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Suppose $f:[0,1]\to\mathbb R$ is continuous, possibly with unbounded variation. We consider sums of the form

$$\sum_{i=1}^n\Big(f(x_i)-f(x_{i-1})\Big)^3$$

where $0=x_0<x_1<x_2<\cdots<x_{n-1}<x_n=1$ is a partition of the unit interval. As the norm $\max_i|x_i-x_{i-1}|$ approaches $0$, do these sums also approach $0$?

For any $p>0$, there are continuous functions without finite $\int_0^1|df(x)|^p$ (for example $x^{1/p}\cos x^{-1}$, or $|\ln x/2|^{-1}\cos x^{-1}$). On the other hand, without the absolute value, we always have $\int_0^1df(x)^1=f(1)-f(0)$. It's not clear whether $\int_0^1df(x)^3$ always exists.

mr_e_man
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1 Answers1

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The problem with $x^{1/3}\cos x^{-1}$ is that the oscillations are too uniform. We want something that increases gradually and decreases sharply, like a sawtooth wave, so that the differences are amplified (by cubing) in one direction more than the other direction. Thus, consider

$$f(x)=x^{1/3}\big(\sin x^{-1}-\tfrac12\sin2x^{-1}\big)=x^{1/3}\sin x^{-1}\big(1-\cos x^{-1}\big),$$

or rather

$$f(x)=\frac4{3\sqrt3}x^{1/3}\Big(\sin\tfrac13\tau x^{-1}-\tfrac12\sin\tfrac23\tau x^{-1}\Big)$$

(where $\tau=2\pi$). Evaluating this function at reciprocals of integers, we get

$$f\left(\frac1{3m}\right)=0,$$ $$f\left(\frac1{3m+1}\right)=\frac1{(3m+1)^{1/3}},$$ $$f\left(\frac1{3m+2}\right)=\frac{-1}{(3m+2)^{1/3}}.$$

Now three consecutive terms in the sum will be

$$\left(0-\frac1{(3m+1)^{1/3}}\right)^3+\left(\frac1{(3m+1)^{1/3}}-\frac{-1}{(3m+2)^{1/3}}\right)^3+\left(\frac{-1}{(3m+2)^{1/3}}-0\right)^3$$ $$=\frac3{(3m+1)^{2/3}(3m+2)^{1/3}}+\frac3{(3m+1)^{1/3}(3m+2)^{2/3}}$$ $$\geq\frac3{(3m+2)^{2/3}(3m+2)^{1/3}}+\frac3{(3m+2)^{1/3}(3m+2)^{2/3}}$$ $$=\frac6{3m+2}$$

which shows that the sum can be made arbitrarily large, since $\sum_m1/m=\infty$. Hence, the integral doesn't exist. (I leave out some details about the partition's norm actually approaching $0$.)

mr_e_man
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