Let $f(x)$ be defined for all real numbers differentiable function of one variable.We know that: $$\int_{-\infty }^{+\infty } |f(x)| \, dx\neq +\infty$$ Problem is to resolve if it is possible or not that: $$\int_{-\infty }^{+\infty } |f'(x)| \, dx= +\infty$$
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Counterexample: take $$ f(x) = \begin{cases} x^2e^{-x^2}\sin(1/x^2) & x \neq 0\\ 0 & x = 0 \end{cases} $$ Note that although $f$ is absolutely integrable, $$ f'(x) = \begin{cases} 2e^{-x^2}\frac{x^2(x^2 - 1)\sin(1/x^2) + \cos(1/x^2)}{x} & x \neq 0\\ 0 & x = 0 \end{cases} $$ Is not.
In this way, I have used the function $e^{-x^2}$ to "play" with David's function and "redefine the tails".
Proof that $f'(x)$ is not absolutely integrable:
Note that $|f'(x)| \geq C \left|\frac{\cos(1/x^2)}{x}\right|$ for some constant $C$. It is thus sufficient to show that the integral $$ \int_{-1}^1 \left|\frac{\cos(1/x^2)}{x}\right| \,dx $$ Diverges. To show that this is the case, note that $$ \int^{1/\sqrt{\pi n}}_{1/\sqrt{\pi(n+1)}} \left|\frac{\cos(1/x^2)}{x}\right| \,dx \\ \geq \int^{1/\sqrt{\pi(n+1/4)}}_{1/\sqrt{\pi(n+3/4)}} \left|\frac{\cos(1/x^2)}{x}\right| \,dx \\ \geq \left(\frac 1{\sqrt{(n + 1/4)\pi}} - \frac 1{\sqrt{(n + 3/4)\pi}}\right)\cdot \sqrt{n \pi} \cdot \frac{\sqrt{2}}{2} $$ Show that this is bounded below by $D/n$ for some constant $D>0$. From there, it follows that $$ \int_{-1}^1 |f'(x)|\,dx \geq \sum_{n=-N}^N D/n $$ From which we conclude that the integral diverges.
Ben Grossmann
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Nicer than what I actually meant (set the function to $0$ off $[-1,1]$ and "smooth it out" around $1$ and $-1$). – David Mitra Feb 28 '14 at 15:11
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@Omnomnomnom Your counterexample is against assumption that $f(x)$ is defined for all real numbers and differentiable. Consider $f(0)$ and $f'(0)$ – Darius Feb 28 '14 at 16:14
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We can fix the condition for $f$ by redefining $f$ at $0$ as $f(0) = \lim_{x \to 0} f(x) = 0$. Not sure about $f'(x)$ though. – Ben Grossmann Feb 28 '14 at 16:16
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In fact, this new $f$ is actually differentiable at $x = 0$. Consider the fact that $$\lim_{h \to 0} \frac{h^2 \sin(1/h^2) - 0}{h} = 0$$ – Ben Grossmann Feb 28 '14 at 16:19
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@Omnomnomnom $$ h(x) = \begin{cases} 2e^{-x^2}\frac{x^2(x^2 - 1)\sin(1/x^2) + \cos(1/x^2)}{x} & |x| \geq 1\ 0 & |x| < 1 \end{cases} $$ is absolutely integrable. So I need some more clarifications. – Darius Feb 28 '14 at 16:51
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That's not $f'(x)$. The function $f'(x)$, which is indeed what I've said it is, is not absolutely integrable between $-1$ and $1$. – Ben Grossmann Feb 28 '14 at 16:55
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@Omnomnomnom Can you prove that, if $$ f'(x) = \begin{cases} 2e^{-x^2}\frac{x^2(x^2 - 1)\sin(1/x^2) + \cos(1/x^2)}{x} & x \neq 0\ 0 & x = 0 \end{cases} $$ then $$\int_{-1}^1 |f'(x)| , dx= +\infty$$ ? – Darius Feb 28 '14 at 16:59
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@Omnomnomnom Note that if $0.465<x<0.47$ then $f'(x)$ is changing its sign, but $\frac{\cos\left(\frac{1}{x^2}\right)}{x}$ not , and its absolute value is not zero. That means that assumption about some positive constant $C$ is wrong. – Darius Feb 28 '14 at 18:42
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Fine. Note instead that $$|f'(x)| \geq B|x^2(x^2-1)\sin(x)/x| + C|\cos(1/x^2)/x|$$ for some constants $B$ and $C$, and that $|x^2(x^2-1)\sin(x)/x|$ is absolutely integrable. We reach the conclusion, once more, that $f'$ is absolutely integrable only if $|\cos(1/x^2)/x|$ is absolutely integrable, which it is not. – Ben Grossmann Feb 28 '14 at 19:34
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Sorry, there was a typo there. I am bounding the integral by the length multiplied by the minimum value of the integrand. – Ben Grossmann Mar 01 '14 at 23:19
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@Omnomnomnom Note that: $$\int_{1/\sqrt{2 \pi +0.75\pi}}^{1/\sqrt{2 \pi +0.25\pi}} \left|\frac{\cos \left(\frac{1}{x^2}\right)}{x}\right| , dx\approx 0.03747674 <\left(\frac{1}{\sqrt{2 \pi +0.25\pi}}-\frac{1}{\sqrt{2 \pi +0.75\pi}}\right)\sqrt{2 \pi }\frac{\sqrt{2}}{2}>0.06$$ It means that your inequality isn't kept when $n=2$. Besides minimum value of the integrand is $0$ when $x=\frac{1}{\sqrt{\pi(n+2/4)}}$. – Darius Mar 02 '14 at 10:25
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I think that trying to use a "single formula function" makes things unnecessarily complicated. Here is a counterexample. It is based on the function $$ t\mapsto (x-a)^2(b-x)^2. $$ This function, on the interval $[a,b]$ is differentiable with both the function and the derivative being zero at the endpoints. Such a feature allows us to put these "lumps" in different places and still get a differentiable function. Let $$ f(t)=\sum_{n=1}^\infty n^8(x-n)^2(n+\frac1{n^2}-x)^2\,1_{[n,n+\frac1{n^2}]}. $$ As $n$ grows, the lumps are thinner and higher. So we can get small areas with big derivatives: $$ \int_{\mathbb R} |f|=\sum_{n=1}^\infty\int_n^{n+1/n^2}n^8(x-n)^2(n+\frac1{n^2}-x)^2 =\sum_{n=1}^\infty\,n^8\,\frac1{30n^{10}}=\sum_{n=1}^\infty\frac1{30n^2}<\infty. $$ And $$ \int_{\mathbb R}|f'|=\sum_{n=1}^\infty\int_n^{n+1/n^2}|2n^8(x-n)(n+1/n^2-x)(2n+1/n^2-2x)|\,dx\\ =\sum_{n=1}^\infty\int_n^{n+1/2n^2}2n^8(x-n)(n+1/n^2-x)(2n+1/n^2-2x)\,dx\\-\int_{n+1/2n^2}^{n+1/n^2}2n^8(x-n)(n+1/n^2-x)(2n+1/n^2-2x)\,dx\\ =\sum_{n=1}^\infty n^8\,\frac1{8n^8}=\infty $$
Martin Argerami
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It looks like created by human , who is counting six-digit numbers products just in his memory. – Darius Mar 01 '14 at 00:39
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Counterexample: $f(x)=\frac{\sin(e^x)}{1+x^2}$ - smooth function $$\int_{-\infty}^{+\infty}\left|\frac{\sin(e^x)}{1+x^2}\right| \,dx\leqslant \int_{-\infty}^{+\infty}\frac{1}{1+x^2} \,dx=\pi$$
Which means that $f(x)\in L^1$ $$f'(x)=\frac{e^x \left(x^2+1\right) \cos \left(e^x\right)-2 x \sin \left(e^x\right)}{\left(x^2+1\right)^2}$$ Note: $\forall_{\varepsilon > 0} \lim_{X\to \infty } \, \int_X^{X+\varepsilon} f'(x) \, dx=0$ $$|f'(x)|=\frac{e^x \left(x^2+1\right) \left|\cos \left(e^x\right)-\frac{2 x}{x^2+1} \frac{\sin \left(e^x\right)}{e^x}\right|}{\left(x^2+1\right)^2}\geqslant \frac{1}{2}\frac{e^x \left|\cos \left(e^x\right)-\frac{2 x}{x^2+1} \frac{\sin \left(e^x\right)}{e^x}\right|}{\left(x^2+1\right)}\geqslant \frac{1}{4}\frac{e^x \left(\cos \left(e^x\right)-\frac{2 x}{x^2+1} \frac{\sin \left(e^x\right)}{e^x}\right)^2}{\left(x^2+1\right)}$$ $$\int_{-\infty}^{+\infty}|f'(x)| \,dx > \frac{1}{4}\int_{0}^{+\infty}\frac{e^x \left(\cos \left(e^x\right)-\frac{2 x}{x^2+1} \frac{\sin \left(e^x\right)}{e^x}\right)^2}{\left(x^2+1\right)} \,dx$$
$\left|\frac{2 x}{x^2+1} \sin \left(e^x\right)\right|\leqslant 1$ which means that if we prove divergence of $\int_{0}^{+\infty}\frac{e^x \cos \left(e^x\right)^2}{\left(x^2+1\right)} \,dx$ we prove divergence of $\int_{-\infty}^{+\infty}|f'(x)| \,dx$. $$\int_0^{+\infty}\frac{e^x \cos \left(e^x\right)^2}{\left(x^2+1\right)} \,dx> \int_0^{+\infty}\cos \left(e^x\right)^2 \,dx=\int_0^{+\infty}\frac{1}{2} \,dx+\frac{1}{2}\int_0^{+\infty}\cos \left(2e^x\right) \,dx$$
Note that:$\left(\frac{e^x}{x^2+1}\right)^{'}=\frac{e^x (-1+x)^2}{\left(1+x^2\right)^2}\geqslant 0$, so finally if $\int_0^{+\infty}\cos \left(2e^x\right) \,dx$ is convergent than $\int_{-\infty}^{+\infty}|f'(x)| \,dx=+\infty$. $$\int_0^{+\infty}\cos \left(2e^x\right) \,dx=\int_2^{+\infty}\frac{\cos \left(v\right)}{v} \,dv$$ $\int_2^{+\infty}\frac{\cos \left(v\right)}{v} \,dv$ is convergent by Dirichlet's test or even Leibniz test with $a_n=\int_{n\pi+0.5\pi}^{(n+1)\pi+0.5\pi}\left|\frac{\cos \left(v\right)}{v}\right| \,dv$ , which completes the proof.
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@Etienne Yes indeed, $$\int_{1}^{+\infty}\cos(x^6)^2 ,dx=\frac{1}{2}\int_{1}^{+\infty} ,dx+\frac{1}{2}\int_{1}^{+\infty}\cos(2 x^6) ,dx$$ $$\int_{1}^{+\infty}\cos(2 x^6) ,dx=\frac{1}{6\sqrt[6]{2}}\int_{2}^{+\infty}\frac{\cos(v)}{v^{\frac{5}{6}}} ,dv$$ $$\int_{2}^{+\infty}\frac{\cos(v)}{v^{\frac{5}{6}}} ,dv=\frac{-\sqrt[6]{2}\sin(2)}{2}+\frac{5}{6}\int_{2}^{+\infty}\frac{\sin(v) \sqrt[6]{v}}{v^2} ,dv\approx -0.486261$$ $$\left(\frac{6 x^5}{(1 + x^2)^2}\right)^{'}=\frac{6 x^4 \left(5+x^2\right)}{\left(1+x^2\right)^3}\geqslant 0$$ – Darius Mar 03 '14 at 12:24
btw.: Do you know if it is possible that $\int_{-\infty}^{+\infty} |h(x)| , dx\neq +\infty$ when $\exists_{x_n:\lim , x_n=+\infty } \lim , |h(x_n)|=+\infty$ ?
– Darius Mar 03 '14 at 15:19