)$ is what I want. – user Sep 22 '15 at 13:21
  • @P.Mike Being Lebesgue integrable implies that the absolute value is integrable, and $$\int_0^1 \frac{\lvert \cos \frac{1}{x}\rvert}{x},dx = +\infty.$$ For $f(x) = \sin (1/x)$, that fails the $C[0,a]$ requirement. – Daniel Fischer Sep 22 '15 at 13:23
  • @DanielFischer Thank you for pointing out. I had forgotten the $C[0,a]$ requirement despite writing question by myself. – user Sep 22 '15 at 13:39
  • @DanielFischer Do you know other example? – user Sep 22 '15 at 13:56
  • @P.Mike Since for $0 < b < a$ you have $$\int_b^a f'(t),dt = f(a) - f(b),$$ the integral $$\int_0^a f'(t),dt$$ always exists as an improper Riemann integral. In order for $f'$ to not be Lebesgue integrable, you therefore need oscillation, and unboundedness of the derivative at $0$. All examples are therefore similar to the one here. Of course, you can take more extreme measures, $x^2 \sin \frac{1}{x^5}$ for example. – Daniel Fischer Sep 22 '15 at 14:07
  • @DanielFischer I see. Thank you! – user Sep 22 '15 at 14:49