Oscillatory Examples That Separate BV From Non-BV

Oscillation, Amplitude, and Variation

A function can approach zero at an endpoint and still have infinite total variation there. The reason is that small oscillations may accumulate: the variation counts the size of each change, not merely the net change between the endpoints. For oscillatory functions, a useful strategy is to compare the amplitude of successive extrema with the number of oscillations occurring near zero.

For a real-valued function \(f\) on \([0,1]\), its total variation is the supremum of the partition sums \(\sum_{j=1}^N |f(t_j)-f(t_{j-1})|\), where \(0=t_0<t_1<\cdots<t_N=1\). The function is of bounded variation, or BV, if this supremum is finite. We study the real-valued function \(f:[0,1]\to\mathbb{R}\) defined, for parameters \(\beta>0\) and \(\alpha\in\mathbb{R}\), by

For every \(x>0\), this formula gives a finite real value, including when \(\alpha\le0\). For \(\alpha>0\), \(f(x)\to0\) as \(x\downarrow0\), so the chosen value at zero makes \(f\) continuous there. When \(\alpha\le0\), continuity at zero is not needed for the variation test; the function is still defined at every point of \([0,1]\).

The two directions use different estimates. When \(\alpha>\beta\), the absolute derivative is integrable near zero and controls every partition sum. When \(\alpha\le\beta\), points at successive positive and negative extrema produce partition sums that grow without bound.

When the Amplitude Is Large Enough

Assume \(\alpha>\beta>0\). On \((0,1]\), ordinary differentiation gives

Since the absolute values of sine and cosine are at most \(1\),

Both powers on the right are integrable over \((0,1]\): their exponents are greater than \(-1\), because \(\alpha>0\) and \(\alpha-\beta>0\). In particular,

Proof of the BV direction. Let \(0=t_0<t_1<\cdots<t_N=1\) be any partition. For \(0<s<t\le1\), the ordinary fundamental theorem of calculus on the compact interval \([s,t]\) gives \(f(t)-f(s)=\int_s^t f'(x)\,dx\). Consequently, \(|f(t)-f(s)|\le\int_s^t|f'(x)|\,dx\). For the first partition interval, apply this inequality with \(s=\varepsilon\) and \(t=t_1\), where \(0<\varepsilon<t_1\), and let \(\varepsilon\downarrow0\). The function satisfies \(f(\varepsilon)\to0=f(0)\), and the displayed integrable bound shows that \(\int_0^{t_1}|f'(x)|\,dx\) is finite. Thus \(|f(t_1)-f(0)|\le\int_0^{t_1}|f'(x)|\,dx\). Applying the same bound to every other partition interval and adding gives \(\sum_{j=1}^N|f(t_j)-f(t_{j-1})|\le\int_0^1|f'(x)|\,dx\le1+\beta/(\alpha-\beta)\). This bound holds for every partition, so taking the supremum proves that \(f\) has bounded variation. \(\square\)

When Alternating Extrema Force Infinite Variation

Now suppose \(\alpha\le\beta\). Set \(q=\alpha/\beta\), so \(q\le1\). The sine attains alternating values \(1,-1,1,-1,\ldots\) at phases \(\theta_n=\pi/2+n\pi=\pi(n+1/2)\), for integers \(n\ge1\). Define \(x_n=\theta_n^{-1/\beta}\). The sequence \(x_n\) decreases to zero. For all sufficiently large \(n\), \(x_n\le1\); at these points, the function values are

Thus the absolute difference of the values at consecutive points is the sum of their amplitudes: \(|f(x_n)-f(x_{n+1})|=\theta_n^{-q}+\theta_{n+1}^{-q}\). A partition containing any finite consecutive run of these points, together with \(0\) and \(1\), includes all those successive differences. Although the points appear in decreasing order \(x_n>x_{n+1}\), a partition lists them in increasing order; the absolute differences are unchanged by reversing the order.

The amplitudes have a divergent sum. If \(q\le0\), then \(\theta_n>1\) and \(\theta_n^{-q}\ge1\) for every \(n\), so the sum diverges. If \(0<q\le1\), then \(n+1/2\le3n/2\) for \(n\ge1\), and hence \(\theta_n\le(3\pi/2)n\). It follows that \(\theta_n^{-q}\ge(3\pi/2)^{-q}n^{-q}\ge(3\pi/2)^{-q}/n\). The harmonic series diverges: in each block \(2^k\le n<2^{k+1}\), its terms sum to at least \(2^k/2^{k+1}=1/2\), so its partial sums are unbounded. This proves the required amplitude divergence in both cases.

Proof of the non-BV direction. Choose \(N\ge1\) large enough that \(x_n\le1\) for all \(n\ge N\). For each integer \(M>N\), form a partition containing \(0,1\), and the points \(x_M,x_{M-1},\ldots,x_N\), listed in increasing order. Its sum includes the consecutive differences between \(f(x_n)\) and \(f(x_{n+1})\), for \(N\le n<M\), and therefore is at least \(\sum_{n=N}^{M-1}\theta_n^{-q}\). The amplitude sum diverges, as just proved, so these finite partition sums have no finite upper bound as \(M\) increases. The supremum of all partition sums is infinite. Hence \(f\) is not of bounded variation. \(\square\)

A Common Pitfall: Net Change Is Not Total Variation

For the borderline function \(f(x)=x\sin(1/x)\), a tempting but invalid argument is to integrate the derivative with its sign and use the finite net change to conclude finite variation. On \([\varepsilon,1]\), the fundamental theorem gives \(\int_\varepsilon^1 f'(x)\,dx=f(1)-f(\varepsilon)\). As \(\varepsilon\downarrow0\), this tends to \(\sin(1)\), a finite number. But this integral records the signed net change: positive and negative changes can cancel. Total variation instead accumulates absolute differences. The extrema calculation shows that its partition sums dominate constant multiples of harmonic partial sums and are unbounded.

Another point requiring care is the direction of the derivative estimate. An integrable absolute derivative supplies an upper bound for every partition sum, as shown when \(\alpha>\beta\). A signed derivative integral does not supply that bound. When \(\alpha\le\beta\), the explicit partitions are what establish infinite variation; merely observing that a particular derivative bound is not integrable would not, by itself, replace that partition argument.